#set document(title: "6.2 Solving Quadratic Equations", 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")) == 6.2#h(0.6em)Solving Quadratic Equations Not every quadratic equation can be solved by factoring or by extraction of roots. For example, the expression #math.equation(block: false, alt: "x squared plus x minus 1")[$x^(2) + x − 1$] cannot be factored, so the equation #math.equation(block: false, alt: "x squared plus x minus 1 equals 0")[$x^(2) + x − 1 = 0$] cannot be solved by factoring. For other equations, factoring may be difficult. In this section we learn two methods that can be used to solve any quadratic equation. === Squares of Binomials In Nonlinear Models we used extraction of roots to solve equations of the form #math.equation(block: true, alt: "a open parenthesis p x plus q close parenthesis squared plus r equals 0")[$a ( p x + q )^(2) + r = 0$] where the left side of the equation includes the square of a binomial, or a #strong[perfect square]. We can write any quadratic equation in this form by #strong[completing the square]. Consider the following squares of binomials. #figure(table( columns: 4, align: left, inset: 6pt, table.header([Square of binomial #math.equation(block: false, alt: "open parenthesis x plus p close parenthesis squared")[$( x + p )^(2)$]], [#math.equation(block: false, alt: "p")[$p$]], [#math.equation(block: false, alt: "2 p")[$2 p$]], [#math.equation(block: false, alt: "p squared")[$p^(2)$]]), [1. #math.equation(block: false, alt: "open parenthesis x plus 5 close parenthesis squared equals x squared plus 10 x plus 25")[$( x + 5 )^(2) = x^(2) + 10 x + 25$]], [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "2 open parenthesis 5 close parenthesis equals 10")[$2 ( 5 ) = 10$]], [#math.equation(block: false, alt: "5 squared equals 25")[$5^(2) = 25$]], [2. #math.equation(block: false, alt: "open parenthesis x minus 3 close parenthesis squared equals x squared minus 6 x plus 9")[$( x − 3 )^(2) = x^(2) − 6 x + 9$]], [#math.equation(block: false, alt: "minus 3")[$− 3$]], [#math.equation(block: false, alt: "2 open parenthesis minus 3 close parenthesis equals minus 6")[$2 ( − 3 ) = − 6$]], [#math.equation(block: false, alt: "minus 3 squared equals 9")[$attach(− 3, t: 2) = 9$]], [3. #math.equation(block: false, alt: "open parenthesis x minus 12 close parenthesis squared equals x squared minus 24 x plus 144")[$( x − 12 )^(2) = x^(2) − 24 x + 144$]], [#math.equation(block: false, alt: "minus 12")[$− 12$]], [#math.equation(block: false, alt: "2 open parenthesis minus 12 close parenthesis equals minus 24")[$2 ( − 12 ) = − 24$]], [#math.equation(block: false, alt: "minus 12 squared equals 144")[$attach(− 12, t: 2) = 144$]], )) In each case, the square of the binomial is a quadratic trinomial, #math.equation(block: true, alt: "open parenthesis x plus p close parenthesis squared equals x squared plus 2 p x plus p squared")[$( x + p )^(2) = x^(2) + 2 p x + p^(2)$] Note that the coefficient of the linear term, #math.equation(block: false, alt: "2 p")[$2 p$], is twice the constant in the binomial, and the constant term of the trinomial, #math.equation(block: false, alt: "p squared")[$p^(2)$], is its square. What is the linear term of #math.equation(block: false, alt: "open parenthesis x plus 6 close parenthesis squared")[$( x + 6 )^(2)$] ? \_\_\_\_\_ #solutionbox[ #math.equation(block: true, alt: "12 x")[$12 x$] ] What is the linear term of #math.equation(block: false, alt: "open parenthesis x plus 6 close parenthesis squared")[$( x + 6 )^(2)$] ? + #math.equation(block: false, alt: "x squared")[$x^(2)$] + #math.equation(block: false, alt: "6 x")[$6 x$] + #math.equation(block: false, alt: "12 x")[$12 x$] + #math.equation(block: false, alt: "36")[$36$] We would like to reverse the process and write a quadratic expression as the square of a binomial. For example, what constant term can we add to #math.equation(block: true, alt: "x squared minus 16 x")[$x^(2) − 16 x$] to produce a perfect square trinomial? Compare the expression to the formula above: #math.equation(block: true, alt: "x squared, plus, 2 p x, plus, p squared, equals, open parenthesis x plus p close parenthesis squared; x squared, minus, 16 x, plus, ?, equals, open parenthesis x plus ? close parenthesis squared")[$x^(2) & + & 2 p x & + & p^(2) & = & ( x + p )^(2) \ x^(2) & − & 16 x & + & "?" & = & ( x + "?" )^(2)$] We see that #math.equation(block: true, alt: "2 p equals minus 16 , so p equals the fraction 1 over 2 open parenthesis minus 16 close parenthesis equals minus 8")[$2 p = − 16 , " " " " " so " " " " " p = frac(1, 2) ( − 16 ) = − 8$] and #math.equation(block: true, alt: "p squared equals open parenthesis minus 8 close parenthesis squared equals 64")[$p^(2) = ( − 8 )^(2) = 64$] We substitute these values for #math.equation(block: false, alt: "p squared")[$p^(2)$] and #math.equation(block: false, alt: "p")[$p$] into the equation to find #math.equation(block: true, alt: "x squared minus 16 x plus 64 equals open parenthesis x minus 8 close parenthesis squared")[$x^(2) − 16 x + 64 = ( x − 8 )^(2)$] Notice that in the resulting trinomial, the constant term is equal to #emph[the square of one-half the coefficient of] #math.equation(block: false, alt: "x")[$x$]. In other words, we can find the constant term by taking one-half the coefficient of #math.equation(block: false, alt: "x")[$x$] and then squaring the result. Adding a constant term obtained in this way is called #strong[completing the square]. #examplebox("Example 1")[][ Complete the square by adding an appropriate constant; write the result as the square of a binomial. + #math.equation(block: false, alt: "x squared minus 12 x plus bar")[$x^(2) − 12 x + underline(#h(3em))$] + #math.equation(block: false, alt: "x squared plus 5 x plus bar")[$x^(2) + 5 x + underline(#h(3em))$] #solutionbox[ + One-half of #math.equation(block: false, alt: "minus 12")[$− 12$] is #math.equation(block: false, alt: "minus 6")[$− 6$], so the constant term is #math.equation(block: false, alt: "open parenthesis minus 6 close parenthesis squared")[$( − 6 )^(2)$], or #math.equation(block: false, alt: "36")[$36$]. We add #math.equation(block: false, alt: "36")[$36$] to obtain #math.equation(block: true, alt: "x squared minus 12 x plus 36 equals open parenthesis x minus 6 close parenthesis squared, p equals the fraction 1 over 2 open parenthesis minus 12 close parenthesis equals minus 6 p squared equals open parenthesis minus 6 close parenthesis squared equals 36")[$x^(2) − 12 x + 36 = ( x − 6 )^(2) & & & limits(limits(zws)^(bold(italic(p)) #h(0.167em) = #h(0.167em) frac(1, 2) ( − 12 ) #h(0.167em) = #h(0.167em) − 6))_(bold(italic(p))^(2) #h(0.167em) = #h(0.167em) ( − 6 )^(2) #h(0.167em) = #h(0.167em) 36 #hide($0$))$] + One-half of #math.equation(block: false, alt: "5")[$5$] is #math.equation(block: false, alt: "the fraction 5 over 2")[$display(frac(5, 2))$], so the constant term is #math.equation(block: false, alt: "open parenthesis the fraction 5 over 2 close parenthesis squared")[$attach(( display(frac(5, 2)) ), t: 2)$], or #math.equation(block: false, alt: "the fraction 25 over 4")[$display(frac(25, 4))$]. We add #math.equation(block: false, alt: "the fraction 25 over 4")[$display(frac(25, 4))$] to obtain #math.equation(block: true, alt: "x squared plus 5 x plus the fraction 25 over 4 equals open parenthesis x plus the fraction 5 over 2 close parenthesis squared, p equals the fraction 1 over 2 open parenthesis 5 close parenthesis equals the fraction 5 over 2 p squared equals open parenthesis the fraction 5 over 2 close parenthesis squared equals the fraction 25 over 4")[$x^(2) + 5 x + frac(25, 4) = attach(( x + frac(5, 2) ), t: 2) & & & limits(limits(zws)^(bold(italic(p)) #h(0.167em) = #h(0.167em) frac(1, 2) ( 5 ) #h(0.167em) = #h(0.167em) frac(5, 2) " " " "))_(bold(italic(p))^(2) #h(0.167em) = #h(0.167em) ( frac(5, 2) )^(2) #h(0.167em) = #h(0.167em) frac(25, 4))$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ You may find it helpful to visualize completing the square geometrically. We can think of the expression #math.equation(block: false, alt: "x squared plus 2 p x")[$x^(2) + 2 p x$] as the area of a rectangle with dimensions #math.equation(block: false, alt: "x")[$x$] and #math.equation(block: false, alt: "x plus 2 p")[$x + 2 p$]. Here is an example. - The rectangle with length #math.equation(block: false, alt: "x plus 10")[$x + 10$] and width #math.equation(block: false, alt: "x")[$x$] has area #math.equation(block: false, alt: "x open parenthesis x plus 10 close parenthesis equals x squared plus 10 x")[$x ( x + 10 ) = x^(2) + 10 x$], as shown in figure (a). - We would like to cut the rectangle into pieces and rearrange them so that we can make a square. - In figure (b), we move half of the #math.equation(block: false, alt: "x")[$x$]-term so that each side of the square has length #math.equation(block: false, alt: "x plus 5")[$x + 5$] (note that #math.equation(block: false, alt: "p equals the fraction 1 over 2 open parenthesis 10 close parenthesis equals 5")[$p = frac(1, 2) ( 10 ) = 5$]). - In figure (c) we see that the missing corner piece has area #math.equation(block: false, alt: "p squared equals 5 squared equals 25")[$p^(2) = 5^(2) = 25$]. #figure(figph[completing the square], alt: "completing the square", caption: none) ] Complete the square by adding an appropriate constant; write the result as the square of a binomial. + #math.equation(block: false, alt: "x squared minus 18 x plus")[$x^(2) − 18 x +$]\_\_\_\_\_#math.equation(block: false, alt: "equals open parenthesis x plus")[$= \( x +$]\_\_\_\_\_ #math.equation(block: false, alt: "close parenthesis squared")[$\)^(2)$] + #math.equation(block: false, alt: "x squared plus 9 x plus")[$x^(2) + 9 x +$]\_\_\_\_\_#math.equation(block: false, alt: "equals open parenthesis x plus")[$= \( x +$]\_\_\_\_\_ #math.equation(block: false, alt: "close parenthesis squared")[$\)^(2)$] For part (a): #math.equation(block: false, alt: "p equals the fraction 1 over 2 open parenthesis minus 18 close parenthesis equals")[$p = frac(1, 2) ( − 18 ) =$]\_\_\_\_\_ , #math.equation(block: false, alt: "p squared equals")[$p^(2) =$] \_\_\_\_\_\_ For part (b): #math.equation(block: false, alt: "p equals the fraction 1 over 2 open parenthesis 9 close parenthesis equals")[$p = frac(1, 2) ( 9 ) =$]\_\_\_\_\_ , #math.equation(block: false, alt: "p squared equals")[$p^(2) =$] \_\_\_\_\_\_ #solutionbox[ + #math.equation(block: false, alt: "x squared minus 18 x plus 81 equals open parenthesis x minus 9 close parenthesis squared")[$x^(2) − 18 x + 81 = ( x − 9 )^(2)$] + #math.equation(block: false, alt: "x squared plus 9 x plus the fraction 81 over 4 equals open parenthesis x plus the fraction 9 over 2 close parenthesis squared")[$x^(2) + 9 x + display(frac(81, 4)) = attach(( x + display(frac(9, 2)) ), t: 2)$] ] Complete the square by adding an appropriate constant; write the result as the square of a binomial. + #math.equation(block: false, alt: "x squared minus 18 x plus bar equals open parenthesis x plus bar close parenthesis squared")[$x^(2) − 18 x + underline(#h(1.5em)) = ( x + underline(#h(1.5em)) )^(2)$] #linebreak() #math.equation(block: false, alt: "p equals the fraction 1 over 2 open parenthesis minus 18 close parenthesis equals bar , p squared equals bar")[$bold(italic(p)) = frac(1, 2) ( − 18 ) = underline(#h(1.5em)) , " " " " bold(italic(p))^(2) = underline(#h(1.5em))$] + #math.equation(block: false, alt: "x squared plus 9 x plus bar equals open parenthesis x plus bar close parenthesis squared")[$x^(2) + 9 x + underline(#h(1.5em)) = ( x + underline(#h(1.5em)) )^(2)$] #linebreak() #math.equation(block: false, alt: "p equals the fraction 1 over 2 open parenthesis 9 close parenthesis equals bar , p squared equals bar")[$bold(italic(p)) = frac(1, 2) ( 9 ) = underline(#h(1.5em)) , " " " " bold(italic(p))^(2) = underline(#h(1.5em))$] #solutionbox[ + #math.equation(block: false, alt: "x squared minus 18 x plus 81 equals open parenthesis x minus 9 close parenthesis squared")[$x^(2) − 18 x + 81 = ( x − 9 )^(2)$] + #math.equation(block: false, alt: "x squared plus 9 x plus the fraction 81 over 4 equals open parenthesis x plus the fraction 9 over 2 close parenthesis squared")[$x^(2) + 9 x + display(frac(81, 4)) = attach(( x + display(frac(9, 2)) ), t: 2)$] ] === Solving Quadratic Equations by Completing the Square Now we will use completing the square to solve quadratic equations. First, we will solve equations in which the coefficient of the squared term is 1. Consider the equation #math.equation(block: true, alt: "x squared minus 6 x minus 7 equals 0")[$x^(2) − 6 x − 7 = 0$] and follow the steps to find the solutions. #emph[Step 1] Begin by moving the constant term to the other side of the equation, to get #math.equation(block: true, alt: "x squared minus 6 x plus bar equals 7")[$x^(2) − 6 x + underline(#h(3em)) = 7$] #emph[Step 2] Now complete the square on the left. Because #math.equation(block: true, alt: "p equals the fraction 1 over 2 open parenthesis minus 6 close parenthesis equals minus 3 and p squared equals open parenthesis minus 3 close parenthesis squared equals 9")[$p = frac(1, 2) ( − 6 ) = − 3 " " " " " " " and " " " " " " " p^(2) = ( − 3 )^(2) = 9$] we add #math.equation(block: false, alt: "9")[$9$] to #emph[both] sides of our equation to get #math.equation(block: true, alt: "x squared minus 6 x plus 9 equals 7 plus 9")[$x^(2) − 6 x + 9 = 7 + 9$] #emph[Step 3] The left side of the equation is now the square of a binomial, namely #math.equation(block: false, alt: "open parenthesis x minus 3 close parenthesis squared")[$( x − 3 )^(2)$]. We write the left side in its square form and simplify the right side, which gives us #math.equation(block: true, alt: "open parenthesis x minus 3 close parenthesis squared equals 16")[$( x − 3 )^(2) = 16$] (You can check that this equation is equivalent to the original one; if you expand the left side and collect like terms, you will return to the original equation.) #emph[Step 4] We can now use extraction of roots to find the solutions. Taking square roots of both sides, we get #math.equation(block: true, alt: "x minus 3, equals 4, or, x minus 3, equals minus 4, Solve each equation.; x, equals 7, or, x, equals minus 1")[$x − 3 & = 4 & "or" & & x − 3 & = − 4 & "Solve each equation." \ x & = 7 & "or" & & x & = − 1$] The solutions are #math.equation(block: false, alt: "7")[$7$] and #math.equation(block: false, alt: "minus 1")[$− 1$]. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #figure(figph[parabola], alt: "parabola", caption: none) The graph of #math.equation(block: false, alt: "y equals x squared minus 6 x minus 7")[$y = x^(2) − 6 x − 7$] is shown in at left. Note that the #math.equation(block: false, alt: "x")[$x$]-intercepts of the graph are #math.equation(block: false, alt: "x equals 7")[$x = 7$] and #math.equation(block: false, alt: "x equals minus 1")[$x = − 1$], and the parabola is symmetric about the vertical line halfway between the intercepts, at #math.equation(block: false, alt: "x equals 3")[$x = 3$]. {"equation":"x^2 - 6\*x - 7 = 0","variable":"x"}7, -1The completing-the-square walkthrough ends at the solutions 7 and −1, and this paragraph reads them off the graph as x-intercepts. At build time a computer algebra system re-solves x² − 6x − 7 = 0 and compares its root set, order-free, against the book's stated pair — a rebuild that no longer reproduces these roots stops the build instead of shipping the disagreement. ] We can also solve #math.equation(block: false, alt: "x squared minus 6 x minus 7 equals 0")[$x^(2) − 6 x − 7 = 0$] by factoring instead of completing the square. Of course, we get the same solutions by either method. In Example, we will solve an equation that cannot be solved by factoring. #examplebox("Example 2")[][ Solve #math.equation(block: false, alt: "x squared minus 4 x minus 3 equals 0")[$" " x^(2) − 4 x − 3 = 0 " "$] by completing the square. #solutionbox[ + First, write the equation with the constant term on the right side. #math.equation(block: true, alt: "x squared minus 4 x plus bar equals 3")[$x^(2) − 4 x + " " underline(#h(3em)) = 3$] + Now complete the square on the left side. The coefficient of #math.equation(block: false, alt: "x")[$x$] is #math.equation(block: false, alt: "minus 4")[$− 4$], so #math.equation(block: true, alt: "p equals the fraction 1 over 2 open parenthesis minus 4 close parenthesis equals minus 2 and p squared equals open parenthesis minus 2 close parenthesis squared equals 4")[$p = frac(1, 2) ( − 4 ) = − 2 " " " " " " " and " " " " " " " p^(2) = ( − 2 )^(2) = 4$] We add #math.equation(block: false, alt: "4")[$4$] to both sides of our equation: #math.equation(block: true, alt: "x squared minus 4 x plus 4 equals 3 plus 4")[$x^(2) − 4 x + 4 = 3 + 4$] + Write the left side as the square of a binomial, and combine terms on the right side: #math.equation(block: true, alt: "open parenthesis x minus 2 close parenthesis squared equals 7")[$( x − 2 )^(2) = 7$] + Finally, use extraction of roots to obtain #linebreak() #math.equation(block: true, alt: "x minus 2, equals the square root of 7, or, x minus 2, equals minus the square root of 7; t e x t S o l v e e a c h e q u a t i o n .; x, equals 2 plus the square root of 7, or, x, equals 2 minus the square root of 7")[$x − 2 & = sqrt(7) & "or" & & x − 2 & = − sqrt(7) \ t e x t S o l v e e a c h e q u a t i o n . \ x & = 2 + sqrt(7) & "or" & & x & = 2 − sqrt(7)$] #linebreak() The solutions are #math.equation(block: false, alt: "2 plus the square root of 7 approximately equals 4.646")[$" " 2 + sqrt(7) ≈ 4.646 " "$] and #math.equation(block: false, alt: "2 minus the square root of 7 approximately equals minus 0.646")[$" " 2 − sqrt(7) ≈ − 0.646 " "$]. The graph of #math.equation(block: false, alt: "y equals x squared minus 4 x minus 3")[$y = x^(2) − 4 x − 3$] is shown below.#figure(figph[parabola showing x-intercepts and line of symmetry], alt: "parabola showing x-intercepts and line of symmetry", caption: none) ] ] What should we add to #math.equation(block: false, alt: "x squared minus 5 x")[$x^(2) − 5 x$] to create a perfect square? \_\_\_\_\_ #solutionbox[ #math.equation(block: true, alt: "the fraction 25 over 4")[$display(frac(25, 4))$] ] What should we add to #math.equation(block: false, alt: "x squared minus 5 x")[$x^(2) − 5 x$] to create a perfect square? + #math.equation(block: false, alt: "25")[$25$] + #math.equation(block: false, alt: "minus 25")[$− 25$] + #math.equation(block: false, alt: "the fraction 25 over 4")[$display(frac(25, 4))$] + #math.equation(block: false, alt: "10")[$10$] + Follow the steps to solve by completing the square: #math.equation(block: false, alt: "x squared minus 1 equals 3 x")[$x^(2) − 1 = 3 x$].+ Write the equation with the constant on the right. #linebreak() \_\_\_\_\_#math.equation(block: false, alt: "equals")[$=$]\_\_\_\_\_ + Complete the square on the left: #linebreak() #math.equation(block: false, alt: "p equals the fraction 1 over 2 open parenthesis minus 3 close parenthesis equals")[$p = display(frac(1, 2)) ( − 3 ) =$]\_\_\_\_\_, #math.equation(block: false, alt: "p squared equals")[$p^(2) =$]\_\_\_\_\_ #linebreak() Add #math.equation(block: false, alt: "p squared")[$p^(2)$] to both sides. #linebreak() \_\_\_\_\_#math.equation(block: false, alt: "equals")[$=$]\_\_\_\_\_ + Write the left side as a perfect square; simplify the right side. #linebreak() #math.equation(block: false, alt: "open parenthesis")[$\($]\_\_\_\_\_#math.equation(block: false, alt: "close parenthesis squared equals")[$\)^(2) =$]\_\_\_\_\_ + Solve by extracting roots. #linebreak() Solutions: #math.equation(block: false, alt: "x equals")[$x =$]\_\_\_\_\_ #linebreak() List all the values that are solutions. Use a comma to separate different solutions. + Find approximations to two decimal places for the solutions. #linebreak() Solutions: #math.equation(block: false, alt: "x equals")[$x =$]\_\_\_\_\_ #linebreak() List all the values that are solutions. Use a comma to separate different solutions. + Graph the parabola #math.equation(block: false, alt: "y equals x squared minus 3 x minus 1")[$y = x^(2) − 3 x − 1$] in the window #math.equation(block: true, alt: "Xmin, equals minus 4.7, Xmax equals 4.7; Ymin, equals minus 5, Ymax equals 5")[$"Xmin" & = − 4.7 & & "Xmax" = 4.7 \ "Ymin" & = − 5 & & "Ymax" = 5$] #solutionbox[ + #math.equation(block: false, alt: "x equals the fraction 3 over 2 plus or minus the square root of the fraction 13 over 4")[$x = display(frac(3, 2)) ± sqrt(frac(13, 4))$] + #math.equation(block: false, alt: "x approximately equals minus 0.30")[$x ≈ − 0.30$] or #math.equation(block: false, alt: "x approximately equals 3.30")[$x ≈ 3.30$] ] #figure(figph[parabola x-squared minus 3x minus 1], alt: "parabola x-squared minus 3x minus 1", caption: none) + Follow the steps to solve by completing the square: #math.equation(block: false, alt: "x squared minus 1 equals 3 x")[$x^(2) − 1 = 3 x$].+ Write the equation with the constant on the right. + Complete the square on the left: #linebreak() #math.equation(block: false, alt: "p equals the fraction 1 over 2 open parenthesis minus 3 close parenthesis equals bar , p squared equals bar")[$p = display(frac(1, 2)) ( − 3 ) = underline(#h(2em)) , " " " " p^(2) = underline(#h(2em))$] #linebreak() Add #math.equation(block: false, alt: "p squared")[$p^(2)$] to both sides. + Write the left side as a perfect square; simplify the right side. + Solve by extracting roots. + Find approximations to two decimal places for the solutions. + Graph the parabola #math.equation(block: false, alt: "y equals x squared minus 3 x minus 1")[$y = x^(2) − 3 x − 1$] in the window #math.equation(block: true, alt: "Xmin, equals minus 4.7, Xmax equals 4.7; Ymin, equals minus 5, Ymax equals 5")[$"Xmin" & = − 4.7 & & "Xmax" = 4.7 \ "Ymin" & = − 5 & & "Ymax" = 5$] #solutionbox[ + #math.equation(block: false, alt: "x equals the fraction 3 over 2 plus or minus the square root of the fraction 13 over 4")[$x = display(frac(3, 2)) ± sqrt(frac(13, 4))$] + #math.equation(block: false, alt: "x approximately equals minus 0.30")[$x ≈ − 0.30$] or #math.equation(block: false, alt: "x approximately equals 3.30")[$x ≈ 3.30$] + #figure(figph[parabola x-squared minus 3x minus 1], alt: "parabola x-squared minus 3x minus 1", caption: none) ] Explain how to tell whether #math.equation(block: false, alt: "x squared plus b x plus c")[$x^(2) + b x + c$] is the square of a binomial. \_\_\_\_\_ Explain how to tell whether #math.equation(block: false, alt: "x squared plus b x plus c")[$x^(2) + b x + c$] is the square of a binomial. === The General Case Our method for completing the square works only if the coefficient of #math.equation(block: false, alt: "x squared")[$x^(2)$] is #math.equation(block: false, alt: "1")[$1$]. If we want to solve a quadratic equation whose lead coefficient is not #math.equation(block: false, alt: "1")[$1$], we first divide each term of the equation by the lead coefficient. #examplebox("Example 3")[][ Solve #math.equation(block: false, alt: "2 x squared minus 6 x minus 5 equals 0")[$" " " " 2 x^(2) − 6 x − 5 = 0$]. #solutionbox[ + Because the coefficient of #math.equation(block: false, alt: "x squared")[$x^(2)$] is #math.equation(block: false, alt: "2")[$2$], we must divide each term of the equation by #math.equation(block: false, alt: "2")[$2$]. #math.equation(block: true, alt: "x squared minus 3 x minus the fraction 5 over 2 equals 0")[$x^(2) − 3 x − frac(5, 2) = 0$] Now we proceed as before. Rewrite the equation with the constant on the right side. #math.equation(block: true, alt: "x squared minus 3 x plus bar equals the fraction 5 over 2")[$x^(2) − 3 x + " " underline(#h(3em)) = frac(5, 2)$] + Complete the square: #math.equation(block: true, alt: "p equals the fraction 1 over 2 open parenthesis minus 3 close parenthesis equals the fraction minus 3 over 2 and p squared equals open parenthesis the fraction minus 3 over 2 close parenthesis squared equals the fraction 9 over 4")[$p = frac(1, 2) ( − 3 ) = frac(− 3, 2) " " " " " " " and " " " " " " " p^(2) = attach(( frac(− 3, 2) ), t: 2) = frac(9, 4)$] Add #math.equation(block: false, alt: "the fraction 9 over 4")[$display(frac(9, 4))$] to both sides of our equation: #math.equation(block: true, alt: "x squared minus 3 x plus the fraction 9 over 4 equals the fraction 5 over 2 plus the fraction 9 over 4")[$x^(2) − 3 x + frac(9, 4) = frac(5, 2) + frac(9, 4)$] + Rewrite the left side as the square of a binomial and simplify the right side to get #math.equation(block: true, alt: "open parenthesis x minus the fraction 3 over 2 close parenthesis squared equals the fraction 19 over 4")[$attach(( x − frac(3, 2) ), t: 2) = frac(19, 4)$] + Finally, extract roots and solve each equation for #math.equation(block: false, alt: "x")[$x$]. #math.equation(block: true, alt: "x minus the fraction 3 over 2 equals the square root of the fraction 19 over 4 or x minus the fraction 3 over 2 equals minus the square root of the fraction 19 over 4")[$x − frac(3, 2) = sqrt(frac(19, 4)) " " " " " " " or " " " " " " " x − frac(3, 2) = − sqrt(frac(19, 4))$] The solutions are #math.equation(block: false, alt: "the fraction 3 over 2 plus the square root of the fraction 19 over 4")[$" " display(frac(3, 2)) + sqrt(display(frac(19, 4))) " "$] and #math.equation(block: false, alt: "the fraction 3 over 2 minus the square root of the fraction 19 over 4")[$" " display(frac(3, 2)) − sqrt(display(frac(19, 4))) " "$]. #linebreak() Using a calculator, we can find decimal approximations for the solutions: #math.equation(block: false, alt: "3.679")[$3.679$] and #math.equation(block: false, alt: "minus 0.679")[$− 0.679$]. ] ] #notebox("Caution", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ In the Example above, it is essential that we first divide each term of the equation by #math.equation(block: false, alt: "2 ,")[$2 ,$] the coefficient of #math.equation(block: false, alt: "x squared")[$x^(2)$]. The following attempt at a solution is #emph[incorrect]. #math.equation(block: true, alt: "2 x squared minus 6 x, equals 5; 2 x squared minus 6 x plus 9, equals 5 plus 9; open parenthesis 2 x minus 3 close parenthesis squared, equals 14 → Incorrect!")[$2 x^(2) − 6 x & = 5 \ 2 x^(2) − 6 x + 9 & = 5 + 9 \ ( 2 x − 3 )^(2) & = 14 " " " " " " → " " " " " " " Incorrect!"$] You can check that #math.equation(block: false, alt: "open parenthesis 2 x minus 3 close parenthesis squared")[$( 2 x − 3 )^(2)$] is not equal to #math.equation(block: false, alt: "2 x squared minus 6 x plus 9")[$2 x^(2) − 6 x + 9$]. We have not written the left side of the equation as a perfect square, so the solutions we obtain by extracting roots will not be correct. ] What is the first step in solving #math.equation(block: false, alt: "3 x squared minus 6 x equals 2")[$3 x^(2) − 6 x = 2$] ? \_\_\_\_\_ #solutionbox[ Divide both sides by 3 ] What is the first step in solving #math.equation(block: false, alt: "3 x squared minus 6 x equals 2")[$3 x^(2) − 6 x = 2$] ? + Divide #math.equation(block: false, alt: "minus 6")[$− 6$] by 2. + Get zero on one side. + Divide both sides by 3. + Factor the left side. + Follow the steps to solve by completing the square: #math.equation(block: true, alt: "minus 4 x squared minus 36 x minus 65 equals 0")[$− 4 x^(2) − 36 x − 65 = 0$].+ Divide each term by #math.equation(block: false, alt: "minus 4")[$− 4$]. Write the equation with the constant on the right. #linebreak() \_\_\_\_\_#math.equation(block: false, alt: "equals")[$=$]\_\_\_\_\_ + Complete the square on the left: #linebreak() #math.equation(block: false, alt: "p equals the fraction 1 over 2 open parenthesis 9 close parenthesis equals")[$p = display(frac(1, 2)) ( 9 ) =$]\_\_\_\_\_, #math.equation(block: false, alt: "p squared equals")[$p^(2) =$]\_\_\_\_\_ #linebreak() Add #math.equation(block: false, alt: "p squared")[$p^(2)$] to both sides. #linebreak() \_\_\_\_\_#math.equation(block: false, alt: "equals")[$=$]\_\_\_\_\_ + Write the left side as a perfect square; simplify the right side. #linebreak() #math.equation(block: false, alt: "open parenthesis")[$\($]\_\_\_\_\_#math.equation(block: false, alt: "close parenthesis squared equals")[$\)^(2) =$]\_\_\_\_\_ + Solve by extracting roots. #linebreak() Solutions: #math.equation(block: false, alt: "x equals")[$x =$]\_\_\_\_\_ #linebreak() List all the values that are solutions. Use a comma to separate different solutions. + Graph #math.equation(block: false, alt: "y equals minus 4 x squared minus 36 x minus 65")[$y = − 4 x^(2) − 36 x − 65$] in the window #math.equation(block: true, alt: "Xmin, equals minus 9.4, Xmax equals 0; Ymin, equals minus 10, Ymax equals 20")[$"Xmin" & = − 9.4 & & "Xmax" = 0 \ "Ymin" & = − 10 & & "Ymax" = 20$] #solutionbox[ #math.equation(block: true, alt: "x equals the fraction minus 13 over 2")[$x = display(frac(− 13, 2))$], #math.equation(block: true, alt: "x equals the fraction minus 5 over 2")[$x = display(frac(− 5, 2))$] ] + Follow the steps to solve by completing the square: #math.equation(block: true, alt: "minus 4 x squared minus 36 x minus 65 equals 0")[$− 4 x^(2) − 36 x − 65 = 0$].+ Divide each term by #math.equation(block: false, alt: "minus 4")[$− 4$]. Write the equation with the constant on the right. + Complete the square on the left: #linebreak() #math.equation(block: false, alt: "p equals the fraction 1 over 2 open parenthesis 9 close parenthesis equals bar , p squared equals bar")[$p = display(frac(1, 2)) ( 9 ) = underline(#h(2em)) , " " " " p^(2) = underline(#h(2em))$] #linebreak() Add #math.equation(block: false, alt: "p squared")[$p^(2)$] to both sides. + Write the left side as a perfect square; simplify the right side. + Solve by extracting roots. + Graph #math.equation(block: false, alt: "y equals minus 4 x squared minus 36 x minus 65")[$y = − 4 x^(2) − 36 x − 65$] in the window #math.equation(block: true, alt: "Xmin, equals minus 9.4, Xmax equals 0; Ymin, equals minus 10, Ymax equals 20")[$"Xmin" & = − 9.4 & & "Xmax" = 0 \ "Ymin" & = − 10 & & "Ymax" = 20$] #solutionbox[ #math.equation(block: true, alt: "x equals the fraction minus 13 over 2")[$x = display(frac(− 13, 2))$], #math.equation(block: true, alt: "x equals the fraction minus 5 over 2")[$x = display(frac(− 5, 2))$] ] Here is a summary of the steps for solving quadratic equations by completing the square. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[To Solve a Quadratic Equation by Completing the Square] + + Write the equation in standard form. + Divide both sides of the equation by the coefficient of the quadratic term, and subtract the constant term from both sides. + Complete the square on the left side: + Multiply the coefficient of the first-degree term by one-half, then square the result. + Add the value obtained in (a) to both sides of the equation. + Write the left side of the equation as the square of a binomial. Simplify the right side. + Use extraction of roots to finish the solution. ] How does creating the square of a binomial help us solve a quadratic equation? \_\_\_\_\_ How does creating the square of a binomial help us solve a quadratic equation? === Quadratic Formula Instead of completing the square every time we solve a new quadratic equation, we can complete the square on the general quadratic equation, #math.equation(block: true, alt: "a x squared plus b x plus c equals 0 , a not equal to 0")[$a x^(2) + b x + c = 0 ", " " " " " " " a ≠ 0$] and obtain a formula for the solutions of any quadratic equation. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[The Quadratic Formula] The solutions of the equation #math.equation(block: false, alt: "a x squared plus b x plus c equals 0 , a not equal to 0")[$a x^(2) + b x + c = 0 ", " " " " " " " a ≠ 0$], are #math.equation(block: true, alt: "x equals the fraction minus b plus or minus the square root of b squared minus 4 a c over 2 a")[$bold(italic(x)) = frac(− bold(italic(b)) ± sqrt(bold(italic(b))^(2) − 4 bold(italic(a)) bold(italic(c))), 2 bold(italic(a)))$] ] Does #math.equation(block: false, alt: "the square root of b squared minus 4 a c equals b minus the square root of 4 a c")[$sqrt(b^(2) − 4 a c) = b − sqrt(4 a c)$] ? Why or why not? \_\_\_\_\_ #solutionbox[ No, #math.equation(block: false, alt: "the square root of x plus y not equal to the square root of x plus the square root of y")[$sqrt(x + y) ≠ sqrt(x) + sqrt(y)$]. ] Does #math.equation(block: false, alt: "the square root of b squared minus 4 a c equals b minus the square root of 4 a c")[$sqrt(b^(2) − 4 a c) = b − sqrt(4 a c)$] ? Why or why not? + No, it should be #math.equation(block: false, alt: "b minus 2 the square root of a c")[$b − 2 sqrt(a c)$]. + Yes, we take the square root of each term. + No, we cannot take the square root of a negative number. + No, #math.equation(block: false, alt: "the square root of x plus y not equal to the square root of x plus the square root of y")[$sqrt(x + y) ≠ sqrt(x) + sqrt(y)$]. This formula expresses the solutions of a quadratic equation in terms of its coefficients. (The proof of the formula is considered in the Homework problems.) The symbol #math.equation(block: false, alt: "plus or minus")[$±$], read "plus or minus," is used to combine the two equations #math.equation(block: true, alt: "x equals the fraction minus b plus the square root of b squared minus 4 a c over 2 a and x equals the fraction minus b minus the square root of b squared minus 4 a c over 2 a")[$x = display(frac(− b + sqrt(b^(2) − 4 a c), 2 a)) " " " " " " " " "and" " " " " " " " " x = display(frac(− b − sqrt(b^(2) − 4 a c), 2 a))$] into a single equation. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Notice that the two solutions can be written as #math.equation(block: true, alt: "x equals the fraction minus b over 2 a plus the fraction the square root of b squared minus 4 a c over 2 a and x equals the fraction minus b over 2 a minus the fraction the square root of b squared minus 4 a c over 2 a")[$x = display(frac(− b, 2 a)) + display(frac(sqrt(b^(2) − 4 a c), 2 a)) " " " " " " " " "and" " " " " " " " " x = display(frac(− b, 2 a)) − display(frac(sqrt(b^(2) − 4 a c), 2 a))$] These expressions show us again that the #math.equation(block: false, alt: "x")[$x$]-intercepts of the graph of #math.equation(block: false, alt: "y equals a x squared plus b x plus c")[$y = a x^(2) + b x + c " " " "$] lie an equal distance on either side of the vertical line #math.equation(block: false, alt: "x equals the fraction minus b over 2 a")[$x = display(frac(− b, 2 a))$], which is the axis of symmetry of the parabola. We'll consider these facts more fully in the next section, Graphing Parabolas. ] To solve a quadratic equation using the quadratic formula, all we have to do is substitute the coefficients #math.equation(block: false, alt: "a")[$a$], #math.equation(block: false, alt: "b")[$b$], and #math.equation(block: false, alt: "c")[$c$] into the formula. {"equation":"2\*x^2 + 1 = 4\*x","variable":"x"}(2 + sqrt(2))/2, (2 - sqrt(2))/2The example substitutes a = 2, b = −4, c = 1 into the quadratic formula and simplifies to x = (2 ± √2)/2. The CAS solves the same equation from scratch at build time and certifies that its exact roots agree with the book's — the substitution, the arithmetic under the radical, and the final simplification are all re-verified on every rebuild.#examplebox("Example 4")[][ Solve #math.equation(block: false, alt: "2 x squared plus 1 equals 4 x")[$" " " " 2 x^(2) + 1 = 4 x$]. #solutionbox[ Write the equation in standard form as #math.equation(block: true, alt: "2 x squared minus 4 x plus 1 equals 0")[$2 x^(2) − 4 x + 1 = 0$] Substitute #math.equation(block: false, alt: "2")[$2$] for #math.equation(block: false, alt: "a")[$a$], #math.equation(block: false, alt: "minus 4")[$− 4$] for #math.equation(block: false, alt: "b")[$b$], and #math.equation(block: false, alt: "1")[$1$] for #math.equation(block: false, alt: "c")[$c$] into the quadratic formula, then simplify. #math.equation(block: true, alt: "x, equals the fraction minus open parenthesis minus 4 close parenthesis plus or minus the square root of open parenthesis minus 4 close parenthesis squared minus 4 open parenthesis 2 close parenthesis open parenthesis 1 close parenthesis over 2 open parenthesis 2 close parenthesis; equals the fraction 4 plus or minus the square root of 8 over 4 equals the fraction 4 plus or minus 2 the square root of 2 over 4 equals the fraction 2 plus or minus the square root of 2 over 2")[$x & = frac(− ( − 4 ) ± sqrt(( − 4 )^(2) − 4 ( 2 ) ( 1 )), 2 ( 2 )) \ & = frac(4 ± sqrt(8), 4) = frac(4 ± 2 sqrt(2), 4) = frac(2 ± sqrt(2), 2)$] (See Appendix A.8 Working with Radicals, to review simplifying radicals.) Using a calculator, we find that the solutions are approximately #math.equation(block: false, alt: "1.707")[$1.707$] and #math.equation(block: false, alt: "0.293")[$0.293$]. We can also verify that the #math.equation(block: false, alt: "x")[$x$]-intercepts of the graph of #math.equation(block: false, alt: "y equals 2 x squared minus 4 x plus 1")[$y = 2 x^(2) − 4 x + 1$] are approximately #math.equation(block: false, alt: "1.707")[$1.707$] and #math.equation(block: false, alt: "0.293")[$0.293$], as shown below. #figure(figph[parabola with irrational x-intercepts], alt: "parabola with irrational x-intercepts", caption: none) ] ] Use the quadratic formula to solve #math.equation(block: false, alt: "x squared minus 3 x equals 1")[$x^(2) − 3 x = 1$]. Solutions: #math.equation(block: false, alt: "x equals")[$x =$]\_\_\_\_\_ List all the values that are solutions. Use a comma to separate different solutions. #solutionbox[ #math.equation(block: true, alt: "x equals the fraction 3 plus or minus the square root of 13 over 2")[$x = display(frac(3 ± sqrt(13), 2))$] ] Use the quadratic formula to solve #math.equation(block: false, alt: "x squared minus 3 x equals 1")[$x^(2) − 3 x = 1$]. #solutionbox[ #math.equation(block: true, alt: "x equals the fraction 3 plus or minus the square root of 13 over 2")[$x = display(frac(3 ± sqrt(13), 2))$] ] Note that the solutions to this equation are the same as the solutions we found in Practice 2. {"equation":"x^2 - 3\*x = 1","variable":"x"}(3 + sqrt(13))/2, (3 - sqrt(13))/2The practice problem's stated answer is x = (3 ± √13)/2, and this note observes it matches an earlier practice — two different methods, one root pair. The CAS gives a third, independent derivation of the same pair at build time, checked as an unordered set against the book's.Note that the solutions to this equation are the same as the solutions we found in Practice 2. How can you decide which of the four methods for solving a quadratic equation you should use? \_\_\_\_\_ How can you decide which of the four methods for solving a quadratic equation you should use? === Applications We have now seen four different algebraic methods for solving quadratic equations: + Factoring + Extraction of roots + Completing the square + Quadratic formula Factoring and extraction of roots are relatively fast and simple, but they do not work on all quadratic equations. The quadratic formula will work on any quadratic equation. What is wrong with this statement of the quadratic formula? #math.equation(block: false, alt: "x equals minus b plus or minus the fraction the square root of b squared minus 4 a c over 2 a")[$x = − b ± display(frac(sqrt(b^(2) − 4 a c), 2 a))$] \_\_\_\_\_ #solutionbox[ #math.equation(block: false, alt: "minus b")[$− b$] should be over #math.equation(block: false, alt: "2 a")[$2 a$]. ] What is wrong with this statement of the quadratic formula? #math.equation(block: false, alt: "x equals minus b plus or minus the fraction the square root of b squared minus 4 a c over 2 a")[$x = − b ± display(frac(sqrt(b^(2) − 4 a c), 2 a))$] + Nothing. + We should simplify the radical. + #math.equation(block: false, alt: "minus b")[$− b$] should be over #math.equation(block: false, alt: "2 a")[$2 a$]. + It should be #math.equation(block: false, alt: "plus or minus b")[$± b$]. #examplebox("Example 5")[][ The owners of a day-care center plan to enclose a divided play area against the back wall of their building, as shown below. They have #math.equation(block: false, alt: "300")[$300$] feet of picket fence and would like the total area of the playground to be #math.equation(block: false, alt: "6000")[$6000$] square feet. Can they enclose the playground with the fence they have, and if so, what should the dimensions of the playground be? #figure(figph[two fenced rectangles against wall], alt: "two fenced rectangles against wall", caption: none) #solutionbox[ Suppose the width of the play area is #math.equation(block: false, alt: "x")[$x$] feet. Because there are three sections of fence along the width of the play area, that leaves #math.equation(block: false, alt: "300 minus 3 x")[$300 − 3 x$] feet of fence for its length. The area of the play area should be #math.equation(block: false, alt: "6000")[$6000$] square feet, so we have the equation #math.equation(block: true, alt: "x open parenthesis 300 minus 3 x close parenthesis equals 6000")[$x ( 300 − 3 x ) = 6000$] This is a quadratic equation. In standard form, #math.equation(block: true, alt: "3 x squared minus 300 x plus 6000, equals 0, Divide each term by 3.; x squared minus 100 x plus 2000, equals 0")[$3 x^(2) − 300 x + 6000 & = 0 & & "Divide each term by 3." \ x^(2) − 100 x + 2000 & = 0$] The left side cannot be factored, so we use the quadratic formula with #math.equation(block: false, alt: "a equals 1")[$a = 1$], #math.equation(block: false, alt: "b equals minus 100")[$b = − 100$], and #math.equation(block: false, alt: "c equals 2000")[$c = 2000$]. #math.equation(block: true, alt: "x, equals the fraction minus open parenthesis minus 100 close parenthesis plus or minus the square root of open parenthesis minus 100 close parenthesis squared minus 4 open parenthesis 1 close parenthesis open parenthesis 2000 close parenthesis over 2 open parenthesis 1 close parenthesis; equals the fraction 100 plus or minus the square root of 2000 over 2 approximately equals the fraction 100 plus or minus 44.7 over 2")[$x & = frac(− ( − 100 ) ± sqrt(( − 100 )^(2) − 4 ( 1 ) ( 2000 )), 2 ( 1 )) \ & = frac(100 ± sqrt(2000), 2) ≈ frac(100 ± 44.7, 2)$] Simplifying the last fraction, we find that #math.equation(block: false, alt: "x approximately equals 72.35")[$x ≈ 72.35$] or #math.equation(block: false, alt: "x approximately equals 27.65")[$x ≈ 27.65$]. Both values give solutions to the problem. - If the width of the play area is #math.equation(block: false, alt: "72.35")[$72.35$] feet, then the length is #math.equation(block: false, alt: "300 minus 3 open parenthesis 72.35 close parenthesis")[$300 − 3 ( 72.35 )$], or #math.equation(block: false, alt: "82.95")[$82.95$] feet. - If the width is #math.equation(block: false, alt: "27.65")[$27.65$] feet, the length is #math.equation(block: false, alt: "300 minus 3 open parenthesis 27.65 close parenthesis")[$300 − 3 ( 27.65 )$], or #math.equation(block: false, alt: "217.05")[$217.05$] feet. ] ] In Height of a Baseball, we considered the height of a baseball, given by the equation #math.equation(block: true, alt: "h equals minus 16 t squared plus 64 t plus 4")[$h = − 16 t^(2) + 64 t + 4$] Find two times when the ball is at a height of #math.equation(block: false, alt: "20")[$20$] feet. Give your answers to two decimal places. Answer: \_\_\_\_\_ sec List all the times that are solutions. Use a comma to separate different times. #solutionbox[ #math.equation(block: false, alt: "0.27")[$0.27$] sec, #math.equation(block: false, alt: "3.73")[$3.73$] sec ] In Height of a Baseball, we considered the height of a baseball, given by the equation #math.equation(block: true, alt: "h equals minus 16 t squared plus 64 t plus 4")[$h = − 16 t^(2) + 64 t + 4$] Find two times when the ball is at a height of #math.equation(block: false, alt: "20")[$20$] feet. Give your answers to two decimal places. #solutionbox[ #math.equation(block: false, alt: "0.27")[$0.27$] sec, #math.equation(block: false, alt: "3.73")[$3.73$] sec ] Sometimes it is useful to solve a quadratic equation for one variable in terms of the others. #examplebox("Example 6")[][ Solve #math.equation(block: false, alt: "x squared minus x y plus y equals 2")[$" " " " x^(2) − x y + y = 2 " " " "$] for #math.equation(block: false, alt: "x")[$x$] in terms of #math.equation(block: false, alt: "y")[$y$]. #solutionbox[ We first write the equation in standard form as a quadratic equation in the variable #math.equation(block: false, alt: "x")[$x$]. #math.equation(block: true, alt: "x squared minus y x plus open parenthesis y minus 2 close parenthesis equals 0")[$x^(2) − y x + ( y − 2 ) = 0$] Expressions in #math.equation(block: false, alt: "y")[$y$] are treated as constants with respect to #math.equation(block: false, alt: "x")[$x$], so that #math.equation(block: false, alt: "a equals 1")[$a = 1$], #math.equation(block: false, alt: "b equals minus y")[$b = − bold(italic(y))$], and #math.equation(block: false, alt: "c equals y minus 2")[$c = bold(italic(y)) − 2$]. Substitute these expressions into the quadratic formula. #math.equation(block: true, alt: "x, equals the fraction minus open parenthesis minus y close parenthesis plus or minus the square root of open parenthesis minus y close parenthesis squared minus 4 open parenthesis 1 close parenthesis open parenthesis y minus 2 close parenthesis over 2 open parenthesis 1 close parenthesis; equals the fraction y plus or minus the square root of y squared minus 4 y plus 8 over 2")[$x & = frac(− ( − bold(italic(y)) ) ± sqrt(( − bold(italic(y)) )^(2) − 4 ( 1 ) ( bold(italic(y)) − 2 )), 2 ( 1 )) \ & = frac(y ± sqrt(y^(2) − 4 y + 8), 2)$] ] ] Solve #math.equation(block: false, alt: "2 x squared plus k x plus k squared equals 1")[$2 x^(2) + k x + k^(2) = 1$] for #math.equation(block: false, alt: "x")[$x$] in terms of #math.equation(block: false, alt: "k")[$k$]. Solutions: #math.equation(block: false, alt: "x equals")[$x =$]\_\_\_\_\_ Use a comma to separate different solutions. #solutionbox[ #math.equation(block: true, alt: "x equals the fraction minus k plus or minus the square root of 8 minus 7 k squared over 4")[$x = display(frac(− k ± sqrt(8 − 7 k^(2)), 4))$] ] Solve #math.equation(block: false, alt: "2 x squared plus k x plus k squared equals 1")[$2 x^(2) + k x + k^(2) = 1$] for #math.equation(block: false, alt: "x")[$x$] in terms of #math.equation(block: false, alt: "k")[$k$]. #solutionbox[ #math.equation(block: true, alt: "x equals the fraction minus k plus or minus the square root of 8 minus 7 k squared over 4")[$x = display(frac(− k ± sqrt(8 − 7 k^(2)), 4))$] ] Explain in words how to evaluate the quadratic formula. \_\_\_\_\_ Explain in words how to evaluate the quadratic formula. === Section Summary ==== Vocabulary Look up the definitions of new terms in the Glossary. - Quadratic trinomial - Complete the square - Quadratic formula ==== CONCEPTS + The square of the binomial #math.equation(block: false, alt: "open parenthesis x plus p close parenthesis")[$( x + p )$] is a #strong[quadratic trinomial], #math.equation(block: true, alt: "open parenthesis x plus p close parenthesis squared equals x squared plus 2 p x plus p squared")[$( x + p )^(2) = x^(2) + 2 p x + p^(2)$] + #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[To Solve a Quadratic Equation by Completing the Square] + + Write the equation in standard form. + Divide both sides of the equation by the coefficient of the quadratic term, and subtract the constant term from both sides. + Complete the square on the left side:+ Multiply the coefficient of the first-degree term by one-half, then square the result. + Add the value obtained in (a) to both sides of the equation. + Write the left side of the equation as the square of a binomial. Simplify the right side. + Use extraction of roots to finish the solution. ] + #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[The Quadratic Formula] The solutions of the equation #math.equation(block: false, alt: "a x squared plus b x plus c equals 0 , a not equal to 0")[$a x^(2) + b x + c = 0 ", " " " " " " " a ≠ 0$], are #math.equation(block: true, alt: "x equals the fraction minus b plus or minus the square root of b squared minus 4 a c over 2 a")[$bold(italic(x)) = frac(− bold(italic(b)) ± sqrt(bold(italic(b))^(2) − 4 bold(italic(a)) bold(italic(c))), 2 bold(italic(a)))$] ] + We have four methods for solving quadratic equations: extracting of roots, factoring, completing the square, and using the quadratic formula. The first two methods are faster, but they don't work on all equations. The last two methods work on any quadratic equation. ==== STUDY QUESTIONS + Name four algebraic methods for solving a quadratic equation. + Give an example of a quadratic trinomial that is the square of a binomial. + What number must be added to #math.equation(block: false, alt: "x squared minus 26 x")[$x^(2) − 26 x$] to make it the square of a binomial? + After completing the square, how do we finish solving the quadratic equation? + What is the first step in solving the equation #math.equation(block: false, alt: "2 x squared minus 6 x equals 5")[$2 x^(2) − 6 x = 5$] by completing the square? ==== SKILLS Practice each skill in the Homework problems listed. + Solve quadratic equations by completing the square: \#3–24 + Solve quadratic equations by using the quadratic formula: \#27–36 + Solve problems by writing and solving quadratic equations: \#37–44 + Solve formulas: \#45–64 === Homework 6.2 For Problems 1–2, complete the square and write the result as the square of a binomial. + #math.equation(block: false, alt: "x squared plus 8 x")[$x^(2) + 8 x$] + #math.equation(block: false, alt: "x squared minus 7 x")[$x^(2) − 7 x$] + #math.equation(block: false, alt: "x squared plus the fraction 3 over 2 x")[$x^(2) + display(frac(3, 2)) x$] + #math.equation(block: false, alt: "x squared minus the fraction 4 over 5 x")[$x^(2) − display(frac(4, 5)) x$] + #math.equation(block: false, alt: "open parenthesis x plus 4 close parenthesis squared")[$( x + 4 )^(2)$] + #math.equation(block: false, alt: "open parenthesis x minus the fraction 7 over 2 close parenthesis squared")[$attach(( x − display(frac(7, 2)) ), t: 2)$] + #math.equation(block: false, alt: "open parenthesis x plus the fraction 3 over 4 close parenthesis squared")[$attach(( x + display(frac(3, 4)) ), t: 2)$] + #math.equation(block: false, alt: "open parenthesis x minus the fraction 2 over 5 close parenthesis squared")[$attach(( x − display(frac(2, 5)) ), t: 2)$] + #math.equation(block: false, alt: "x squared minus 14 x")[$x^(2) − 14 x$] + #math.equation(block: false, alt: "x squared plus 3 x")[$x^(2) + 3 x$] + #math.equation(block: false, alt: "x squared minus the fraction 5 over 2 x")[$x^(2) − display(frac(5, 2)) x$] + #math.equation(block: false, alt: "x squared plus the fraction 2 over 3 x")[$x^(2) + display(frac(2, 3)) x$] For Problems 3-18, solve by completing the square. #math.equation(block: true, alt: "x squared minus 2 x plus 1 equals 0")[$x^(2) − 2 x + 1 = 0$] #math.equation(block: true, alt: "1")[$1$] #math.equation(block: true, alt: "x squared plus 4 x plus 4 equals 0")[$x^(2) + 4 x + 4 = 0$] #math.equation(block: true, alt: "x squared plus 9 x plus 20 equals 0")[$x^(2) + 9 x + 20 = 0$] #math.equation(block: true, alt: "minus 4")[$− 4$], #math.equation(block: true, alt: "minus 5")[$" " − 5$] #math.equation(block: true, alt: "x squared minus x minus 20 equals 0")[$x^(2) − x − 20 = 0$] #math.equation(block: true, alt: "x squared equals 3 minus 3 x")[$x^(2) = 3 − 3 x$] #math.equation(block: true, alt: "the fraction 3 over 2 plus or minus the square root of the fraction 21 over 4 equals the fraction minus 3 plus or minus the square root of 21 over 2")[$display(frac(3, 2)) ± sqrt(display(frac(21, 4))) = display(frac(− 3 ± sqrt(21), 2))$] #math.equation(block: true, alt: "x squared equals 5 minus 5 x")[$x^(2) = 5 − 5 x$] #math.equation(block: true, alt: "2 x squared plus 4 x minus 3 equals 0")[$2 x^(2) + 4 x − 3 = 0$] #math.equation(block: true, alt: "minus 1 plus or minus the square root of the fraction 5 over 2")[$− 1 ± sqrt(display(frac(5, 2)))$] #math.equation(block: true, alt: "3 x squared plus 12 x plus 2 equals 0")[$3 x^(2) + 12 x + 2 = 0$] #math.equation(block: true, alt: "3 x squared plus x equals 4")[$3 x^(2) + x = 4$] #math.equation(block: true, alt: "the fraction minus 4 over 3")[$display(frac(− 4, 3))$], #math.equation(block: true, alt: "1")[$" " 1$] #math.equation(block: true, alt: "4 x squared plus 6 x equals 3")[$4 x^(2) + 6 x = 3$] #math.equation(block: true, alt: "4 x squared minus 3 equals 2 x")[$4 x^(2) − 3 = 2 x$] #math.equation(block: true, alt: "the fraction 1 over 4 plus or minus the square root of the fraction 13 over 16 equals the fraction 1 plus or minus the square root of 13 over 4")[$display(frac(1, 4)) ± sqrt(display(frac(13, 16))) = display(frac(1 ± sqrt(13), 4))$] #math.equation(block: true, alt: "2 x squared minus 5 equals 3 x")[$2 x^(2) − 5 = 3 x$] #math.equation(block: true, alt: "3 x squared minus x minus 4 equals 0")[$3 x^(2) − x − 4 = 0$] #math.equation(block: true, alt: "minus 1")[$− 1$], #math.equation(block: true, alt: "the fraction 4 over 3")[$display(frac(4, 3))$] #math.equation(block: true, alt: "2 x squared minus x minus 3 equals 0")[$2 x^(2) − x − 3 = 0$] #math.equation(block: true, alt: "5 x squared plus 8 x equals 4")[$5 x^(2) + 8 x = 4$] #math.equation(block: true, alt: "minus 2")[$− 2$], #math.equation(block: true, alt: "the fraction 2 over 5")[$display(frac(2, 5))$] #math.equation(block: true, alt: "9 x squared minus 12 x minus 5 equals 0")[$9 x^(2) − 12 x − 5 = 0$] For Problems 19-24, solve by completing the square. Your answers will involve #math.equation(block: false, alt: "a")[$a$], #math.equation(block: false, alt: "b")[$b$], or #math.equation(block: false, alt: "c")[$c$]. #math.equation(block: true, alt: "x squared plus 2 x plus c equals 0")[$x^(2) + 2 x + c = 0$] #math.equation(block: true, alt: "minus 1 plus or minus the square root of 1 minus c")[$− 1 ± sqrt(1 − c)$] #math.equation(block: true, alt: "x squared minus 4 x plus c equals 0")[$x^(2) − 4 x + c = 0$] #math.equation(block: true, alt: "x squared plus b x plus 1 equals 0")[$x^(2) + b x + 1 = 0$] #math.equation(block: true, alt: "minus the fraction b over 2 plus or minus the square root of the fraction b squared minus 4 over 4 equals the fraction minus b plus or minus the square root of b squared minus 4 over 2")[$− display(frac(b, 2)) ± sqrt(display(frac(b^(2) − 4, 4))) = display(frac(− b ± sqrt(b^(2) − 4), 2))$] #math.equation(block: true, alt: "x squared plus b x minus 4 equals 0")[$x^(2) + b x − 4 = 0$] #math.equation(block: true, alt: "a x squared plus 2 x minus 4 equals 0")[$a x^(2) + 2 x − 4 = 0$] #math.equation(block: true, alt: "the fraction minus 1 plus or minus the square root of 4 a plus 1 over a")[$display(frac(− 1 ± sqrt(4 a + 1), a))$] #math.equation(block: true, alt: "a x squared minus 4 x plus 9 equals 0")[$a x^(2) − 4 x + 9 = 0$] + Write an expression for the area of the square in the figure. + Express the area as a polynomial. + Divide the square into four pieces whose areas are given by the terms of your answer to part (b). #figure(figph[square], alt: "square", caption: none) + #math.equation(block: false, alt: "A equals open parenthesis x plus y close parenthesis squared")[$A = ( x + y )^(2)$] + #math.equation(block: false, alt: "A equals x squared plus 2 x y plus y squared")[$A = x^(2) + 2 x y + y^(2)$] + #math.equation(block: false, alt: "x squared")[$x^(2)$], #math.equation(block: false, alt: "x y")[$x y$], #math.equation(block: false, alt: "x y")[$x y$], #math.equation(block: false, alt: "y squared")[$y^(2)$]#figure(figph[square], alt: "square", caption: none) + Write an expression for the area of the shaded region in the figure. + Express the area in factored form. + By making one cut in the shaded region, rearrange the pieces into a rectangle whose area is given by your answer to part (b). #figure(figph[shaded square], alt: "shaded square", caption: none) For Problems 23-36, solve using the quadratic formula. Round your answers to three decimal places. #math.equation(block: true, alt: "x squared minus x minus 1 equals 0")[$x^(2) − x − 1 = 0$] #math.equation(block: true, alt: "1.618")[$1.618$], #math.equation(block: true, alt: "minus 0.618")[$" " − 0.618$] #math.equation(block: true, alt: "x squared plus x minus 1 equals 0")[$x^(2) + x − 1 = 0$] #math.equation(block: true, alt: "y squared plus 2 y equals 5")[$y^(2) + 2 y = 5$] #math.equation(block: true, alt: "1.449")[$1.449$], #math.equation(block: true, alt: "minus 3.449")[$" " − 3.449$] #math.equation(block: true, alt: "y squared minus 4 y equals 4")[$y^(2) − 4 y = 4$] #math.equation(block: true, alt: "3 z squared equals 4.2 z plus 1.5")[$3 z^(2) = 4.2 z + 1.5$] #math.equation(block: true, alt: "1.695")[$1.695$], #math.equation(block: true, alt: "minus 0.295")[$" " − 0.295$] #math.equation(block: true, alt: "2 z squared equals 7.5 z minus 6.3")[$2 z^(2) = 7.5 z − 6.3$] #math.equation(block: true, alt: "0 equals x squared minus the fraction 5 over 3 x plus the fraction 1 over 3")[$0 = x^(2) − display(frac(5, 3)) x + display(frac(1, 3))$] #math.equation(block: true, alt: "1.434")[$1.434$], #math.equation(block: true, alt: "0.232")[$" " 0.232$] #math.equation(block: true, alt: "0 equals minus x squared plus the fraction 5 over 2 x minus the fraction 1 over 2")[$0 = − x^(2) + display(frac(5, 2)) x − display(frac(1, 2))$] #math.equation(block: true, alt: "minus 5.2 z squared plus 176 z plus 1218 equals 0")[$− 5.2 z^(2) + 176 z + 1218 = 0$] #math.equation(block: true, alt: "minus 5.894")[$− 5.894$], #math.equation(block: true, alt: "39.740")[$" " 39.740$] #math.equation(block: true, alt: "15 z squared minus 18 z minus 2750 equals 0")[$15 z^(2) − 18 z − 2750 = 0$] A car traveling at #math.equation(block: false, alt: "s")[$s$] miles per hour on a dry road surface requires approximately #math.equation(block: false, alt: "d")[$d$] feet to stop, where #math.equation(block: false, alt: "d")[$d$] is given by the function #math.equation(block: true, alt: "d equals f open parenthesis s close parenthesis equals the fraction s squared over 24 plus the fraction s over 2")[$d = f ( s ) = display(frac(s^(2), 24)) + display(frac(s, 2))$] + Make a table showing the stopping distance, #math.equation(block: false, alt: "d")[$d$], for speeds of #math.equation(block: false, alt: "10")[$10$], #math.equation(block: false, alt: "20")[$20$], #math.equation(block: false, alt: "and so on")[$…$] , #math.equation(block: false, alt: "100")[$100$] miles per hour. (Use the #strong[Table] feature of your calculator.) + Graph the function for #math.equation(block: false, alt: "d")[$d$] in terms of #math.equation(block: false, alt: "s")[$s$]. Use your table values to help you choose appropriate window settings. + Write and solve an equation to answer the question: If a car must be able to stop in #math.equation(block: false, alt: "50")[$50$] feet, what is the maximum safe speed it can travel? Verify your answer on your graph. + #figure(table( columns: 11, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "s")[$s$]], [#math.equation(block: false, alt: "10")[$10$]], [#math.equation(block: false, alt: "20")[$20$]], [#math.equation(block: false, alt: "30")[$30$]], [#math.equation(block: false, alt: "40")[$40$]], [#math.equation(block: false, alt: "50")[$50$]], [#math.equation(block: false, alt: "60")[$60$]], [#math.equation(block: false, alt: "70")[$70$]], [#math.equation(block: false, alt: "80")[$80$]], [#math.equation(block: false, alt: "90")[$90$]], [#math.equation(block: false, alt: "100")[$100$]]), [#math.equation(block: false, alt: "d")[$d$]], [#math.equation(block: false, alt: "9")[$9$]], [#math.equation(block: false, alt: "27")[$27$]], [#math.equation(block: false, alt: "53")[$53$]], [#math.equation(block: false, alt: "87")[$87$]], [#math.equation(block: false, alt: "129")[$129$]], [#math.equation(block: false, alt: "180")[$180$]], [#math.equation(block: false, alt: "239")[$239$]], [#math.equation(block: false, alt: "307")[$307$]], [#math.equation(block: false, alt: "383")[$383$]], [#math.equation(block: false, alt: "467")[$467$]], )) + #figure(figph[parabola], alt: "parabola", caption: none) + #math.equation(block: false, alt: "the fraction s squared over 24 plus the fraction s over 2 equals 50")[$display(frac(s^(2), 24)) + display(frac(s, 2)) = 50$]; #math.equation(block: false, alt: "29.16")[$" " 29.16$] mph A car traveling at #math.equation(block: false, alt: "s")[$s$] miles per hour on a wet road surface requires approximately #math.equation(block: false, alt: "d")[$d$] feet to stop, where #math.equation(block: false, alt: "d")[$d$] is given by the function #math.equation(block: true, alt: "d equals f open parenthesis s close parenthesis equals the fraction s squared over 12 plus the fraction s over 2")[$d = f ( s ) = display(frac(s^(2), 12)) + display(frac(s, 2))$] + Make a table showing the stopping distance, #math.equation(block: false, alt: "d")[$d$], for speeds of #math.equation(block: false, alt: "10")[$10$], #math.equation(block: false, alt: "20")[$20$], #math.equation(block: false, alt: "and so on")[$…$] , #math.equation(block: false, alt: "100")[$100$] miles per hour. (Use the #strong[Table] feature of your calculator.) + Graph the function for #math.equation(block: false, alt: "d")[$d$] in terms of #math.equation(block: false, alt: "s")[$s$]. Use your table values to help you choose appropriate window settings. + Insurance investigators at the scene of an accident find skid marks #math.equation(block: false, alt: "100")[$100$] feet long leading up to the point of impact. Write and solve an equation to discover how fast the car was traveling when it put on the brakes. Verify your answer on your graph. A skydiver jumps out of an airplane at #math.equation(block: false, alt: "11 , 000")[$11 , 000$] feet. While she is in free-fall, her altitude in feet #math.equation(block: false, alt: "t")[$t$] seconds after jumping is given by the function #math.equation(block: true, alt: "h equals f open parenthesis t close parenthesis equals minus 16 t squared minus 16 t plus 11 , 000")[$h = f ( t ) = − 16 t^(2) − 16 t + 11 , 000$] + Make a table of values showing the skydiver's altitude at #math.equation(block: false, alt: "5")[$5$]-second intervals after she jumps from the airplane. (Use the #strong[Table] feature of your calculator.) + Graph the function. Use your table of values to choose appropriate window settings. + If the skydiver must open her parachute at an altitude of #math.equation(block: false, alt: "1000")[$1000$] feet, how long can she free-fall? Write and solve an equation to find the answer. + If the skydiver drops a marker just before she opens her parachute, how long will it take the marker to hit the ground? (#emph[Hint]: The marker continues to fall according to the equation given above.) + Find points on your graph that correspond to your answers to parts (c) and (d). + #figure(table( columns: 7, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "t")[$t$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "10")[$10$]], [#math.equation(block: false, alt: "15")[$15$]], [#math.equation(block: false, alt: "20")[$20$]], [#math.equation(block: false, alt: "25")[$25$]]), [#math.equation(block: false, alt: "h")[$h$]], [#math.equation(block: false, alt: "11 , 000")[$11 , 000$]], [#math.equation(block: false, alt: "10 , 520")[$10 , 520$]], [#math.equation(block: false, alt: "9240")[$9240$]], [#math.equation(block: false, alt: "7160")[$7160$]], [#math.equation(block: false, alt: "4280")[$4280$]], [#math.equation(block: false, alt: "600")[$600$]], )) + #figure(figph[parabola], alt: "parabola", caption: none) + #math.equation(block: false, alt: "minus 16 t squared minus 16 t plus 11 , 000 equals 1000")[$− 16 t^(2) − 16 t + 11 , 000 = 1000$]; #math.equation(block: false, alt: "24.5")[$24.5$] sec + #math.equation(block: false, alt: "1.2")[$1.2$] sec A high diver jumps from the #math.equation(block: false, alt: "10")[$10$]-meter springboard. His height in meters above the water #math.equation(block: false, alt: "t")[$t$] seconds after leaving the board is given by the function #math.equation(block: true, alt: "h equals f open parenthesis t close parenthesis equals minus 4.9 t squared plus 8 t plus 10")[$h = f ( t ) = − 4.9 t^(2) + 8 t + 10$] + Make a table of values showing the diver's altitude at #math.equation(block: false, alt: "0.25")[$0.25$]-second intervals after he jumps from the airplane. (Use the #strong[Table] feature of your calculator.) + Graph the function. Use your table of values to choose appropriate window settings. + How long is it before the diver passes the board on the way down? + How long is it before the diver hits the water? + Find points on your graph that correspond to your answers to parts (c) and (d). A dog trainer has #math.equation(block: false, alt: "100")[$100$] meters of chain link fence. She wants to enclose #math.equation(block: false, alt: "250")[$250$] square meters in three pens of equal size, as shown in the figure. #figure(figph[three adjacent pens], alt: "three adjacent pens", caption: none) + Let #math.equation(block: false, alt: "l")[$l$] and #math.equation(block: false, alt: "w")[$w$] represent the length and width, respectively, of the entire area. Write an equation about the amount of chain link fence. + Solve your equation for #math.equation(block: false, alt: "l")[$l$] in terms #math.equation(block: false, alt: "w")[$w$]. + Write and solve an equation in #math.equation(block: false, alt: "w")[$w$] for the total area enclosed. + Find the dimensions of each pen. + #math.equation(block: false, alt: "2 l plus 4 w equals 100")[$2 l + 4 w = 100$] + #math.equation(block: false, alt: "l equals 50 minus 2 w")[$l = 50 − 2 w$] + #math.equation(block: false, alt: "w open parenthesis 50 minus 2 w close parenthesis equals 250")[$w ( 50 − 2 w ) = 250$]; #math.equation(block: false, alt: "w equals 6.91")[$" " w = 6.91$], #math.equation(block: false, alt: "18.09")[$" " 18.09$] + #math.equation(block: false, alt: "12.06")[$12.06$] m by #math.equation(block: false, alt: "6.91")[$6.91$] m, or #math.equation(block: false, alt: "4.61")[$4.61$] m by #math.equation(block: false, alt: "18.09")[$18.09$] m An architect is planning to include a rectangular window topped by a semicircle in his plans for a new house, as shown in the figure. In order to admit enough light, the window should have an area of #math.equation(block: false, alt: "120")[$120$] square feet. The architect wants the rectangular portion of the window to be #math.equation(block: false, alt: "2")[$2$] feet wider than it is tall. #figure(figph[rectangle surmounted by semicircle], alt: "rectangle surmounted by semicircle", caption: none) + Let #math.equation(block: false, alt: "x")[$x$] stand for the horizontal width of the window. Write expressions for the height of the rectangular portion and for the radius of the semicircular portion. + Write an expression for the total area of the window. + Write and solve an equation to find the width and overall height of the window. When you look down from a height, say a tall building or a mountain peak, your line of sight is tangent to the Earth at the horizon, as shown in the figure. #figure(figph[tangent to earth], alt: "tangent to earth", caption: none) + Suppose you are standing on top of the Petronas Tower in Kuala Lumpur, #math.equation(block: false, alt: "1483")[$1483$] feet high. How far can you see on a clear day? (You will need to use the Pythagorean theorem and the fact that the radius of the Earth is #math.equation(block: false, alt: "3960")[$3960$] miles. Do not forget to convert the height of the Petronas Tower to miles.) + How tall a building should you stand on in order to see #math.equation(block: false, alt: "100")[$100$] miles? + #math.equation(block: false, alt: "47.2")[$47.2$] mi + #math.equation(block: false, alt: "1.26")[$1.26$] mi + If the radius of the Earth is #math.equation(block: false, alt: "6370")[$6370$] kilometers, how far can you see from an airplane at an altitude of #math.equation(block: false, alt: "10 , 000")[$10 , 000$] meters? (#emph[Hint]: See Problem 43.) + b. How high would the airplane have to be in order for you to see a distance of #math.equation(block: false, alt: "10")[$10$] kilometers? For Problems 45-52, use the quadratic formula to solve each equation for the indicated variable. #math.equation(block: false, alt: "A equals 2 w squared plus 4 l w ,")[$A = 2 w^(2) + 4 l w , " " " "$] for #math.equation(block: false, alt: "w")[$w$] #math.equation(block: true, alt: "w equals the fraction minus 4 l plus or minus the square root of 16 l squared plus 8 A over 4 equals the fraction minus 2 l plus or minus the square root of 4 l squared plus 2 A over 2")[$w = display(frac(− 4 l ± sqrt(16 l^(2) + 8 A), 4)) = display(frac(− 2 l ± sqrt(4 l^(2) + 2 A), 2))$] #math.equation(block: false, alt: "A equals π r squared plus π r s ,")[$A = π r^(2) + π r s , " " " "$] for #math.equation(block: false, alt: "r")[$r$] #math.equation(block: false, alt: "h equals 4 t minus 16 t squared ,")[$h = 4 t − 16 t^(2) , " " " "$] for #math.equation(block: false, alt: "t")[$t$] #math.equation(block: true, alt: "t equals the fraction 4 plus or minus the square root of 16 plus 64 h over 32 equals the fraction 1 plus or minus the square root of 1 plus 4 h over 8")[$t = display(frac(4 ± sqrt(16 + 64 h), 32)) = display(frac(1 ± sqrt(1 + 4 h), 8))$] #math.equation(block: false, alt: "P equals I E minus R I squared ,")[$P = I E − R I^(2) , " " " "$] for #math.equation(block: false, alt: "I")[$I$] #math.equation(block: false, alt: "s equals v t minus the fraction 1 over 2 a t squared ,")[$s = v t − display(frac(1, 2)) a t^(2) , " " " "$] for #math.equation(block: false, alt: "t")[$t$] #math.equation(block: true, alt: "t equals the fraction v plus or minus the square root of v squared minus 2 a s over a")[$t = display(frac(v ± sqrt(v^(2) − 2 a s), a))$] #math.equation(block: false, alt: "S equals the fraction n squared plus n over 2 ,")[$S = display(frac(n^(2) + n, 2)) , " " " "$] for #math.equation(block: false, alt: "n")[$n$] #math.equation(block: false, alt: "3 x squared plus x y plus y squared equals 2 ,")[$3 x^(2) + x y + y^(2) = 2 , " " " "$] for #math.equation(block: false, alt: "y")[$y$] #math.equation(block: true, alt: "y equals the fraction minus x plus or minus the square root of 8 minus 11 x squared over 2")[$y = display(frac(− x ± sqrt(8 − 11 x^(2)), 2))$] #math.equation(block: false, alt: "y squared minus 3 x y plus x squared equals 3 ,")[$y^(2) − 3 x y + x^(2) = 3 , " " " "$] for #math.equation(block: false, alt: "x")[$x$] For Problems 53-60, solve for #math.equation(block: false, alt: "y")[$y$] in terms of #math.equation(block: false, alt: "x")[$x$]. Use whichever method of solution seems easiest. #math.equation(block: true, alt: "x squared y minus y squared equals 0")[$x^(2) y − y^(2) = 0$] #math.equation(block: true, alt: "0 , x squared")[$0 , " " x^(2)$] #math.equation(block: true, alt: "x squared y squared minus y equals 0")[$x^(2) y^(2) − y = 0$] #math.equation(block: true, alt: "open parenthesis 2 y plus 3 x close parenthesis squared equals 9")[$( 2 y + 3 x )^(2) = 9$] #math.equation(block: true, alt: "the fraction minus 3 x plus or minus 3 over 2")[$display(frac(− 3 x ± 3, 2))$] #math.equation(block: true, alt: "open parenthesis 3 y minus 2 x close parenthesis squared equals 4")[$( 3 y − 2 x )^(2) = 4$] #math.equation(block: true, alt: "4 x squared minus 9 y squared equals 36")[$4 x^(2) − 9 y^(2) = 36$] #math.equation(block: true, alt: "the fraction plus or minus the square root of 4 x squared minus 36 over 3 equals the fraction plus or minus 2 the square root of x squared minus 9 over 3")[$display(frac(± sqrt(4 x^(2) − 36), 3)) = display(frac(± 2 sqrt(x^(2) − 9), 3))$] #math.equation(block: true, alt: "9 x squared plus 4 y squared equals 36")[$9 x^(2) + 4 y^(2) = 36$] #math.equation(block: true, alt: "4 x squared minus 25 y squared equals 0")[$4 x^(2) − 25 y^(2) = 0$] #math.equation(block: true, alt: "the fraction plus or minus 2 x over 5")[$display(frac(± 2 x, 5))$] #math.equation(block: true, alt: "open parenthesis 2 x minus 5 y close parenthesis squared equals 0")[$( 2 x − 5 y )^(2) = 0$] For Problems 61-66, solve the formula for the indicated variable. #math.equation(block: false, alt: "V equals π open parenthesis r minus 3 close parenthesis squared h ,")[$V = π ( r − 3 )^(2) h , " " " "$] for #math.equation(block: false, alt: "r")[$r$] #math.equation(block: true, alt: "3 plus or minus the square root of the fraction V over π h")[$3 ± sqrt(display(frac(V, π h)))$] #math.equation(block: false, alt: "A equals P open parenthesis 1 plus r close parenthesis squared ,")[$A = P ( 1 + r )^(2) , " " " "$] for #math.equation(block: false, alt: "r")[$r$] #math.equation(block: false, alt: "E equals the fraction 1 over 2 m v squared plus m g h ,")[$E = display(frac(1, 2)) m v^(2) + m g h , " " " "$] for #math.equation(block: false, alt: "v")[$v$] #math.equation(block: true, alt: "plus or minus the square root of the fraction 2 open parenthesis E minus m g h close parenthesis over m")[$± sqrt(display(frac(2 ( E − m g h ), m)))$] #math.equation(block: false, alt: "h equals the fraction 1 over 2 g t squared plus d l ,")[$h = display(frac(1, 2)) g t^(2) + d l , " " " "$] for #math.equation(block: false, alt: "t")[$t$] #math.equation(block: false, alt: "V equals 2 open parenthesis s squared plus t squared close parenthesis w ,")[$V = 2 ( s^(2) + t^(2) ) w , " " " "$] for #math.equation(block: false, alt: "t")[$t$] #math.equation(block: true, alt: "plus or minus the square root of the fraction V over 2 w minus s squared")[$± sqrt(display(frac(V, 2 w)) − s^(2))$] #math.equation(block: false, alt: "V equals π open parenthesis r squared plus R squared close parenthesis h ,")[$V = π ( r^(2) + R^(2) ) h , " " " "$] for #math.equation(block: false, alt: "R")[$R$] What is the sum of the two solutions of the quadratic equation #math.equation(block: false, alt: "a x squared plus b x plus c equals 0")[$a x^(2) + b x + c = 0$]? The two solutions are given by the quadratic formula. #math.equation(block: true, alt: "the fraction minus b over a")[$display(frac(− b, a))$] What is the product of the two solutions of the quadratic equation #math.equation(block: false, alt: "a x squared plus b x plus c equals 0")[$a x^(2) + b x + c = 0$]? Do #emph[not] try to multiply the two solutions given by the quadratic formula! Think about the factored form of the equation In Problems 69 and 70, we prove the quadratic formula. Complete the square to find the solutions of the equation #math.equation(block: false, alt: "x squared plus b x plus c equals 0")[$x^(2) + b x + c = 0$]. (Your answers will be expressions in #math.equation(block: false, alt: "b")[$b$] and #math.equation(block: false, alt: "c")[$c$].) #math.equation(block: true, alt: "the fraction minus b plus or minus the square root of b squared minus 4 c over 2")[$display(frac(− b ± sqrt(b^(2) − 4 c), 2))$] Complete the square to find the solutions of the equation #math.equation(block: false, alt: "a x squared plus b x plus c equals 0")[$a x^(2) + b x + c = 0$]. (Your answers will be expressions in #math.equation(block: false, alt: "a")[$a$], #math.equation(block: false, alt: "b")[$b$], and #math.equation(block: false, alt: "c")[$c$].)