#set document(title: "10.2 Solve Quadratic Equations by Completing the Square", author: "OpenStax / XYZ Homework") #set page(width: 8.5in, height: auto, margin: 1in) #import "@preview/cetz:0.5.2" #set text(font: ("STIX Two Text", "Libertinus Serif", "New Computer Modern"), size: 10.5pt, lang: "en") #show math.equation: set text(font: ("STIX Two Math", "New Computer Modern Math")) #set par(justify: true, leading: 0.62em, spacing: 0.9em) #set enum(spacing: 1.1em) // room between list items so tall inline fractions don't collide #set list(spacing: 1.1em) #set table(stroke: 0.5pt + rgb("#c7ccd3")) #let BLUE = rgb("#183B6F") // brand navy — section bars + example/solution labels (white on navy 11.09:1) #let ORANGE = rgb("#A94509") // brand primary-700 — AA-safe deep orange for TEXT (5.93:1 on white; raw brand #F37021 is 2.94:1 and must never carry text) #let RED = rgb("#DC2626") // brand error-600 #let GREEN = rgb("#059669") // brand success-600 (decoration only; small green text uses green-text #007942) #show heading.where(level: 1): it => block(width: 100%, above: 0pt, below: 16pt, fill: gradient.linear(BLUE, rgb("#2C5AA0")), inset: (x: 14pt, y: 12pt), radius: 3pt, text(fill: white, weight: "bold", size: 19pt, it.body)) #show heading.where(level: 2): it => block(width: 100%, above: 18pt, below: 10pt, fill: BLUE, inset: (x: 10pt, y: 6pt), radius: 2pt, text(fill: white, weight: "bold", size: 12pt, it.body)) #show heading.where(level: 3): it => text(fill: ORANGE, weight: "bold", size: 12.5pt, it.body) #show heading.where(level: 4): it => text(fill: BLUE, weight: "bold", size: 10.5pt, it.body) #let examplebox(label, title, body) = block(width: 100%, breakable: true, fill: rgb("#EFF1F5"), stroke: 0.5pt + rgb("#CFDDF0"), radius: 4pt, inset: 10pt, above: 12pt, below: 12pt)[ #block(below: 6pt)[#box(fill: BLUE, inset: (x: 6pt, y: 2pt), radius: 2pt, text(fill: white, weight: "bold", size: 8.5pt, label)) #h(0.4em) #strong[#title]] #body] // rail = decorative left rule (raw brand token); labelcolor = AA-safe label text shade #let notebox(label, rail, labelcolor, tint, body) = block(width: 100%, breakable: true, fill: tint, stroke: (left: 3pt + rail), inset: (left: 10pt, rest: 8pt), radius: (right: 4pt), above: 11pt, below: 11pt)[ #text(fill: labelcolor, weight: "bold", size: 7.5pt, tracking: 0.5pt)[#upper(label)] #linebreak() #body] #let solutionbox(body) = block(above: 4pt, below: 8pt)[ #text(fill: BLUE, weight: "bold", size: 8.5pt)[Solution] #linebreak() #body] #let figph(msg) = block(width: 100%, height: 60pt, fill: rgb("#f6f7f9"), stroke: (paint: rgb("#c7ccd3"), dash: "dashed"), radius: 4pt, inset: 10pt)[ #align(center + horizon, text(fill: rgb("#889"), style: "italic", size: 9pt, msg))] // Standardize inlined figure sizes: measure the natural CeTZ canvas, then scale to a // consistent envelope (aspect-aware; see build_typst.py FIG_* constants). Unlike the // print preamble, dimensions are FLOORED: in an editor a user can trim a figure to a // degenerate 1-D shape (a bare line), and w/h or tw/w would then divide by zero. #let _STD_W = 3.5 #let _WIDE_W = 5.6 #let _MAX_H = 3.4 #let _ASPECT_WIDE = 2.2 #let _UPSCALE_MAX = 1.15 #let stdfig(body) = context { let m = measure(body) let w = calc.max(m.width / 1in, 0.01) let h = calc.max(m.height / 1in, 0.01) let tw = if w / h > _ASPECT_WIDE { _WIDE_W } else { _STD_W } let s = calc.min(tw / w, _MAX_H / h, _UPSCALE_MAX) align(center, box(scale(x: s * 100%, y: s * 100%, reflow: true, body))) } #show figure: set block(breakable: false) #set figure(gap: 8pt) #show figure.caption: set text(size: 8.5pt, fill: rgb("#555")) == 10.2#h(0.6em)Solve Quadratic Equations by Completing the Square #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Before you get started, take this readiness quiz. If you miss a problem, go back to the section listed and review the material. Simplify #math.equation(block: false, alt: "open parenthesis x plus 12 close parenthesis squared")[$attach(( x + 12 ), t: 2)$]. #linebreak() If you missed this problem, review . #solutionbox[ #math.equation(block: true, alt: "x squared plus 24 x plus 144")[$x^(2) + 24 x + 144$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Factor #math.equation(block: false, alt: "y squared minus 18 y plus 81")[$y^(2) − 18 y + 81$]. #linebreak() If you missed this problem, review . #solutionbox[ #math.equation(block: true, alt: "y minus 9 squared")[$y − 9^(2)$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Factor #math.equation(block: false, alt: "5 n squared plus 40 n plus 80")[$5 n^(2) + 40 n + 80$]. #linebreak() If you missed this problem, review . #solutionbox[ #math.equation(block: true, alt: "5 open parenthesis n plus 4 close parenthesis squared")[$5 ( n + 4 )^(2)$] ] ] So far, we have solved quadratic equations by factoring and using the Square Root Property. In this section, we will solve quadratic equations by a process called ‘completing the square.’ === Complete The Square of a Binomial Expression In the last section, we were able to use the Square Root Property to solve the equation #math.equation(block: false, alt: "open parenthesis y minus 7 close parenthesis squared equals 12")[$attach(( y − 7 ), t: 2) = 12$] because the left side was a perfect square. #math.equation(block: true, alt: "open parenthesis y minus 7 close parenthesis squared, equals, 12; y minus 7, equals, plus or minus the square root of 12; y minus 7, equals, plus or minus 2 the square root of 3; y, equals, 7 plus or minus 2 the square root of 3")[$attach(( y − 7 ), t: 2) & = & 12 \ y − 7 & = & ± #h(0.2em) sqrt(12) \ y − 7 & = & ± #h(0.2em) 2 sqrt(3) \ y & = & 7 ± 2 sqrt(3)$]We also solved an equation in which the left side was a perfect square trinomial, but we had to rewrite it the form #math.equation(block: false, alt: "open parenthesis x minus k close parenthesis squared")[$attach(( x − k ), t: 2)$] in order to use the square root property. #math.equation(block: true, alt: "x squared minus 10 x plus 25, equals, 18; open parenthesis x minus 5 close parenthesis squared, equals, 18")[$x^(2) − 10 x + 25 & = & 18 \ attach(( x − 5 ), t: 2) & = & 18$]What happens if the variable is not part of a perfect square? Can we use algebra to make a perfect square? Let’s study the binomial square pattern we have used many times. We will look at two examples. #math.equation(block: true, alt: "open parenthesis x plus 9 close parenthesis squared; open parenthesis x plus 9 close parenthesis open parenthesis x plus 9 close parenthesis; x squared plus 9 x plus 9 x plus 81; x squared plus 18 x plus 81, open parenthesis y minus 7 close parenthesis squared; open parenthesis y minus 7 close parenthesis open parenthesis y minus 7 close parenthesis; y squared minus 7 y minus 7 y plus 49; y squared minus 14 y plus 49")[$attach(( x + 9 ), t: 2) \ ( x + 9 ) ( x + 9 ) \ x^(2) + 9 x + 9 x + 81 \ x^(2) + 18 x + 81 & & & #h(2em) attach(( y − 7 ), t: 2) \ ( y − 7 ) #h(0.2em) ( y − 7 ) \ y^(2) − 7 y − 7 y + 49 \ y^(2) − 14 y + 49$]#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Binomial Squares Pattern] If #math.equation(block: false, alt: "a , b")[$a , b$] are real numbers, #math.equation(block: true, alt: "open parenthesis a plus b close parenthesis squared equals a squared plus 2 a b plus b squared")[$attach(( a + b ), t: 2) = a^(2) + 2 a b + b^(2)$]#figure(figph[Algebraic expansion of (a+b)^2, demonstrating that a binomial squared equals the square of the first term, plus twice the product of the terms, plus the square of the second term.], alt: "Algebraic expansion of (a+b)^2, demonstrating that a binomial squared equals the square of the first term, plus twice the product of the terms, plus the square of the second term.", caption: none) #math.equation(block: true, alt: "open parenthesis a minus b close parenthesis squared equals a squared minus 2 a b plus b squared")[$attach(( a − b ), t: 2) = a^(2) − 2 a b + b^(2)$]#figure(figph[An algebraic identity showing the square of a binomial (a-b)² equals the square of the first term minus twice the product of the terms plus the square of the second term.], alt: "An algebraic identity showing the square of a binomial (a-b)² equals the square of the first term minus twice the product of the terms plus the square of the second term.", caption: none) ] We can use this pattern to “make” a perfect square. We will start with the expression #math.equation(block: false, alt: "x squared plus 6 x")[$x^(2) + 6 x$]. Since there is a plus sign between the two terms, we will use the #math.equation(block: false, alt: "open parenthesis a plus b close parenthesis squared")[$attach(( a + b ), t: 2)$] pattern. #math.equation(block: true, alt: "a squared plus 2 a b plus b squared equals open parenthesis a plus b close parenthesis squared")[$a^(2) + 2 a b + b^(2) = attach(( a + b ), t: 2)$]Notice that the first term of #math.equation(block: false, alt: "x squared plus 6 x")[$x^(2) + 6 x$] is a square, #math.equation(block: false, alt: "x squared")[$x^(2)$]. We now know #math.equation(block: false, alt: "a equals x")[$a = x$]. What number can we add to #math.equation(block: false, alt: "x squared plus 6 x")[$x^(2) + 6 x$] to make a perfect square trinomial? #figure(figph[The image shows the expression a squared plus two a b plus b squared. Below it is the expression x squared plus six x plus a blank space. The x squared is below the a squared, the six x is below two a b and the blank is below the b squared.], alt: "The image shows the expression a squared plus two a b plus b squared. Below it is the expression x squared plus six x plus a blank space. The x squared is below the a squared, the six x is below two a b and the blank is below the b squared.", caption: none) The middle term of the Binomial Squares Pattern, #math.equation(block: false, alt: "2 a b")[$2 a b$], is twice the product of the two terms of the binomial. This means twice the product of #math.equation(block: false, alt: "x")[$x$] and some number is #math.equation(block: false, alt: "6 x")[$6 x$]. So, two times some number must be six. The number we need is #math.equation(block: false, alt: "the fraction 1 over 2 times 6 equals 3 .")[$frac(1, 2) · 6 = 3 .$] The second term in the binomial, #math.equation(block: false, alt: "b ,")[$b ,$] must be 3. #figure(figph[The image is similar to the image above. It shows the expression a squared plus two a b plus b squared. Below it is the expression x squared plus two times three times x plus a blank space. The x squared is below the a squared, the two times three times x is below two a b and the blank is below the b squared.], alt: "The image is similar to the image above. It shows the expression a squared plus two a b plus b squared. Below it is the expression x squared plus two times three times x plus a blank space. The x squared is below the a squared, the two times three times x is below two a b and the blank is below the b squared.", caption: none) We now know #math.equation(block: false, alt: "b equals 3")[$b = 3$]. Now, we just square the second term of the binomial to get the last term of the perfect square trinomial, so we square three to get the last term, nine. #figure(figph[The image shows the expression a squared plus two a b plus b squared. Below it is the expression x squared plus six x plus nine.], alt: "The image shows the expression a squared plus two a b plus b squared. Below it is the expression x squared plus six x plus nine.", caption: none) We can now factor to #figure(figph[The image shows the expression quantity a plus b squared. Below it is the expression quantity x plus three squared.], alt: "The image shows the expression quantity a plus b squared. Below it is the expression quantity x plus three squared.", caption: none) So, we found that adding nine to #math.equation(block: false, alt: "x squared plus 6 x")[$x^(2) + 6 x$] ‘completes the square,’ and we write it as #math.equation(block: false, alt: "open parenthesis x plus 3 close parenthesis squared")[$attach(( x + 3 ), t: 2)$]. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Complete a square.] To complete the square of #math.equation(block: false, alt: "x squared plus b x")[$x^(2) + b x$]: + Identify #math.equation(block: false, alt: "b")[$b$], the coefficient of #math.equation(block: false, alt: "x")[$x$]. + Find #math.equation(block: false, alt: "open parenthesis the fraction 1 over 2 b close parenthesis squared")[$attach(( frac(1, 2) b ), t: 2)$], the number to complete the square. + Add the #math.equation(block: false, alt: "open parenthesis the fraction 1 over 2 b close parenthesis squared")[$attach(( frac(1, 2) b ), t: 2)$] to #math.equation(block: false, alt: "x squared plus b x")[$x^(2) + b x$]. ] #examplebox("Example 1")[][ Complete the square to make a perfect square trinomial. Then, write the result as a binomial square. #math.equation(block: true, alt: "x squared plus 14 x")[$x^(2) + 14 x$] #solutionbox[ #figure(table( columns: 2, align: left, inset: 6pt, table.header([The coefficient of #emph[x] is 14.], [#figure(figph[Two mathematical expressions are displayed: x^2 + bx in red, followed by x^2 + 14x in black, illustrating a comparison or specific case where b=14.], alt: "Two mathematical expressions are displayed: x^2 + bx in red, followed by x^2 + 14x in black, illustrating a comparison or specific case where b=14.", caption: none)]), [#math.equation(block: false, alt: "Find open parenthesis the fraction 1 over 2 b close parenthesis squared .; open parenthesis the fraction 1 over 2 times 14 close parenthesis squared; open parenthesis 7 close parenthesis squared; 49")[$"Find" #h(0.2em) attach(( frac(1, 2) b ), t: 2) . \ attach(( frac(1, 2) ⋅ 14 ), t: 2) \ attach(( 7 ), t: 2) \ 49$]], [], [Add 49 to the binomial to complete the square.], [#math.equation(block: false, alt: "x squared plus 14 x plus 49")[$x^(2) + 14 x + 49$]], [Rewrite as a binomial square.], [#math.equation(block: false, alt: "open parenthesis x plus 7 close parenthesis squared")[$attach(( x + 7 ), t: 2)$]], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Complete the square to make a perfect square trinomial. Write the result as a binomial square. #math.equation(block: true, alt: "y squared plus 12 y")[$y^(2) + 12 y$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis y plus 6 close parenthesis squared")[$attach(( y + 6 ), t: 2)$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Complete the square to make a perfect square trinomial. Write the result as a binomial square. #math.equation(block: true, alt: "z squared plus 8 z")[$z^(2) + 8 z$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis z plus 4 close parenthesis squared")[$attach(( z + 4 ), t: 2)$] ] ] #examplebox("Example 2")[][ Complete the square to make a perfect square trinomial. Then, write the result as a binomial squared. #math.equation(block: false, alt: "m squared minus 26 m")[$m^(2) − 26 m$] #solutionbox[ #figure(table( columns: 2, align: left, inset: 6pt, table.header([The coefficient of #emph[m] is −26.], [#figure(figph[The image shows the expression m squared minus 26 m with x squared plus b x written above it. The coefficient of m is negative 26 so b is negative 26. Find half of b and square it. Half of negative 26 is negative 13 and negative 13 squared is 169. Add 169 to the binomial to complete the square and get the expression m squared minus 26 m plus 169 which is the quantity m minus 13 squared.], alt: "The image shows the expression m squared minus 26 m with x squared plus b x written above it. The coefficient of m is negative 26 so b is negative 26. Find half of b and square it. Half of negative 26 is negative 13 and negative 13 squared is 169. Add 169 to the binomial to complete the square and get the expression m squared minus 26 m plus 169 which is the quantity m minus 13 squared.", caption: none)]), [#math.equation(block: false, alt: "Find open parenthesis the fraction 1 over 2 b close parenthesis squared .; open parenthesis the fraction 1 over 2 times open parenthesis − 26 close parenthesis close parenthesis squared; open parenthesis − 13 close parenthesis squared; 169")[$"Find" #h(0.2em) attach(( frac(1, 2) b ), t: 2) . \ attach(( frac(1, 2) ⋅ ( "−" 26 ) ), t: 2) \ attach(( "−" 13 ), t: 2) \ 169$]], [], [Add 169 to the binomial to complete the square.], [#math.equation(block: false, alt: "m squared minus 26 m plus 169")[$m^(2) − 26 m + 169$]], [Rewrite as a binomial square.], [#math.equation(block: false, alt: "open parenthesis m minus 13 close parenthesis squared")[$attach(( m − 13 ), t: 2)$]], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Complete the square to make a perfect square trinomial. Write the result as a binomial square. #math.equation(block: true, alt: "a squared minus 20 a")[$a^(2) − 20 a$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis a minus 10 close parenthesis squared")[$attach(( a − 10 ), t: 2)$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Complete the square to make a perfect square trinomial. Write the result as a binomial square. #math.equation(block: true, alt: "b squared minus 4 b")[$b^(2) − 4 b$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis b minus 2 close parenthesis squared")[$attach(( b − 2 ), t: 2)$] ] ] #examplebox("Example 3")[][ Complete the square to make a perfect square trinomial. Then, write the result as a binomial squared. #math.equation(block: true, alt: "u squared minus 9 u")[$u^(2) − 9 u$] #solutionbox[ #figure(table( columns: 2, align: left, inset: 6pt, table.header([The coefficient of #emph[u] is −9.], [#figure(figph[Two algebraic expressions are shown: 'x^2 + bx' in red, and 'u^2 - 9u' in grey, representing quadratic terms or parts of polynomial equations.], alt: "Two algebraic expressions are shown: 'x^2 + bx' in red, and 'u^2 - 9u' in grey, representing quadratic terms or parts of polynomial equations.", caption: none)]), [#math.equation(block: false, alt: "Find open parenthesis the fraction 1 over 2 b close parenthesis squared .; open parenthesis the fraction 1 over 2 times open parenthesis − 9 close parenthesis close parenthesis squared; open parenthesis minus the fraction 9 over 2 close parenthesis squared; the fraction 81 over 4")[$"Find" #h(0.2em) attach(( frac(1, 2) b ), t: 2) . \ attach(( frac(1, 2) ⋅ ( "−" 9 ) ), t: 2) \ attach(( − frac(9, 2) ), t: 2) \ frac(81, 4)$]], [], [Add #math.equation(block: false, alt: "the fraction 81 over 4")[$frac(81, 4)$] to the binomial to complete the square.], [#math.equation(block: false, alt: "u squared minus 9 u plus the fraction 81 over 4")[$u^(2) − 9 u + frac(81, 4)$]], [Rewrite as a binomial square.], [#math.equation(block: false, alt: "open parenthesis u minus the fraction 9 over 2 close parenthesis squared")[$attach(( u − frac(9, 2) ), t: 2)$]], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Complete the square to make a perfect square trinomial. Write the result as a binomial square. #math.equation(block: true, alt: "m squared minus 5 m")[$m^(2) − 5 m$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis m minus the fraction 5 over 2 close parenthesis squared")[$attach(( m − frac(5, 2) ), t: 2)$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Complete the square to make a perfect square trinomial. Write the result as a binomial square. #math.equation(block: true, alt: "n squared plus 13 n")[$n^(2) + 13 n$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis n plus the fraction 13 over 2 close parenthesis squared")[$attach(( n + frac(13, 2) ), t: 2)$] ] ] #examplebox("Example 4")[][ Complete the square to make a perfect square trinomial. Then, write the result as a binomial squared. #math.equation(block: true, alt: "p squared plus the fraction 1 over 2 p")[$p^(2) + frac(1, 2) p$] #solutionbox[ #figure(table( columns: 2, align: left, inset: 6pt, table.header([The coefficient of #emph[p] is #math.equation(block: false, alt: "the fraction 1 over 2")[$frac(1, 2)$].], [#figure(figph[Two quadratic expressions, x×2 + bx and p×2 + (1/2)p, illustrating terms often involved in completing the square.], alt: "Two quadratic expressions, x×2 + bx and p×2 + (1/2)p, illustrating terms often involved in completing the square.", caption: none)]), [#math.equation(block: false, alt: "Find open parenthesis the fraction 1 over 2 b close parenthesis squared .; open parenthesis the fraction 1 over 2 times the fraction 1 over 2 close parenthesis squared; open parenthesis the fraction 1 over 4 close parenthesis squared; the fraction 1 over 16")[$"Find" #h(0.2em) attach(( frac(1, 2) b ), t: 2) . \ attach(( frac(1, 2) ⋅ frac(1, 2) ), t: 2) \ attach(( frac(1, 4) ), t: 2) \ frac(1, 16)$]], [], [Add #math.equation(block: false, alt: "the fraction 1 over 16")[$frac(1, 16)$] to the binomial to complete the square.], [#math.equation(block: false, alt: "p squared plus the fraction 1 over 2 p plus the fraction 1 over 16")[$p^(2) + frac(1, 2) p + frac(1, 16)$]], [Rewrite as a binomial square.], [#math.equation(block: false, alt: "open parenthesis p plus the fraction 1 over 4 close parenthesis squared")[$attach(( p + frac(1, 4) ), t: 2)$]], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Complete the square to make a perfect square trinomial. Write the result as a binomial square. #math.equation(block: true, alt: "p squared plus the fraction 1 over 4 p")[$p^(2) + frac(1, 4) p$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis p plus the fraction 1 over 8 close parenthesis squared")[$attach(( p + frac(1, 8) ), t: 2)$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Complete the square to make a perfect square trinomial. Write the result as a binomial square. #math.equation(block: true, alt: "q squared minus the fraction 2 over 3 q")[$q^(2) − frac(2, 3) q$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis q minus the fraction 1 over 3 close parenthesis squared")[$attach(( q − frac(1, 3) ), t: 2)$] ] ] === Solve Quadratic Equations of the Form #emph[x]#super[2] + #emph[bx] + #emph[c] = 0 by completing the square In solving equations, we must always do the same thing to both sides of the equation. This is true, of course, when we solve a quadratic equation by #strong[completing the square], too. When we add a term to one side of the equation to make a perfect square trinomial, we must also add the same term to the other side of the equation. For example, if we start with the equation #math.equation(block: false, alt: "x squared plus 6 x equals 40")[$x^(2) + 6 x = 40$] and we want to complete the square on the left, we will add nine to both sides of the equation. #figure(figph[The image shows the equation x squared plus six x equals 40. Below that the equation is rewritten as x squared plus six x plus blank space equals 40 plus blank space. Below that the equation is rewritten again as x squared plus six x plus nine equals 40 plus nine.], alt: "The image shows the equation x squared plus six x equals 40. Below that the equation is rewritten as x squared plus six x plus blank space equals 40 plus blank space. Below that the equation is rewritten again as x squared plus six x plus nine equals 40 plus nine.", caption: none) Then, we factor on the left and simplify on the right. #math.equation(block: true, alt: "open parenthesis x plus 3 close parenthesis squared equals 49")[$attach(( x + 3 ), t: 2) = 49$]Now the equation is in the form to solve using the Square Root Property. Completing the square is a way to transform an equation into the form we need to be able to use the Square Root Property. #examplebox("Example 5")[How To Solve a Quadratic Equation of the Form #math.equation(block: false, alt: "x squared plus b x plus c equals 0")[$x^(2) + b x + c = 0$] by Completing the Square][ Solve #math.equation(block: false, alt: "x squared plus 8 x equals 48")[$x^(2) + 8 x = 48$] by completing the square. #solutionbox[ #figure(figph[The image shows the steps to solve the equation x squared plus eight x equals 48. Step one is to isolate the variable terms on one side and the constant terms on the other. The equation already has all the variables on the left.], alt: "The image shows the steps to solve the equation x squared plus eight x equals 48. Step one is to isolate the variable terms on one side and the constant terms on the other. The equation already has all the variables on the left.", caption: none) #figure(figph[Step two is to find the quantity half of b squared, the number to complete the square and add it to both sides of the equation. The coefficient of x is eight so b is eight. Take half of eight, which is four and square it to get 16. Add 16 to both sides of the equation to get x squared plus eight x plus 16 equals 48 plus 16.], alt: "Step two is to find the quantity half of b squared, the number to complete the square and add it to both sides of the equation. The coefficient of x is eight so b is eight. Take half of eight, which is four and square it to get 16. Add 16 to both sides of the equation to get x squared plus eight x plus 16 equals 48 plus 16.", caption: none) #figure(figph[Step three is to factor the perfect square trinomial as a binomial square. The left side is the perfect square trinomial x squared plus eight x plus 16 which factors to the quantity x plus four squared. Adding on the right side 48 plus 16 is 64. The equation is now the quantity x plus four squared equals 64.], alt: "Step three is to factor the perfect square trinomial as a binomial square. The left side is the perfect square trinomial x squared plus eight x plus 16 which factors to the quantity x plus four squared. Adding on the right side 48 plus 16 is 64. The equation is now the quantity x plus four squared equals 64.", caption: none) #figure(figph[Step four is to use the square root property to make the equation x plus four equals plus or minus the square root of 64.], alt: "Step four is to use the square root property to make the equation x plus four equals plus or minus the square root of 64.", caption: none) #figure(figph[Step five is to simplify the radical and then solve the two resulting equations. The square root of 64 is eight. The equation can be written as two equations: x plus four equals eight and x plus four equals negative eight. Solving each equation gives x equals four or negative 12.], alt: "Step five is to simplify the radical and then solve the two resulting equations. The square root of 64 is eight. The equation can be written as two equations: x plus four equals eight and x plus four equals negative eight. Solving each equation gives x equals four or negative 12.", caption: none) #figure(figph[Step six is to check the solutions. To check the solutions put each answer in the original equation. Substituting x equals four in the original equation to get four squared plus eight times four equals 48. The left side simplifies to 16 plus 32 which is 48. Substituting x equals negative 12 in the original equation to get negative 12 squared plus eight times negative 12 equals 48. The left side simplifies to 144 minus 96 which is 48.], alt: "Step six is to check the solutions. To check the solutions put each answer in the original equation. Substituting x equals four in the original equation to get four squared plus eight times four equals 48. The left side simplifies to 16 plus 32 which is 48. Substituting x equals negative 12 in the original equation to get negative 12 squared plus eight times negative 12 equals 48. The left side simplifies to 144 minus 96 which is 48.", caption: none) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve #math.equation(block: false, alt: "c squared plus 4 c equals 5")[$c^(2) + 4 c = 5$] by completing the square. #solutionbox[ #math.equation(block: true, alt: "c equals −5 , c equals 1")[$c = −5 , c = 1$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve #math.equation(block: false, alt: "d squared plus 10 d equals −9")[$d^(2) + 10 d = −9$] by completing the square. #solutionbox[ #math.equation(block: true, alt: "d equals −9 , d equals −1")[$d = −9 , d = −1$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Solve a quadratic equation of the form #math.equation(block: false, alt: "x squared plus b x plus c equals 0")[$x^(2) + b x + c = 0$] by completing the square.] + Isolate the variable terms on one side and the constant terms on the other. + Find #math.equation(block: false, alt: "open parenthesis the fraction 1 over 2 times b close parenthesis squared")[$attach(( frac(1, 2) · b ), t: 2)$], the number to complete the square. Add it to both sides of the equation. + Factor the perfect square trinomial as a binomial square. + Use the Square Root Property. + Simplify the radical and then solve the two resulting equations. + Check the solutions. ] #examplebox("Example 6")[][ Solve #math.equation(block: false, alt: "y squared minus 6 y equals 16")[$y^(2) − 6 y = 16$] by completing the square. #solutionbox[ #figure(table( columns: 2, align: left, inset: 6pt, table.header([The variable terms are on the left side.], [#figure(figph[Two mathematical expressions are shown: 'x^2 - bx' and 'c' in red text, and 'y^2 - 6y = 16' in black text.], alt: "Two mathematical expressions are shown: 'x^2 - bx' and 'c' in red text, and 'y^2 - 6y = 16' in black text.", caption: none)]), [Take half of #math.equation(block: false, alt: "minus 6")[$− 6$] and square it. #math.equation(block: false, alt: "open parenthesis the fraction 1 over 2 open parenthesis − 6 close parenthesis close parenthesis squared equals 9")[$( frac(1, 2) ( "−" 6 ) )^(2) = 9$]], [#figure(figph[A mathematical equation y^2 - 6y + blank space over ((1/2)\*(-6))^2 = 16, illustrating the process of completing the square with a missing term. The term to be added is shown as ((1/2)\*(-6))^2.], alt: "A mathematical equation y^2 - 6y + blank space over ((1/2)*(-6))^2 = 16, illustrating the process of completing the square with a missing term. The term to be added is shown as ((1/2)*(-6))^2.", caption: none)], [Add 9 to both sides.], [#figure(figph[Mathematical equation y^2 - 6y + 9 = 16 + 9, where the number 9 is highlighted in red on both sides, illustrating the completion of the square method.], alt: "Mathematical equation y^2 - 6y + 9 = 16 + 9, where the number 9 is highlighted in red on both sides, illustrating the completion of the square method.", caption: none)], [Factor the perfect square trinomial as a binomial square.], [#figure(figph[The image shows the mathematical equation (y-3)^2 = 25, which is a quadratic equation where the square of the difference between y and 3 is equal to 25.], alt: "The image shows the mathematical equation (y-3)^2 = 25, which is a quadratic equation where the square of the difference between y and 3 is equal to 25.", caption: none)], [Use the Square Root Property.], [#figure(figph[Equation solving for y: y minus 3 equals plus or minus the square root of 25. The solution will involve considering both the positive and negative square roots of 25.], alt: "Equation solving for y: y minus 3 equals plus or minus the square root of 25. The solution will involve considering both the positive and negative square roots of 25.", caption: none)], [Simplify the radical.], [#figure(figph[A mathematical equation is displayed, showing 'y - 3 = '\> with '±5' to the right of the equals sign. The equation is centered on a white background.], alt: "A mathematical equation is displayed, showing 'y - 3 = '> with '±5' to the right of the equals sign. The equation is centered on a white background.", caption: none)], [Solve for #emph[y].], [#figure(figph[A mathematical expression reads 'y = 3 ± 5', indicating that y can be either 3 plus 5 or 3 minus 5.], alt: "A mathematical expression reads 'y = 3 ± 5', indicating that y can be either 3 plus 5 or 3 minus 5.", caption: none)], [Rewrite to show two solutions.], [#figure(figph[Two mathematical equations are displayed on a white background. The first equation is y = 3 + 5, and the second is y = 3 - 5.], alt: "Two mathematical equations are displayed on a white background. The first equation is y = 3 + 5, and the second is y = 3 - 5.", caption: none)], [Solve the equations.], [#figure(figph[The image displays mathematical equations 'y = 8' and 'y = -2' written in a dark grey font on a plain white background.], alt: "The image displays mathematical equations 'y = 8' and 'y = -2' written in a dark grey font on a plain white background.", caption: none)], [Check. #linebreak() #linebreak() #figure(figph[Verification of solutions for the quadratic equation y^2 - 6y = 16. It shows that substituting y=8 and y=-2 into the equation results in a true statement (16=16), confirming they are valid roots.], alt: "Verification of solutions for the quadratic equation y^2 - 6y = 16. It shows that substituting y=8 and y=-2 into the equation results in a true statement (16=16), confirming they are valid roots.", caption: none)], [], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve #math.equation(block: false, alt: "r squared minus 4 r equals 12")[$r^(2) − 4 r = 12$] by completing the square. #solutionbox[ #math.equation(block: true, alt: "r equals −2 , r equals 6")[$r = −2 , r = 6$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve #math.equation(block: false, alt: "t squared minus 10 t equals 11")[$t^(2) − 10 t = 11$] by completing the square. #solutionbox[ #math.equation(block: true, alt: "t equals −1 , t equals 11")[$t = −1 , t = 11$] ] ] #examplebox("Example 7")[][ Solve #math.equation(block: false, alt: "x squared plus 4 x equals −21")[$x^(2) + 4 x = −21$] by completing the square. #solutionbox[ #figure(table( columns: 2, align: left, inset: 6pt, table.header([The variable terms are on the left side.], [#figure(figph[A general algebraic expression, x^2 + bx c (in red), is displayed above a specific quadratic equation, x^2 + 4x = -21 (in black), for comparison of coefficients.], alt: "A general algebraic expression, x^2 + bx c (in red), is displayed above a specific quadratic equation, x^2 + 4x = -21 (in black), for comparison of coefficients.", caption: none)]), [Take half of #math.equation(block: false, alt: "4")[$4$] and square it. #math.equation(block: false, alt: "open parenthesis the fraction 1 over 2 open parenthesis 4 close parenthesis close parenthesis squared equals 4")[$( frac(1, 2) ( 4 ) )^(2) = 4$]], [#figure(figph[A mathematical equation illustrating the process of completing the square: x^2 + 4x + (1/2 \* 4)^2 = -21. The term to be added for completing the square is highlighted in red.], alt: "A mathematical equation illustrating the process of completing the square: x^2 + 4x + (1/2 * 4)^2 = -21. The term to be added for completing the square is highlighted in red.", caption: none)], [Add 4 to both sides.], [#figure(figph[A quadratic equation x² + 4x + 4 = -21 + 4 is shown, with the number 4, added to both sides, highlighted in red, as a step in completing the square.], alt: "A quadratic equation x² + 4x + 4 = -21 + 4 is shown, with the number 4, added to both sides, highlighted in red, as a step in completing the square.", caption: none)], [Factor the perfect square trinomial as a binomial square.], [#figure(figph[The equation (x + 2)² = -17 is displayed, representing a quadratic equation with no real solutions since a squared term cannot equal a negative number.], alt: "The equation (x + 2)² = -17 is displayed, representing a quadratic equation with no real solutions since a squared term cannot equal a negative number.", caption: none)], [Use the Square Root Property.], [#figure(figph[A mathematical equation is displayed on a white background: 'x + 2 = ±√-17'. The equation involves a variable x, the number 2, an equality sign, and the positive or negative square root of -17.], alt: "A mathematical equation is displayed on a white background: 'x + 2 = ±√-17'. The equation involves a variable x, the number 2, an equality sign, and the positive or negative square root of -17.", caption: none)], [We cannot take the square root of a negative number.], [There is no real solution.], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve #math.equation(block: false, alt: "y squared minus 10 y equals −35")[$y^(2) − 10 y = −35$] by completing the square. #solutionbox[ no real solution ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve #math.equation(block: false, alt: "z squared plus 8 z equals −19")[$z^(2) + 8 z = −19$] by completing the square. #solutionbox[ no real solution ] ] In the previous example, there was no real solution because #math.equation(block: false, alt: "open parenthesis x plus k close parenthesis squared")[$attach(( x + k ), t: 2)$] was equal to a negative number. #examplebox("Example 8")[][ Solve #math.equation(block: false, alt: "p squared minus 18 p equals −6")[$p^(2) − 18 p = −6$] by completing the square. #solutionbox[ #figure(table( columns: 2, align: left, inset: 6pt, table.header([The variable terms are on the left side.], [#figure(figph[Mathematical notation featuring the quadratic equation P^2 - 18P = -6, with general quadratic terms x^2 + bx and constant c highlighted in red above it.], alt: "Mathematical notation featuring the quadratic equation P^2 - 18P = -6, with general quadratic terms x^2 + bx and constant c highlighted in red above it.", caption: none)]), [Take half of #math.equation(block: false, alt: "− 18")[$"−" 18$] and square it. #math.equation(block: false, alt: "open parenthesis the fraction 1 over 2 open parenthesis − 18 close parenthesis close parenthesis squared equals 81")[$( frac(1, 2) ( "−" 18 ) )^(2) = 81$]], [#figure(figph[Mathematical equation demonstrating completing the square: P^2 - 18p + (1/2 \* -18)^2 = -6.], alt: "Mathematical equation demonstrating completing the square: P^2 - 18p + (1/2 * -18)^2 = -6.", caption: none)], [Add 81 to both sides.], [#figure(figph[A mathematical equation shows p squared minus 18p plus 81 equals negative 6 plus 81.], alt: "A mathematical equation shows p squared minus 18p plus 81 equals negative 6 plus 81.", caption: none)], [Factor the perfect square trinomial as a binomial square.], [#figure(figph[A mathematical equation is displayed, showing (p - 9)^2 = 75. The equation is centered on a white background, suggesting a problem from a textbook or test.], alt: "A mathematical equation is displayed, showing (p - 9)^2 = 75. The equation is centered on a white background, suggesting a problem from a textbook or test.", caption: none)], [Use the Square Root Property.], [#figure(figph[Equation showing p minus 9 equals positive or negative square root of 75.], alt: "Equation showing p minus 9 equals positive or negative square root of 75.", caption: none)], [Simplify the radical.], [#figure(figph[A mathematical equation displays p - 9 = ±5√3 on a white background, representing an algebraic step involving a variable, a constant, and a radical expression with positive and negative roots.], alt: "A mathematical equation displays p - 9 = ±5√3 on a white background, representing an algebraic step involving a variable, a constant, and a radical expression with positive and negative roots.", caption: none)], [Solve for #emph[p.]], [#figure(figph[A mathematical equation showing p equals 9 plus or minus 5 times the square root of 3, written in a clear, standard mathematical notation on a white background.], alt: "A mathematical equation showing p equals 9 plus or minus 5 times the square root of 3, written in a clear, standard mathematical notation on a white background.", caption: none)], [Rewrite to show two solutions.], [#figure(figph[Two mathematical equations are displayed, showing two possible values for 'p': p = 9 + 5√3 and p = 9 - 5√3.], alt: "Two mathematical equations are displayed, showing two possible values for 'p': p = 9 + 5√3 and p = 9 - 5√3.", caption: none)], [Check. #linebreak() #figure(figph[The image displays the verification of two solutions, (9 + 5sqrt(3)) and (9 - 5sqrt(3)), for the quadratic equation p^2 - 18p = -6, with both calculations confirming the equality -6 = -6.], alt: "The image displays the verification of two solutions, (9 + 5sqrt(3)) and (9 - 5sqrt(3)), for the quadratic equation p^2 - 18p = -6, with both calculations confirming the equality -6 = -6.", caption: none)], [], )) Another way to check this would be to use a calculator. Evaluate #math.equation(block: false, alt: "p squared minus 18 p")[$p^(2) − 18 p$] for both of the solutions. The answer should be #math.equation(block: false, alt: "−6")[$−6$]. ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve #math.equation(block: false, alt: "x squared minus 16 x equals −16")[$x^(2) − 16 x = −16$] by completing the square. #solutionbox[ #math.equation(block: true, alt: "x equals 8 plus or minus 4 the square root of 3")[$x = 8 ± 4 sqrt(3)$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve #math.equation(block: false, alt: "y squared plus 8 y equals 11")[$y^(2) + 8 y = 11$] by completing the square. #solutionbox[ #math.equation(block: true, alt: "y equals −4 plus or minus 3 the square root of 3")[$y = −4 ± 3 sqrt(3)$] ] ] We will start the next example by isolating the variable terms on the left side of the equation. #examplebox("Example 9")[][ Solve #math.equation(block: false, alt: "x squared plus 10 x plus 4 equals 15")[$x^(2) + 10 x + 4 = 15$] by completing the square. #solutionbox[ #figure(table( columns: 2, align: left, inset: 6pt, table.header([The variable terms are on the left side.], [#figure(figph[The image shows a quadratic equation: x^2 + 10x + 4 = 15. The equation is presented in a clear, digital font against a white background.], alt: "The image shows a quadratic equation: x^2 + 10x + 4 = 15. The equation is presented in a clear, digital font against a white background.", caption: none)]), [Subtract #math.equation(block: false, alt: "4")[$4$] to get the constant terms on the right side.], [#figure(figph[A quadratic equation is shown: x^2 + 10x = 11.], alt: "A quadratic equation is shown: x^2 + 10x = 11.", caption: none)], [Take half of 10 and square it. #math.equation(block: false, alt: "open parenthesis the fraction 1 over 2 open parenthesis 10 close parenthesis close parenthesis squared equals 25")[$( frac(1, 2) ( 10 ) )^(2) = 25$]], [#figure(figph[An example of completing the square for x^2 + 10x + \_\_ = 11. The term to be added is shown as (1/2 \* 10)^2, which is 25.], alt: "An example of completing the square for x^2 + 10x + __ = 11. The term to be added is shown as (1/2 * 10)^2, which is 25.", caption: none)], [Add 25 to both sides.], [#figure(figph[The quadratic equation x^2 + 10x + 25 = 11 + 25 is shown, illustrating the technique of adding a constant to both sides to complete the square, enabling factorization of the left side.], alt: "The quadratic equation x^2 + 10x + 25 = 11 + 25 is shown, illustrating the technique of adding a constant to both sides to complete the square, enabling factorization of the left side.", caption: none)], [Factor the perfect square trinomial as a binomial square.], [#figure(figph[A mathematical equation is displayed against a white background, reading '(x + 5) squared = 36'.], alt: "A mathematical equation is displayed against a white background, reading '(x + 5) squared = 36'.", caption: none)], [Use the Square Root Property.], [#figure(figph[A mathematical equation is displayed on a white background: x + 5 = plus or minus the square root of 36.], alt: "A mathematical equation is displayed on a white background: x + 5 = plus or minus the square root of 36.", caption: none)], [Simplify the radical.], [#figure(figph[A mathematical equation is displayed, showing 'x + 5 = 6' with a plus-minus sign before the 6.], alt: "A mathematical equation is displayed, showing 'x + 5 = 6' with a plus-minus sign before the 6.", caption: none)], [Solve for #emph[x.]], [#figure(figph[A mathematical equation is displayed against a white background, reading 'x = -5 ×1 6'.], alt: "A mathematical equation is displayed against a white background, reading 'x = -5 ×1 6'.", caption: none)], [Rewrite to show two equations.], [#figure(figph[Two mathematical equations are displayed: x = -5 + 6 and x = -5 - 6.], alt: "Two mathematical equations are displayed: x = -5 + 6 and x = -5 - 6.", caption: none)], [Solve the equations.], [#figure(figph[A mathematical expression on a white background states 'x = 1, x = -11'.], alt: "A mathematical expression on a white background states 'x = 1, x = -11'.", caption: none)], [Check. #linebreak() #figure(figph[Verification of solutions for the quadratic equation x^2 + 10x + 4 = 15. The calculations demonstrate that both x=1 and x=-11 satisfy the equation, resulting in 15 = 15.], alt: "Verification of solutions for the quadratic equation x^2 + 10x + 4 = 15. The calculations demonstrate that both x=1 and x=-11 satisfy the equation, resulting in 15 = 15.", caption: none)], [], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve #math.equation(block: false, alt: "a squared plus 4 a plus 9 equals 30")[$a^(2) + 4 a + 9 = 30$] by completing the square. #solutionbox[ #math.equation(block: true, alt: "a equals −7 , a equals 3")[$a = −7 , a = 3$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve #math.equation(block: false, alt: "b squared plus 8 b minus 4 equals 16")[$b^(2) + 8 b − 4 = 16$] by completing the square. #solutionbox[ #math.equation(block: true, alt: "b equals −10 , b equals 2")[$b = −10 , b = 2$] ] ] To solve the next equation, we must first collect all the variable terms to the left side of the equation. Then, we proceed as we did in the previous examples. #examplebox("Example 10")[][ Solve #math.equation(block: false, alt: "n squared equals 3 n plus 11")[$n^(2) = 3 n + 11$] by completing the square. #solutionbox[ #figure(table( columns: 2, align: left, inset: 6pt, table.header([], [#figure(figph[The image shows a mathematical equation: n squared equals 3n plus 11.], alt: "The image shows a mathematical equation: n squared equals 3n plus 11.", caption: none)]), [Subtract 3#emph[n] to get the variable terms on the left side.], [#figure(figph[A mathematical equation shows 'n squared minus three n equals eleven' in black text on a white background.], alt: "A mathematical equation shows 'n squared minus three n equals eleven' in black text on a white background.", caption: none)], [Take half of #math.equation(block: false, alt: "− 3")[$"−" 3$] and square it. #math.equation(block: false, alt: "open parenthesis the fraction 1 over 2 open parenthesis − 3 close parenthesis close parenthesis squared equals the fraction 9 over 4")[$( frac(1, 2) ( "−" 3 ) )^(2) = frac(9, 4)$]], [#figure(figph[A mathematical equation n^2 - 3n + \_\_\_\_ = 11, illustrating the completing the square method. The blank space indicates where (1/2 \* (-3))^2 should be placed.], alt: "A mathematical equation n^2 - 3n + ____ = 11, illustrating the completing the square method. The blank space indicates where (1/2 * (-3))^2 should be placed.", caption: none)], [Add #math.equation(block: false, alt: "the fraction 9 over 4")[$frac(9, 4)$] to both sides.], [#figure(figph[A mathematical equation showing n squared minus 3n plus the fraction 9/4 equals 11 plus the fraction 9/4. The fraction 9/4 is highlighted in red on both sides of the equation.], alt: "A mathematical equation showing n squared minus 3n plus the fraction 9/4 equals 11 plus the fraction 9/4. The fraction 9/4 is highlighted in red on both sides of the equation.", caption: none)], [Factor the perfect square trinomial as a binomial square.], [#figure(figph[A mathematical equation shows (n - 3/2)^2 = 44/4 + 9/4. The equation involves a variable 'n', fractions, and an exponent.], alt: "A mathematical equation shows (n - 3/2)^2 = 44/4 + 9/4. The equation involves a variable 'n', fractions, and an exponent.", caption: none)], [Add the fractions on the right side.], [#figure(figph[A mathematical equation showing the quantity (n minus three-halves) squared equals fifty-three fourths.], alt: "A mathematical equation showing the quantity (n minus three-halves) squared equals fifty-three fourths.", caption: none)], [Use the Square Root Property.], [#figure(figph[Equation showing n minus three-halves equals plus or minus square root of fifty-three over four, representing the solutions for n in an algebraic equation.], alt: "Equation showing n minus three-halves equals plus or minus square root of fifty-three over four, representing the solutions for n in an algebraic equation.", caption: none)], [Simplify the radical.], [#figure(figph[A mathematical equation showing 'n minus three halves equals plus or minus the square root of fifty-three over two'. The equation is presented in a clear, standard mathematical notation on a white background.], alt: "A mathematical equation showing 'n minus three halves equals plus or minus the square root of fifty-three over two'. The equation is presented in a clear, standard mathematical notation on a white background.", caption: none)], [Solve for #emph[n.]], [#figure(figph[A mathematical equation showing 'n' equals three halves plus or minus the square root of fifty-three divided by two. This expresses the two possible solutions for 'n'.], alt: "A mathematical equation showing 'n' equals three halves plus or minus the square root of fifty-three divided by two. This expresses the two possible solutions for 'n'.", caption: none)], [Rewrite to show two equations.], [#figure(figph[Two mathematical equations for 'n' are shown, where n equals three-halves plus the square root of 53 over two, and n also equals three-halves minus the square root of 53 over two.], alt: "Two mathematical equations for 'n' are shown, where n equals three-halves plus the square root of 53 over two, and n also equals three-halves minus the square root of 53 over two.", caption: none)], [Check. We leave the check for you!], [], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve #math.equation(block: false, alt: "p squared equals 5 p plus 9")[$p^(2) = 5 p + 9$] by completing the square. #solutionbox[ #math.equation(block: true, alt: "p equals the fraction 5 over 2 plus or minus the fraction the square root of 61 over 2")[$p = frac(5, 2) ± frac(sqrt(61), 2)$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve #math.equation(block: false, alt: "q squared equals 7 q minus 3")[$q^(2) = 7 q − 3$] by completing the square. #solutionbox[ #math.equation(block: true, alt: "q equals the fraction 7 over 2 plus or minus the fraction the square root of 37 over 2")[$q = frac(7, 2) ± frac(sqrt(37), 2)$] ] ] Notice that the left side of the next equation is in factored form. But the right side is not zero, so we cannot use the Zero Product Property. Instead, we multiply the factors and then put the equation into the standard form to solve by completing the square. #examplebox("Example 11")[][ Solve #math.equation(block: false, alt: "open parenthesis x minus 3 close parenthesis open parenthesis x plus 5 close parenthesis equals 9")[$( x − 3 ) ( x + 5 ) = 9$] by completing the square. #solutionbox[ #figure(table( columns: 2, align: left, inset: 6pt, table.header([], [#figure(figph[A mathematical equation is displayed, reading '(x - 3)(x + 5) = 9' in black text on a white background.], alt: "A mathematical equation is displayed, reading '(x - 3)(x + 5) = 9' in black text on a white background.", caption: none)]), [We multiply binomials on the left.], [#figure(figph[A quadratic equation is displayed, showing 'x^2 + 2x - 15 = 9'.], alt: "A quadratic equation is displayed, showing 'x^2 + 2x - 15 = 9'.", caption: none)], [Add 15 to get the variable terms on the left side.], [#figure(figph[The image displays the quadratic equation x^2 + 2x = 24.], alt: "The image displays the quadratic equation x^2 + 2x = 24.", caption: none)], [Take half of 2 and square it. #math.equation(block: false, alt: "open parenthesis the fraction 1 over 2 open parenthesis 2 close parenthesis close parenthesis squared equals 1")[$( frac(1, 2) ( 2 ) )^(2) = 1$]], [#figure(figph[A mathematical equation x^2 + 2x + blank = 24, with the expression (1/2 \* 2)^2 written in red below the blank, illustrating the 'completing the square' method for solving quadratic equations.], alt: "A mathematical equation x^2 + 2x + blank = 24, with the expression (1/2 * 2)^2 written in red below the blank, illustrating the 'completing the square' method for solving quadratic equations.", caption: none)], [Add 1 to both sides.], [#figure(figph[A mathematical equation shows x squared plus 2x plus 1 equals 24 plus 1, demonstrating the first step in completing the square where 1 is added to both sides.], alt: "A mathematical equation shows x squared plus 2x plus 1 equals 24 plus 1, demonstrating the first step in completing the square where 1 is added to both sides.", caption: none)], [Factor the perfect square trinomial as a binomial square.], [#figure(figph[A mathematical equation is displayed with a white background, reading (x + 1)^2 = 25. The numbers and symbols are in a dark gray or black font.], alt: "A mathematical equation is displayed with a white background, reading (x + 1)^2 = 25. The numbers and symbols are in a dark gray or black font.", caption: none)], [Use the Square Root Property.], [#figure(figph[A mathematical equation showing x + 1 equals plus or minus the square root of 25. This sets up a problem to solve for x, indicating two possible solutions.], alt: "A mathematical equation showing x + 1 equals plus or minus the square root of 25. This sets up a problem to solve for x, indicating two possible solutions.", caption: none)], [Solve for #emph[x].], [#figure(figph[A mathematical equation is displayed, showing 'X = -1 ± 5' in a clear, digital font against a white background.], alt: "A mathematical equation is displayed, showing 'X = -1 ± 5' in a clear, digital font against a white background.", caption: none)], [Rewite to show two solutions.], [#figure(figph[Two equations are presented: x = -1 + 5 and x = -1 - 5, showing two distinct solutions for the variable x.], alt: "Two equations are presented: x = -1 + 5 and x = -1 - 5, showing two distinct solutions for the variable x.", caption: none)], [Simplify.], [#figure(figph[The image displays the solutions for a variable, showing 'x = 4, x = -6' in a plain white background with black text.], alt: "The image displays the solutions for a variable, showing 'x = 4, x = -6' in a plain white background with black text.", caption: none)], [Check. We leave the check for you!], [], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve #math.equation(block: false, alt: "open parenthesis c minus 2 close parenthesis open parenthesis c plus 8 close parenthesis equals 7")[$( c − 2 ) ( c + 8 ) = 7$] by completing the square. #solutionbox[ #math.equation(block: true, alt: "c equals −3 plus or minus 4 the square root of 2")[$c = −3 ± 4 sqrt(2)$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve #math.equation(block: false, alt: "open parenthesis d minus 7 close parenthesis open parenthesis d plus 3 close parenthesis equals 56")[$( d − 7 ) ( d + 3 ) = 56$] by completing the square. #solutionbox[ #math.equation(block: true, alt: "d equals −7 , d equals 11")[$d = −7 , d = 11$] ] ] === Solve Quadratic Equations of the form #emph[ax]#super[2] + #emph[bx] + #emph[c] = 0 by completing the square The process of completing the square works best when the leading coefficient is one, so the left side of the equation is of the form #math.equation(block: false, alt: "x squared plus b x plus c")[$x^(2) + b x + c$]. If the #math.equation(block: false, alt: "x squared")[$x^(2)$] term has a coefficient, we take some preliminary steps to make the coefficient equal to one. Sometimes the coefficient can be factored from all three terms of the trinomial. This will be our strategy in the next example. #examplebox("Example 12")[][ Solve #math.equation(block: false, alt: "3 x squared minus 12 x minus 15 equals 0")[$3 x^(2) − 12 x − 15 = 0$] by completing the square. #solutionbox[ To complete the square, we need the coefficient of #math.equation(block: false, alt: "x squared")[$x^(2)$] to be one. If we factor out the coefficient of #math.equation(block: false, alt: "x squared")[$x^(2)$] as a common factor, we can continue with solving the equation by completing the square. #figure(table( columns: 2, align: left, inset: 6pt, table.header([], [#figure(figph[The image displays the quadratic equation 3x² - 12x - 15 = 0 against a white background.], alt: "The image displays the quadratic equation 3x² - 12x - 15 = 0 against a white background.", caption: none)]), [Factor out the greatest common factor.], [#figure(figph[A mathematical equation, 3(x^2 - 4x - 5) = 0, is displayed in black text against a white background.], alt: "A mathematical equation, 3(x^2 - 4x - 5) = 0, is displayed in black text against a white background.", caption: none)], [Divide both sides by 3 to isolate the trinomial.], [#figure(figph[A mathematical equation is displayed, showing 3 multiplied by the quantity x squared minus 4x minus 5, all divided by 3, which is set equal to 0 divided by 3.], alt: "A mathematical equation is displayed, showing 3 multiplied by the quantity x squared minus 4x minus 5, all divided by 3, which is set equal to 0 divided by 3.", caption: none)], [Simplify.], [#figure(figph[A mathematical equation is displayed on a white background, reading 'x^2 - 4x - 5 = 0' in black text.], alt: "A mathematical equation is displayed on a white background, reading 'x^2 - 4x - 5 = 0' in black text.", caption: none)], [Subtract 5 to get the constant terms on the right.], [#figure(figph[A mathematical equation is displayed, showing 'x squared minus 4x equals 5'.], alt: "A mathematical equation is displayed, showing 'x squared minus 4x equals 5'.", caption: none)], [Take half of 4 and square it. #math.equation(block: false, alt: "open parenthesis the fraction 1 over 2 open parenthesis 4 close parenthesis close parenthesis squared equals 4")[$( frac(1, 2) ( 4 ) )^(2) = 4$]], [#figure(figph[A mathematical equation, x^2 - 4x + \[blank\] = 5, demonstrating the process of completing the square. The term to add, (1/2 \* 4)^2, is shown below the blank space.], alt: "A mathematical equation, x^2 - 4x + [blank] = 5, demonstrating the process of completing the square. The term to add, (1/2 * 4)^2, is shown below the blank space.", caption: none)], [Add 4 to both sides.], [#figure(figph[An equation showing x^2 - 4x + 4 = 5 + 4, where the added ' + 4' terms on both sides are highlighted in red, indicating a step in completing the square.], alt: "An equation showing x^2 - 4x + 4 = 5 + 4, where the added ' + 4' terms on both sides are highlighted in red, indicating a step in completing the square.", caption: none)], [Factor the perfect square trinomial as a binomial square.], [#figure(figph[A mathematical equation is displayed with the expression (x-2)^2 = 9 centered against a white background.], alt: "A mathematical equation is displayed with the expression (x-2)^2 = 9 centered against a white background.", caption: none)], [Use the Square Root Property.], [#figure(figph[The equation x - 2 = plus or minus the square root of 9 is shown on a white background.], alt: "The equation x - 2 = plus or minus the square root of 9 is shown on a white background.", caption: none)], [Solve for #emph[x.]], [#figure(figph[A simple algebraic equation displays 'x - 2 = +/- 3' on a white background.], alt: "A simple algebraic equation displays 'x - 2 = +/- 3' on a white background.", caption: none)], [Rewrite to show 2 solutions.], [#figure(figph[Two mathematical equations are displayed: x = 2 + 3, and x = 2 - 3.], alt: "Two mathematical equations are displayed: x = 2 + 3, and x = 2 - 3.", caption: none)], [Simplify.], [#figure(figph[The image displays mathematical notation showing two possible values for 'x': x equals 5 and x equals -1, presented against a plain white background.], alt: "The image displays mathematical notation showing two possible values for 'x': x equals 5 and x equals -1, presented against a plain white background.", caption: none)], [Check. #linebreak() #figure(figph[This image demonstrates checking if x=5 and x=-1 are valid solutions for the equation 3x^2 - 12x - 15 = 0. Both substitutions lead to 0=0, confirming they are correct roots.], alt: "This image demonstrates checking if x=5 and x=-1 are valid solutions for the equation 3x^2 - 12x - 15 = 0. Both substitutions lead to 0=0, confirming they are correct roots.", caption: none)], [], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve #math.equation(block: false, alt: "2 m squared plus 16 m minus 8 equals 0")[$2 m^(2) + 16 m − 8 = 0$] by completing the square. #solutionbox[ #math.equation(block: true, alt: "m equals −4 plus or minus 2 the square root of 5")[$m = −4 ± 2 sqrt(5)$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve #math.equation(block: false, alt: "4 n squared minus 24 n minus 56 equals 8")[$4 n^(2) − 24 n − 56 = 8$] by completing the square. #solutionbox[ #math.equation(block: true, alt: "n equals −2 , 8")[$n = −2 , 8$] ] ] To complete the square, the leading coefficient must be one. When the leading coefficient is not a factor of all the terms, we will divide both sides of the equation by the leading coefficient. This will give us a fraction for the second coefficient. We have already seen how to complete the square with fractions in this section. #examplebox("Example 13")[][ Solve #math.equation(block: false, alt: "2 x squared minus 3 x equals 20")[$2 x^(2) − 3 x = 20$] by completing the square. #solutionbox[ Again, our first step will be to make the coefficient of #math.equation(block: false, alt: "x squared")[$x^(2)$] be one. By dividing both sides of the equation by the coefficient of #math.equation(block: false, alt: "x squared")[$x^(2)$], we can then continue with solving the equation by completing the square. #figure(table( columns: 2, align: left, inset: 6pt, table.header([], [#figure(figph[A mathematical equation, 2x^2 - 3x = 20, is displayed in black text against a white background.], alt: "A mathematical equation, 2x^2 - 3x = 20, is displayed in black text against a white background.", caption: none)]), [Divide both sides by 2 to get the coefficient of #math.equation(block: false, alt: "x squared")[$x^(2)$] to be 1.], [#figure(figph[A mathematical equation is displayed: the fraction 2x squared minus 3x all over 2, is equal to the fraction 20 over 2.], alt: "A mathematical equation is displayed: the fraction 2x squared minus 3x all over 2, is equal to the fraction 20 over 2.", caption: none)], [Simplify.], [#figure(figph[A mathematical equation is displayed, reading x squared minus three-halves x equals 10.], alt: "A mathematical equation is displayed, reading x squared minus three-halves x equals 10.", caption: none)], [Take half of #math.equation(block: false, alt: "minus the fraction 3 over 2")[$− frac(3, 2)$] and square it. #math.equation(block: false, alt: "open parenthesis the fraction 1 over 2 open parenthesis minus the fraction 3 over 2 close parenthesis close parenthesis squared equals the fraction 9 over 16")[$( frac(1, 2) ( − frac(3, 2) ) )^(2) = frac(9, 16)$]], [#figure(figph[A quadratic equation showing the step for completing the square: x^2 - (3/2)x + (1/2 \* (-3/2))^2 = 10, highlighting the term added to both sides in red.], alt: "A quadratic equation showing the step for completing the square: x^2 - (3/2)x + (1/2 * (-3/2))^2 = 10, highlighting the term added to both sides in red.", caption: none)], [Add #math.equation(block: false, alt: "the fraction 9 over 16")[$frac(9, 16)$] to both sides.], [#figure(figph[A quadratic equation is displayed: x squared minus three halves x plus nine sixteenths equals ten plus nine sixteenths.], alt: "A quadratic equation is displayed: x squared minus three halves x plus nine sixteenths equals ten plus nine sixteenths.", caption: none)], [Factor the perfect square trinomial as a binomial square.], [#figure(figph[A mathematical equation is displayed against a white background: (x - 3/4)^2 = 160/16 + 9/16. It shows a binomial squared on the left and a sum of fractions on the right.], alt: "A mathematical equation is displayed against a white background: (x - 3/4)^2 = 160/16 + 9/16. It shows a binomial squared on the left and a sum of fractions on the right.", caption: none)], [Add the fractions on the right side.], [#figure(figph[A mathematical equation is presented, showing (x minus 3/4) squared equals 169/16.], alt: "A mathematical equation is presented, showing (x minus 3/4) squared equals 169/16.", caption: none)], [Use the Square Root Property.], [#figure(figph[A mathematical equation shows x - 3/4 = +/- sqrt(169/16)], alt: "A mathematical equation shows x - 3/4 = +/- sqrt(169/16)", caption: none)], [Simplify the radical.], [#figure(figph[A mathematical equation shows 'x minus three-fourths equals plus or minus thirteen-fourths.'], alt: "A mathematical equation shows 'x minus three-fourths equals plus or minus thirteen-fourths.'", caption: none)], [Solve for #emph[x.]], [#figure(figph[The equation X equals three-fourths plus or minus thirteen-fourths.], alt: "The equation X equals three-fourths plus or minus thirteen-fourths.", caption: none)], [Rewrite to show 2 solutions.], [#figure(figph[Two mathematical equations are displayed horizontally, showing 'x' equals '3/4 + 13/4' and 'x' equals '3/4 - 13/4'.], alt: "Two mathematical equations are displayed horizontally, showing 'x' equals '3/4 + 13/4' and 'x' equals '3/4 - 13/4'.", caption: none)], [Simplify.], [#figure(figph[A mathematical expression showing two possible values for x: x = 4 and x = -5/2.], alt: "A mathematical expression showing two possible values for x: x = 4 and x = -5/2.", caption: none)], [Check. We leave the check for you.], [], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve #math.equation(block: false, alt: "3 r squared minus 2 r equals 21")[$3 r^(2) − 2 r = 21$] by completing the square. #solutionbox[ #math.equation(block: true, alt: "r equals minus the fraction 7 over 3 , r equals 3")[$r = − frac(7, 3) , r = 3$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve #math.equation(block: false, alt: "4 t squared plus 2 t equals 20")[$4 t^(2) + 2 t = 20$] by completing the square. #solutionbox[ #math.equation(block: true, alt: "t equals minus the fraction 5 over 2 , t equals 2")[$t = − frac(5, 2) , t = 2$] ] ] #examplebox("Example 14")[][ Solve #math.equation(block: false, alt: "3 x squared plus 2 x equals 4")[$3 x^(2) + 2 x = 4$] by completing the square. #solutionbox[ Again, our first step will be to make the coefficient of #math.equation(block: false, alt: "x squared")[$x^(2)$] be one. By dividing both sides of the equation by the coefficient of #math.equation(block: false, alt: "x squared")[$x^(2)$], we can then continue with solving the equation by completing the square. #figure(table( columns: 2, align: left, inset: 6pt, table.header([], [#figure(figph[A mathematical equation is displayed on a white background: 3x^2 + 2x = 4.], alt: "A mathematical equation is displayed on a white background: 3x^2 + 2x = 4.", caption: none)]), [Divide both sides by 3 to make the coefficient of #math.equation(block: false, alt: "x squared")[$x^(2)$] equal 1.], [#figure(figph[A mathematical equation displays (3x^2 + 2x) / 3 = 4/3, representing a quadratic expression set equal to a fraction.], alt: "A mathematical equation displays (3x^2 + 2x) / 3 = 4/3, representing a quadratic expression set equal to a fraction.", caption: none)], [Simplify.], [#figure(figph[A quadratic equation is displayed, showing x squared plus two-thirds x equals four-thirds. The equation is x^2 + (2/3)x = 4/3.], alt: "A quadratic equation is displayed, showing x squared plus two-thirds x equals four-thirds. The equation is x^2 + (2/3)x = 4/3.", caption: none)], [Take half of #math.equation(block: false, alt: "the fraction 2 over 3")[$frac(2, 3)$] and square it. #math.equation(block: false, alt: "open parenthesis the fraction 1 over 2 times the fraction 2 over 3 close parenthesis squared equals the fraction 1 over 9")[$( frac(1, 2) ⋅ frac(2, 3) )^(2) = frac(1, 9)$]], [#figure(figph[A mathematical equation showing x squared plus two-thirds x, plus a blank line over the quantity one-half times two-thirds, all squared, equals four-thirds. This represents a step in completing the square.], alt: "A mathematical equation showing x squared plus two-thirds x, plus a blank line over the quantity one-half times two-thirds, all squared, equals four-thirds. This represents a step in completing the square.", caption: none)], [Add #math.equation(block: false, alt: "the fraction 1 over 9")[$frac(1, 9)$] to both sides.], [#figure(figph[A mathematical equation displays x squared plus two-thirds x plus one-ninth equals four-thirds plus one-ninth. The one-ninth terms on both sides are highlighted in red.], alt: "A mathematical equation displays x squared plus two-thirds x plus one-ninth equals four-thirds plus one-ninth. The one-ninth terms on both sides are highlighted in red.", caption: none)], [Factor the perfect square trinomial as a binomial square.], [#figure(figph[A mathematical equation shows (x + 1/3)^2 = 12/9 + 1/9.], alt: "A mathematical equation shows (x + 1/3)^2 = 12/9 + 1/9.", caption: none)], [Use the Square Root Property.], [#figure(figph[Equation showing x plus one-third equals positive or negative square root of thirteen-ninths.], alt: "Equation showing x plus one-third equals positive or negative square root of thirteen-ninths.", caption: none)], [Simplify the radical.], [#figure(figph[A mathematical equation, 'x + 1/3 = plus or minus sqrt(13) / 3'.], alt: "A mathematical equation, 'x + 1/3 = plus or minus sqrt(13) / 3'.", caption: none)], [Solve for #emph[x].], [#figure(figph[The image shows the mathematical equation X = -1/3 ± sqrt(13)/3, representing the solution for a variable X.], alt: "The image shows the mathematical equation X = -1/3 ± sqrt(13)/3, representing the solution for a variable X.", caption: none)], [Rewrite to show 2 solutions.], [#figure(figph[The image shows two solutions for x: x = -1/3 + sqrt(13)/3 and x = -1/3 - sqrt(13)/3, expressed as fractions with a common denominator.], alt: "The image shows two solutions for x: x = -1/3 + sqrt(13)/3 and x = -1/3 - sqrt(13)/3, expressed as fractions with a common denominator.", caption: none)], [Check. We leave the check for you.], [], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve #math.equation(block: false, alt: "4 x squared plus 3 x equals 12")[$4 x^(2) + 3 x = 12$] by completing the square. #solutionbox[ #math.equation(block: true, alt: "x equals minus the fraction 3 over 8 plus or minus the fraction the square root of 201 over 8")[$x = − frac(3, 8) ± frac(sqrt(201), 8)$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve #math.equation(block: false, alt: "5 y squared plus 3 y equals 10")[$5 y^(2) + 3 y = 10$] by completing the square. #solutionbox[ #math.equation(block: true, alt: "y equals minus the fraction 3 over 10 plus or minus the fraction the square root of 209 over 10")[$y = − frac(3, 10) ± frac(sqrt(209), 10)$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Access these online resources for additional instruction and practice with solving quadratic equations by completing the square: - #link("https://openstax.org/l/25Completethesq")[Introduction to the method of completing the square] - #link("https://openstax.org/l/25Solvebycompsq")[How to Solve By Completing the Square] ] === Key Concepts - Binomial Squares Pattern If #math.equation(block: false, alt: "a , b")[$a , b$] are real numbers, #linebreak() #math.equation(block: false, alt: "open parenthesis a plus b close parenthesis squared equals a squared plus 2 a b plus b squared")[$attach(( a + b ), t: 2) = a^(2) + 2 a b + b^(2)$] #linebreak() #figure(figph[Algebraic identity: (a + b)^2 = a^2 + 2ab + b^2. This illustrates that squaring a binomial results in the square of the first term, plus twice the product of the terms, plus the square of the second term.], alt: "Algebraic identity: (a + b)^2 = a^2 + 2ab + b^2. This illustrates that squaring a binomial results in the square of the first term, plus twice the product of the terms, plus the square of the second term.", caption: none) #linebreak() #math.equation(block: false, alt: "open parenthesis a minus b close parenthesis squared equals a squared minus 2 a b plus b squared")[$attach(( a − b ), t: 2) = a^(2) − 2 a b + b^(2)$] #linebreak() #figure(figph[An algebraic identity showing the expansion of (a-b)^2, which equals the square of the first term, minus twice the product of terms, plus the square of the second term.], alt: "An algebraic identity showing the expansion of (a-b)^2, which equals the square of the first term, minus twice the product of terms, plus the square of the second term.", caption: none) - Complete a Square #linebreak() To complete the square of #math.equation(block: false, alt: "x squared plus b x")[$x^(2) + b x$]: + Identify #math.equation(block: false, alt: "b")[$b$], the coefficient of #math.equation(block: false, alt: "x")[$x$]. + Find #math.equation(block: false, alt: "open parenthesis the fraction 1 over 2 b close parenthesis squared")[$attach(( frac(1, 2) b ), t: 2)$], the number to complete the square. + Add the #math.equation(block: false, alt: "open parenthesis the fraction 1 over 2 b close parenthesis squared")[$attach(( frac(1, 2) b ), t: 2)$] to #math.equation(block: false, alt: "x squared plus b x")[$x^(2) + b x$]. ==== Practice Makes Perfect #strong[Complete the Square of a Binomial Expression] In the following exercises, complete the square to make a perfect square trinomial. Then, write the result as a binomial squared. #math.equation(block: true, alt: "a squared plus 10 a")[$a^(2) + 10 a$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis a plus 5 close parenthesis squared")[$attach(( a + 5 ), t: 2)$] ] #math.equation(block: true, alt: "b squared plus 12 b")[$b^(2) + 12 b$] #math.equation(block: true, alt: "m squared plus 18 m")[$m^(2) + 18 m$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis m plus 9 close parenthesis squared")[$attach(( m + 9 ), t: 2)$] ] #math.equation(block: true, alt: "n squared plus 16 n")[$n^(2) + 16 n$] #math.equation(block: true, alt: "m squared minus 24 m")[$m^(2) − 24 m$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis m minus 12 close parenthesis squared")[$attach(( m − 12 ), t: 2)$] ] #math.equation(block: true, alt: "n squared minus 16 n")[$n^(2) − 16 n$] #math.equation(block: true, alt: "p squared minus 22 p")[$p^(2) − 22 p$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis p minus 11 close parenthesis squared")[$attach(( p − 11 ), t: 2)$] ] #math.equation(block: true, alt: "q squared minus 6 q")[$q^(2) − 6 q$] #math.equation(block: true, alt: "x squared minus 9 x")[$x^(2) − 9 x$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis x minus the fraction 9 over 2 close parenthesis squared")[$attach(( x − frac(9, 2) ), t: 2)$] ] #math.equation(block: true, alt: "y squared plus 11 y")[$y^(2) + 11 y$] #math.equation(block: true, alt: "p squared minus the fraction 1 over 3 p")[$p^(2) − frac(1, 3) p$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis p minus the fraction 1 over 6 close parenthesis squared")[$attach(( p − frac(1, 6) ), t: 2)$] ] #math.equation(block: true, alt: "q squared plus the fraction 3 over 4 q")[$q^(2) + frac(3, 4) q$] #strong[Solve Quadratic Equations of the Form] #math.equation(block: false, alt: "x squared plus b x plus c equals 0")[$x^(2) + b x + c = 0$] #strong[by Completing the Square] In the following exercises, solve by completing the square. #math.equation(block: true, alt: "v squared plus 6 v equals 40")[$v^(2) + 6 v = 40$] #solutionbox[ #math.equation(block: true, alt: "v equals −10 , v equals 4")[$v = −10 , v = 4$] ] #math.equation(block: true, alt: "w squared plus 8 w equals 65")[$w^(2) + 8 w = 65$] #math.equation(block: true, alt: "u squared plus 2 u equals 3")[$u^(2) + 2 u = 3$] #solutionbox[ #math.equation(block: true, alt: "u equals −3 , u equals 1")[$u = −3 , u = 1$] ] #math.equation(block: true, alt: "z squared plus 12 z equals −11")[$z^(2) + 12 z = −11$] #math.equation(block: true, alt: "c squared minus 12 c equals 13")[$c^(2) − 12 c = 13$] #solutionbox[ #math.equation(block: true, alt: "c equals −1 , c equals 13")[$c = −1 , c = 13$] ] #math.equation(block: true, alt: "d squared minus 8 d equals 9")[$d^(2) − 8 d = 9$] #math.equation(block: true, alt: "x squared minus 20 x equals 21")[$x^(2) − 20 x = 21$] #solutionbox[ #math.equation(block: true, alt: "x equals −1 , x equals 21")[$x = −1 , x = 21$] ] #math.equation(block: true, alt: "y squared minus 2 y equals 8")[$y^(2) − 2 y = 8$] #math.equation(block: true, alt: "m squared plus 4 m equals −44")[$m^(2) + 4 m = −44$] #solutionbox[ no real solution ] #math.equation(block: true, alt: "n squared minus 2 n equals −3")[$n^(2) − 2 n = −3$] #math.equation(block: true, alt: "r squared plus 6 r equals −11")[$r^(2) + 6 r = −11$] #solutionbox[ no real solution ] #math.equation(block: true, alt: "t squared minus 14 t equals −50")[$t^(2) − 14 t = −50$] #math.equation(block: true, alt: "a squared minus 10 a equals −5")[$a^(2) − 10 a = −5$] #solutionbox[ #math.equation(block: true, alt: "a equals 5 plus or minus 2 the square root of 5")[$a = 5 ± 2 sqrt(5)$] ] #math.equation(block: true, alt: "b squared plus 6 b equals 41")[$b^(2) + 6 b = 41$] #math.equation(block: true, alt: "u squared minus 14 u plus 12 equals −1")[$u^(2) − 14 u + 12 = −1$] #solutionbox[ #math.equation(block: true, alt: "u equals 1 , u equals 13")[$u = 1 , u = 13$] ] #math.equation(block: true, alt: "z squared plus 2 z minus 5 equals 2")[$z^(2) + 2 z − 5 = 2$] #math.equation(block: true, alt: "v squared equals 9 v plus 2")[$v^(2) = 9 v + 2$] #solutionbox[ #math.equation(block: true, alt: "v equals the fraction 9 over 2 plus or minus the fraction the square root of 89 over 2")[$v = frac(9, 2) ± frac(sqrt(89), 2)$] ] #math.equation(block: true, alt: "w squared equals 5 w minus 1")[$w^(2) = 5 w − 1$] #math.equation(block: true, alt: "open parenthesis x plus 6 close parenthesis open parenthesis x minus 2 close parenthesis equals 9")[$( x + 6 ) ( x − 2 ) = 9$] #solutionbox[ #math.equation(block: true, alt: "x equals −7 , x equals 3")[$x = −7 , x = 3$] ] #math.equation(block: true, alt: "open parenthesis y plus 9 close parenthesis open parenthesis y plus 7 close parenthesis equals 79")[$( y + 9 ) ( y + 7 ) = 79$] #strong[Solve Quadratic Equations of the Form] #math.equation(block: false, alt: "a x squared plus b x plus c equals 0")[$a x^(2) + b x + c = 0$] #strong[by Completing the Square] In the following exercises, solve by completing the square. #math.equation(block: true, alt: "3 m squared plus 30 m minus 27 equals 6")[$3 m^(2) + 30 m − 27 = 6$] #solutionbox[ #math.equation(block: true, alt: "m equals −11 , m equals 1")[$m = −11 , m = 1$] ] #math.equation(block: true, alt: "2 n squared plus 4 n minus 26 equals 0")[$2 n^(2) + 4 n − 26 = 0$] #math.equation(block: true, alt: "2 c squared plus c equals 6")[$2 c^(2) + c = 6$] #solutionbox[ #math.equation(block: true, alt: "c equals −2 , c equals the fraction 3 over 2")[$c = −2 , c = frac(3, 2)$] ] #math.equation(block: true, alt: "3 d squared minus 4 d equals 15")[$3 d^(2) − 4 d = 15$] #math.equation(block: true, alt: "2 p squared plus 7 p equals 14")[$2 p^(2) + 7 p = 14$] #solutionbox[ #math.equation(block: true, alt: "p equals minus the fraction 7 over 4 plus or minus the fraction the square root of 161 over 4")[$p = − frac(7, 4) ± frac(sqrt(161), 4)$] ] #math.equation(block: true, alt: "3 q squared minus 5 q equals 9")[$3 q^(2) − 5 q = 9$] ==== Everyday Math Rafi is designing a rectangular playground to have an area of 320 square feet. He wants one side of the playground to be four feet longer than the other side. Solve the equation #math.equation(block: false, alt: "p squared plus 4 p equals 320")[$p^(2) + 4 p = 320$] for #math.equation(block: false, alt: "p")[$p$], the length of one side of the playground. What is the length of the other side? #solutionbox[ 16 feet, 20 feet ] Yvette wants to put a square swimming pool in the corner of her backyard. She will have a 3 foot deck on the south side of the pool and a 9 foot deck on the west side of the pool. She has a total area of 1080 square feet for the pool and two decks. Solve the equation #math.equation(block: false, alt: "open parenthesis s plus 3 close parenthesis open parenthesis s plus 9 close parenthesis equals 1080")[$( s + 3 ) ( s + 9 ) = 1080$] for #math.equation(block: false, alt: "s")[$s$], the length of a side of the pool. ==== Writing Exercises Solve the equation #math.equation(block: false, alt: "x squared plus 10 x equals −25")[$x^(2) + 10 x = −25$] ⓐ by using the Square Root Property and ⓑ by completing the square. ⓒ Which method do you prefer? Why? #solutionbox[ ⓐ #math.equation(block: false, alt: "−5")[$−5$] ⓑ #math.equation(block: false, alt: "−5")[$−5$] ⓒ Answers will vary. ] Solve the equation #math.equation(block: false, alt: "y squared plus 8 y equals 48")[$y^(2) + 8 y = 48$] by completing the square and explain all your steps. ==== Self Check ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section. #figure(figph[This table has four rows and four columns. The first row is a header row and it labels each column. The first column is labeled “I can ...”, the second “Confidently”, the third “With some help” and the last “No–I don’t get it”. In the “I can...” column the next row reads “complete the square of a binomial expression.” The next row reads “solve quadratic equations of the form x squared plus b x plus c equals zero by completing the square.” and the last row reads “solve quadratic equations of the form a x squared plus b x plus c equals zero by completing the square.” The remaining columns are blank.], alt: "This table has four rows and four columns. The first row is a header row and it labels each column. The first column is labeled “I can ...”, the second “Confidently”, the third “With some help” and the last “No–I don’t get it”. In the “I can...” column the next row reads “complete the square of a binomial expression.” The next row reads “solve quadratic equations of the form x squared plus b x plus c equals zero by completing the square.” and the last row reads “solve quadratic equations of the form a x squared plus b x plus c equals zero by completing the square.” The remaining columns are blank.", caption: none) ⓑ After reviewing this checklist, what will you do to become confident for all objectives?