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== 9.2#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.

Expand: #math.equation(block: false, alt: "open parenthesis x plus 9 close parenthesis squared .")[$attach(( x + 9 ), t: 2) .$] #linebreak() 
If you missed this problem, review .

#solutionbox[
#math.equation(block: true, alt: "x squared plus 18 x plus 81")[$x^(2) + 18 x + 81$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Factor #math.equation(block: false, alt: "y squared minus 14 y plus 49 .")[$y^(2) − 14 y + 49 .$] #linebreak() 
If you missed this problem, review .

#solutionbox[
#math.equation(block: true, alt: "open parenthesis y minus 7 close parenthesis squared")[$( y − 7 )^(2)$]
]
]

#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 #strong[completing the square], which is important for our work on conics later.


=== Complete the Square of a Binomial Expression

In the last section, we were able to use the #strong[Square Root Property] to solve the equation (#emph[y] − 7)#super[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")[$attach(( y − 7 ), t: 2) & = & 12 \ y − 7 & = & ± sqrt(12) \ y − 7 & = & ± 2 sqrt(3) \ y & = & 7 ± 2 sqrt(3)$]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 look at two examples to help us recognize the patterns.

#math.equation(block: true, alt: "open parenthesis x plus 9 close parenthesis squared, open parenthesis y minus 7 close parenthesis squared; open parenthesis x plus 9 close parenthesis open parenthesis x plus 9 close parenthesis, open parenthesis y minus 7 close parenthesis open parenthesis y minus 7 close parenthesis; x squared plus 9 x plus 9 x plus 81, y squared minus 7 y minus 7 y plus 49; x squared plus 18 x plus 81, y squared minus 14 y plus 49")[$attach(( x + 9 ), t: 2) &  &  & #h(3em) attach(( y − 7 ), t: 2) \ ( x + 9 ) ( x + 9 ) &  &  & #h(3em) ( y − 7 ) ( y − 7 ) \ x^(2) + 9 x + 9 x + 81 &  &  & #h(3em) y^(2) − 7 y − 7 y + 49 \ x^(2) + 18 x + 81 &  &  & #h(3em) y^(2) − 14 y + 49$]We restate the patterns here for reference.

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Binomial Squares Pattern]

If #emph[a] and #emph[b] are real numbers,

#figure(figph[Quantity a plus b squared equals a squared plus 2 a b plus b2 where the binomial squared equals the first term squared plus 2 times the product of terms plus the second term squared. Quantity a minus b squared equals a squared minus 2 a b plus b2 where the binomial squared equals the first term squared minus 2 times the product of terms plus the second term squared.], alt: "Quantity a plus b squared equals a squared plus 2 a b plus b2 where the binomial squared equals the first term squared plus 2 times the product of terms plus the second term squared. Quantity a minus b squared equals a squared minus 2 a b plus b2 where the binomial squared equals the first term squared minus 2 times the product of terms plus the second term squared.", caption: none)
]

We can use this pattern to “make” a perfect square.

We will start with the expression #emph[x]#super[2] + 6#emph[x]. Since there is a plus sign between the two terms, we will use the (#emph[a] + #emph[b])#super[2] pattern, #emph[a]#super[2] + 2#emph[ab] + #emph[b]#super[2] = (#emph[a] + #emph[b])#super[2].

#figure(figph[The perfect square expression a squared plus 2 a b plus b squared is shown above the expression x squared plus 6x plus an unknown to allow a comparison of the corresponding terms of the expressions.], alt: "The perfect square expression a squared plus 2 a b plus b squared is shown above the expression x squared plus 6x plus an unknown to allow a comparison of the corresponding terms of the expressions.", caption: none)

We ultimately need to find the last term of this trinomial that will make it a perfect square trinomial. To do that we will need to find #emph[b]. But first we start with determining #emph[a]. Notice that the first term of #emph[x]#super[2] + 6#emph[x] is a square, #emph[x]#super[2]. This tells us that #emph[a] = #emph[x].

#figure(figph[The perfect square expression a squared plus 2 a b plus b squared is shown above the expression x squared plus 2 x b + b squared. Note that x has been substituted for a in the second equation and compare corresponding terms.], alt: "The perfect square expression a squared plus 2 a b plus b squared is shown above the expression x squared plus 2 x b + b squared. Note that x has been substituted for a in the second equation and compare corresponding terms.", caption: none)

What number, #emph[b,] when multiplied with 2#emph[x] gives 6#emph[x]? It would have to be 3, which is #math.equation(block: false, alt: "the fraction 1 over 2 open parenthesis 6 close parenthesis .")[$frac(1, 2) ( 6 ) .$] So #emph[b] = 3.

#figure(figph[The perfect square expression a squared plus 2 a b plus b squared is shown above the expression x squared plus 2 times 3 times x plus an unknown value to help compare terms.], alt: "The perfect square expression a squared plus 2 a b plus b squared is shown above the expression x squared plus 2 times 3 times x plus an unknown value to help compare terms.", caption: none)

Now to complete the perfect square trinomial, we will find the last term by squaring #emph[b], which is 3#super[2] = 9.

#figure(figph[The perfect square expression a squared plus 2 a b plus b squared is shown above the expression x squared plus 6 x plus 9.], alt: "The perfect square expression a squared plus 2 a b plus b squared is shown above the expression x squared plus 6 x plus 9.", caption: none)

We can now factor.

#figure(figph[The factored expression, the square of a plus b, is shown over the square of the expression x + 3.], alt: "The factored expression, the square of a plus b, is shown over the square of the expression x + 3.", caption: none)

So we found that adding 9 to #emph[x]#super[2] + 6#emph[x] ‘completes the square’, and we write it as (#emph[x] + 3)#super[2].

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Complete a square of #math.equation(block: false, alt: "x squared plus b x .")[$x^(2) + b x .$]]

+ Identify #emph[b], the coefficient of #emph[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 #emph[x]#super[2] + #emph[bx].
+ Factor the perfect square trinomial, writing it as a binomial squared.
]

#examplebox("Example 1")[][
Complete the square to make a perfect square trinomial. Then write the result as a binomial squared.

ⓐ #math.equation(block: false, alt: "x squared minus 26 x")[$x^(2) − 26 x$] ⓑ #math.equation(block: false, alt: "y squared minus 9 y")[$y^(2) − 9 y$] ⓒ #math.equation(block: false, alt: "n squared plus the fraction 1 over 2 n")[$n^(2) + frac(1, 2) n$]

#solutionbox[
ⓐ  #linebreak()

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#figure(figph[Two algebraic expressions are displayed: 'x² - bx' in red text, followed by 'x² - 26x' in black text directly below it.], alt: "Two algebraic expressions are displayed: 'x² - bx' in red text, followed by 'x² - 26x' in black text directly below it.", caption: none)]),
  [The coefficient of #math.equation(block: false, alt: "x")[$x$] is −26.], [],
  [#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.], [#figure(figph[The image shows the algebraic expression x^2 - 26x + 169, which is a quadratic trinomial. This expression is a perfect square trinomial, factoring to (x - 13)^2.], alt: "The image shows the algebraic expression x^2 - 26x + 169, which is a quadratic trinomial. This expression is a perfect square trinomial, factoring to (x - 13)^2.", caption: none)],
  [Factor the perfect square trinomial, writing it as #linebreak() a binomial squared.], [#figure(figph[A mathematical expression showing the quantity (x-13) raised to the power of 2, enclosed in parentheses.], alt: "A mathematical expression showing the quantity (x-13) raised to the power of 2, enclosed in parentheses.", caption: none)],
))

ⓑ  #linebreak()

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#figure(figph[Two algebraic expressions are shown: x^2 - bx in red, and y^2 - 9y in black, stacked vertically.], alt: "Two algebraic expressions are shown: x^2 - bx in red, and y^2 - 9y in black, stacked vertically.", caption: none)]),
  [The coefficient of #math.equation(block: false, alt: "y")[$y$] is #math.equation(block: false, alt: "−9")[$−9$].], [],
  [#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 − 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.], [#figure(figph[A mathematical expression shown in black text on a white background, which reads y squared minus 9y plus 81 over 4.], alt: "A mathematical expression shown in black text on a white background, which reads y squared minus 9y plus 81 over 4.", caption: none)],
  [Factor the perfect square trinomial, writing it as #linebreak() a binomial squared.], [#figure(figph[(y - 9/2)^2], alt: "(y - 9/2)^2", caption: none)],
))

ⓒ  #linebreak()

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#figure(figph[Two lines of algebraic expressions are shown on a white background. The top line reads 'x^2 + bx' in red text. The bottom line reads 'n^2 + (1/2)n' in black text.], alt: "Two lines of algebraic expressions are shown on a white background. The top line reads 'x^2 + bx' in red text. The bottom line reads 'n^2 + (1/2)n' in black text.", caption: none)]),
  [The coefficient of #math.equation(block: false, alt: "n")[$n$] is #math.equation(block: false, alt: "the fraction 1 over 2 .")[$frac(1, 2) .$]], [],
  [#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.], [#figure(figph[A mathematical expression displays n squared plus one-half n plus one-sixteenth, set against a plain white background.], alt: "A mathematical expression displays n squared plus one-half n plus one-sixteenth, set against a plain white background.", caption: none)],
  [Rewrite as a binomial square.], [#figure(figph[A mathematical expression showing the quantity 'n plus one-fourth' enclosed in parentheses, all raised to the power of two: (n + 1/4)^2.], alt: "A mathematical expression showing the quantity 'n plus one-fourth' enclosed in parentheses, all raised to the power of two: (n + 1/4)^2.", caption: none)],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Complete the square to make a perfect square trinomial. Then write the result as a binomial squared.

ⓐ #math.equation(block: false, alt: "a squared minus 20 a")[$a^(2) − 20 a$] ⓑ #math.equation(block: false, alt: "m squared minus 5 m")[$m^(2) − 5 m$] ⓒ #math.equation(block: false, alt: "p squared plus the fraction 1 over 4 p")[$p^(2) + frac(1, 4) p$]

#solutionbox[
ⓐ #math.equation(block: false, alt: "open parenthesis a minus 10 close parenthesis squared")[$attach(( a − 10 ), t: 2)$] ⓑ #math.equation(block: false, alt: "open parenthesis b minus the fraction 5 over 2 close parenthesis squared")[$attach(( b − frac(5, 2) ), t: 2)$] #linebreak() ⓒ #math.equation(block: false, 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. Then write the result as a binomial squared.

ⓐ #math.equation(block: false, alt: "b squared minus 4 b")[$b^(2) − 4 b$] ⓑ #math.equation(block: false, alt: "n squared plus 13 n")[$n^(2) + 13 n$] ⓒ #math.equation(block: false, alt: "q squared minus the fraction 2 over 3 q")[$q^(2) − frac(2, 3) q$]

#solutionbox[
ⓐ #math.equation(block: false, alt: "open parenthesis b minus 2 close parenthesis squared")[$attach(( b − 2 ), t: 2)$] ⓑ #math.equation(block: false, alt: "open parenthesis n plus the fraction 13 over 2 close parenthesis squared")[$attach(( n + frac(13, 2) ), t: 2)$] #linebreak() ⓒ #math.equation(block: false, 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 #strong[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 #emph[x]#super[2] + 6#emph[x] = 40, and we want to complete the square on the left, we will add 9 to both sides of the equation.

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#figure(figph[The image displays the quadratic equation x squared plus 6x equals 40.], alt: "The image displays the quadratic equation x squared plus 6x equals 40.", caption: none)]),
  [], [#figure(figph[An incomplete quadratic equation is shown, x squared plus 6x plus an empty blank space on the left side, which equals 40 plus another empty blank space on the right side.], alt: "An incomplete quadratic equation is shown, x squared plus 6x plus an empty blank space on the left side, which equals 40 plus another empty blank space on the right side.", caption: none)],
  [], [#figure(figph[A mathematical equation, x^2 + 6x + 9 = 40 + 9, with the number 9 highlighted in red, indicating it has been added to both sides to complete the square.], alt: "A mathematical equation, x^2 + 6x + 9 = 40 + 9, with the number 9 highlighted in red, indicating it has been added to both sides to complete the square.", caption: none)],
  [Add 9 to both sides to complete the square.], [#figure(figph[A mathematical equation is displayed: (x + 3)'  = 49. The equation involves an algebraic expression squared equal to a numerical value, representing a quadratic equation in a specific format.], alt: "A mathematical equation is displayed: (x + 3)' = 49. The equation involves an algebraic expression squared equal to a numerical value, representing a quadratic equation in a specific format.", caption: none)],
))

Now the equation is in the form to solve using the #strong[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 2")[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 by completing the square: #math.equation(block: false, alt: "x squared plus 8 x equals 48 .")[$x^(2) + 8 x = 48 .$]

#solutionbox[
#figure(figph[Step 1 is to isolate the variable terms on one side and the constant terms on the other. This equation, x squared plus 8 x equals 48 already has all variable terms on the left. Note that the leading coefficient is 1, so b equals 8.], alt: "Step 1 is to isolate the variable terms on one side and the constant terms on the other. This equation, x squared plus 8 x equals 48 already has all variable terms on the left. Note that the leading coefficient is 1, so b equals 8.", caption: none)

#figure(figph[In step 2, find the expression one half times b, squared, the number needed to complete the square. Add this value to both sides of the equation. Take half of 8 and square it. The square of one half times 8 equals 16, so add 16 to BOTH sides of the equation. The equation becomes x squared plus 8 x plus 16 equals 48 plus 16.], alt: "In step 2, find the expression one half times b, squared, the number needed to complete the square. Add this value to both sides of the equation. Take half of 8 and square it. The square of one half times 8 equals 16, so add 16 to BOTH sides of the equation. The equation becomes x squared plus 8 x plus 16 equals 48 plus 16.", caption: none)

#figure(figph[In step 3, factor the perfect square trinomial, writing it as a binomial squared on the left and simplify by adding the terms on the right. Factor x squared plus 8 x plus 16 on the left side. Add 48+16 on the right side. The equation becomes the square of x plus 4 equals 64.], alt: "In step 3, factor the perfect square trinomial, writing it as a binomial squared on the left and simplify by adding the terms on the right. Factor x squared plus 8 x plus 16 on the left side. Add 48+16 on the right side. The equation becomes the square of x plus 4 equals 64.", caption: none)

#figure(figph[Step 4 is to use the Square Root Property. Take the square root of both sides of the equation to yield x plus 4 equals the positive or negative square root of 64.], alt: "Step 4 is to use the Square Root Property. Take the square root of both sides of the equation to yield x plus 4 equals the positive or negative square root of 64.", caption: none)

#figure(figph[In step 5, simplify the radical and then solve the two resulting equations. X plus 4 equals positive 8 or negative 8. If x plus 4 equals 8, then x equals 4. If x plus 4 equals negative 8, then x equals negative 12.], alt: "In step 5, simplify the radical and then solve the two resulting equations. X plus 4 equals positive 8 or negative 8. If x plus 4 equals 8, then x equals 4. If x plus 4 equals negative 8, then x equals negative 12.", caption: none)

#figure(figph[Finally, step 6, check the solutions. Put each answer in the original equation to check. First substitute x equals 4. We need to show that 4 squared plus 8 times 4 equals 48. Simplify. The expression 4 squared plus 8 times 4 is equivalent to 16 plus 32, or 48. X equals 4 is a solution. Next substitute x equals negative 12 into the original equation, x squared plus 8 x equals 48. The square of negative 12 plus 8 times negative 12 equals 144 minus 96, or 48. X equals negative 12 is also a solution.], alt: "Finally, step 6, check the solutions. Put each answer in the original equation to check. First substitute x equals 4. We need to show that 4 squared plus 8 times 4 equals 48. Simplify. The expression 4 squared plus 8 times 4 is equivalent to 16 plus 32, or 48. X equals 4 is a solution. Next substitute x equals negative 12 into the original equation, x squared plus 8 x equals 48. The square of negative 12 plus 8 times negative 12 equals 144 minus 96, or 48. X equals negative 12 is also a solution.", caption: none)
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve by completing the square: #math.equation(block: false, alt: "x squared plus 4 x equals 5 .")[$x^(2) + 4 x = 5 .$]

#solutionbox[
#math.equation(block: true, alt: "x equals −5 , x equals 1")[$x = −5 , x = 1$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve by completing the square: #math.equation(block: false, alt: "y squared minus 10 y equals −9 .")[$y^(2) − 10 y = −9 .$]

#solutionbox[
#math.equation(block: true, alt: "y equals 1 , y equals 9")[$y = 1 , y = 9$]
]
]

The steps to solve a quadratic equation by completing the square are listed here.

#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 ,")[$attach(( frac(1, 2) #h(0.2em) · #h(0.2em) b ), t: 2) ,$] the number needed to complete the square. Add it to both sides of the equation.
+ Factor the perfect square trinomial, writing it as a binomial squared on the left and simplify by adding the terms on the right
+ Use the Square Root Property.
+ Simplify the radical and then solve the two resulting equations.
+ Check the solutions.
]

When we solve an equation by completing the square, the answers will not always be integers.

#examplebox("Example 3")[][
Solve by completing the square: #math.equation(block: false, alt: "x squared plus 4 x equals −21 .")[$x^(2) + 4 x = −21 .$]

#solutionbox[
#figure(table(
  columns: 3, align: left, inset: 6pt,
  [], [], [#figure(figph[Two math expressions are shown: the general quadratic form x² + bx + c in red, and a specific quadratic equation x² + 4x = -21 in black on a white background.], alt: "Two math expressions are shown: the general quadratic form x² + bx + c in red, and a specific quadratic equation x² + 4x = -21 in black on a white background.", caption: none)],
  [The variable terms are on the left side. #linebreak() 
Take half of 4 and square it.], [#figure(figph[An algebraic equation showing the initial step of completing the square: x^2 + 4x + \[blank line with (1/2 \* 4)^2 in red below it\] = -21.], alt: "An algebraic equation showing the initial step of completing the square: x^2 + 4x + [blank line with (1/2 * 4)^2 in red below it] = -21.", caption: none)], [],
  [#math.equation(block: false, alt: "open parenthesis the fraction 1 over 2 open parenthesis 4 close parenthesis close parenthesis squared equals 4")[$attach(( frac(1, 2) ( 4 ) ), t: 2) = 4$]], [], [],
  [Add 4 to both sides.], [#figure(figph[A mathematical equation showing the step of adding 4 to both sides: x^2 + 4x + 4 = -21 + 4, as part of completing the square to solve for x.], alt: "A mathematical equation showing the step of adding 4 to both sides: x^2 + 4x + 4 = -21 + 4, as part of completing the square to solve for x.", caption: none)], [],
  [Factor the perfect square trinomial, #linebreak() writing it as a binomial squared.], [#figure(figph[The image displays the quadratic equation (x + 2)^2 = -17, which has no real solutions.], alt: "The image displays the quadratic equation (x + 2)^2 = -17, which has no real solutions.", caption: none)], [],
  [Use the Square Root Property.], [#figure(figph[A mathematical equation is displayed on a white background: x + 2 = ±sqrt(-17). This equation involves a variable 'x', constants, an equality sign, and the plus-minus square root of a negative number, indicating complex solutions.], alt: "A mathematical equation is displayed on a white background: x + 2 = ±sqrt(-17). This equation involves a variable 'x', constants, an equality sign, and the plus-minus square root of a negative number, indicating complex solutions.", caption: none)], [],
  [Simplify using complex numbers.], [#figure(figph[Equation showing x plus 2 equals positive or negative square root of 17 times i. This represents complex solutions for x.], alt: "Equation showing x plus 2 equals positive or negative square root of 17 times i. This represents complex solutions for x.", caption: none)], [],
  [Subtract 2 from each side.], [#figure(figph[The image displays a mathematical equation for x: x = -2 ± sqrt(17)i, showing a solution with both real and imaginary components.], alt: "The image displays a mathematical equation for x: x = -2 ± sqrt(17)i, showing a solution with both real and imaginary components.", caption: none)], [],
  [Rewrite to show two solutions.], [#figure(figph[Two complex conjugate roots are displayed: x = -2 + sqrt(17)i and x = -2 - sqrt(17)i.], alt: "Two complex conjugate roots are displayed: x = -2 + sqrt(17)i and x = -2 - sqrt(17)i.", caption: none)], [],
  [We leave the check to you.], [], [],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve by completing the square: #math.equation(block: false, alt: "y squared minus 10 y equals −35 .")[$y^(2) − 10 y = −35 .$]

#solutionbox[
#math.equation(block: true, alt: "y equals 5 plus or minus the square root of 10 i")[$y = 5 ± sqrt(10) i$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve by completing the square: #math.equation(block: false, alt: "z squared plus 8 z equals −19 .")[$z^(2) + 8 z = −19 .$]

#solutionbox[
#math.equation(block: true, alt: "z equals −4 plus the square root of 3 i , z equals −4 minus the square root of 3 i")[$z = −4 + sqrt(3) i , #h(0.2em) "z" = −4 − sqrt(3) i$]
]
]

In the previous example, our solutions were complex numbers. In the next example, the solutions will be irrational numbers.

#examplebox("Example 4")[][
Solve by completing the square: #math.equation(block: false, alt: "y squared minus 18 y equals −6 .")[$y^(2) − 18 y = −6 .$]

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  [], [#figure(figph[Two algebraic expressions are shown: one on the top reads 'x^2 - bx c' with 'c' appearing slightly separate, and the other below it reads 'y^2 - 18y = -6'.], alt: "Two algebraic expressions are shown: one on the top reads 'x^2 - bx c' with 'c' appearing slightly separate, and the other below it reads 'y^2 - 18y = -6'.", caption: none)],
  [The variable terms are on the left side. #linebreak() 
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")[$attach(( frac(1, 2) ( −18 ) ), t: 2) = 81$]], [#figure(figph[A mathematical equation y^2 - 18y + blank = -6 is shown, with a red hint below the blank indicating (1/2 \* -18)^2, illustrating the process of completing the square to solve a quadratic equation.], alt: "A mathematical equation y^2 - 18y + blank = -6 is shown, with a red hint below the blank indicating (1/2 * -18)^2, illustrating the process of completing the square to solve a quadratic equation.", caption: none)],
  [Add 81 to both sides.], [#figure(figph[An equation: y^2 - 18y + 81 = -6 + 81. The value 81 is highlighted in red, indicating its addition to both sides, a common step in solving quadratic equations by completing the square.], alt: "An equation: y^2 - 18y + 81 = -6 + 81. The value 81 is highlighted in red, indicating its addition to both sides, a common step in solving quadratic equations by completing the square.", caption: none)],
  [Factor the perfect square trinomial, #linebreak() writing it as a binomial squared.], [#figure(figph[A mathematical equation is displayed, showing (y-9)^2 = 75. The equation involves a variable y, subtraction, exponentiation, and equality to a constant.], alt: "A mathematical equation is displayed, showing (y-9)^2 = 75. The equation involves a variable y, subtraction, exponentiation, and equality to a constant.", caption: none)],
  [Use the Square Root Property.], [#figure(figph[A mathematical equation is displayed, showing y - 9 = +/- sqrt(75).], alt: "A mathematical equation is displayed, showing y - 9 = +/- sqrt(75).", caption: none)],
  [Simplify the radical.], [#figure(figph[A mathematical equation shows 'y minus 9 equals plus or minus 5 times the square root of 3'.], alt: "A mathematical equation shows 'y minus 9 equals plus or minus 5 times the square root of 3'.", caption: none)],
  [Solve for #math.equation(block: false, alt: "y")[$y$].], [#figure(figph[A mathematical expression showing y equals 9 plus or minus 5 times the square root of 3.], alt: "A mathematical expression showing y equals 9 plus or minus 5 times the square root of 3.", caption: none)],
  [Check. #linebreak() 
#figure(figph[Verifying solutions for the quadratic equation y² - 18y = -6 by substituting y with (9 + 5√3) and (9 - 5√3), both yielding -6 = -6.], alt: "Verifying solutions for the quadratic equation y² - 18y = -6 by substituting y with (9 + 5√3) and (9 - 5√3), both yielding -6 = -6.", caption: none)], [],
))

Another way to check this would be to use a calculator. Evaluate #math.equation(block: false, alt: "y squared minus 18 y")[$y^(2) − 18 y$]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 by completing the square: #math.equation(block: false, alt: "x squared minus 16 x equals −16 .")[$x^(2) − 16 x = −16 .$]

#solutionbox[
#math.equation(block: true, alt: "x equals 8 plus 4 the square root of 3 , x equals 8 minus 4 the square root of 3")[$x = 8 + 4 sqrt(3) , #h(0.2em) #h(0.2em) x = 8 − 4 sqrt(3)$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve by completing the square: #math.equation(block: false, alt: "y squared plus 8 y equals 11 .")[$y^(2) + 8 y = 11 .$]

#solutionbox[
#math.equation(block: true, alt: "y equals −4 plus 3 the square root of 3 , y equals −4 minus 3 the square root of 3")[$y = −4 + 3 sqrt(3) , #h(0.2em) "y" = −4 − 3 sqrt(3)$]
]
]

We will start the next example by isolating the variable terms on the left side of the equation.

#examplebox("Example 5")[][
Solve by completing the square: #math.equation(block: false, alt: "x squared plus 10 x plus 4 equals 15 .")[$x^(2) + 10 x + 4 = 15 .$]

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  [], [#figure(figph[A mathematical equation is displayed on a white background. The equation reads: x squared plus 10x plus 4 equals 15.], alt: "A mathematical equation is displayed on a white background. The equation reads: x squared plus 10x plus 4 equals 15.", caption: none)],
  [Isolate the variable terms on the left side. #linebreak() Subtract 4 to get the constant terms on the right side.], [#figure(figph[A quadratic equation displayed against a plain white background, reading 'x^2 + 10x = 11'.], alt: "A quadratic equation displayed against a plain white background, reading '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")[$attach(( frac(1, 2) ( 10 ) ), t: 2) = 25$]], [#figure(figph[A mathematical equation illustrating a step in completing the square: x^2 - 10x + blank = 11, where the blank is represented by a fraction bar with an empty numerator and a denominator of (1/2 \* 10)^2 in red.], alt: "A mathematical equation illustrating a step in completing the square: x^2 - 10x + blank = 11, where the blank is represented by a fraction bar with an empty numerator and a denominator of (1/2 * 10)^2 in red.", caption: none)],
  [Add 25 to both sides.], [#figure(figph[A quadratic equation shown as x squared plus 10x plus 25 equals 11 plus 25, with the number 25 highlighted in red on both sides of the equation.], alt: "A quadratic equation shown as x squared plus 10x plus 25 equals 11 plus 25, with the number 25 highlighted in red on both sides of the equation.", caption: none)],
  [Factor the perfect square trinomial, writing it as #linebreak() a binomial squared.], [#figure(figph[A mathematical equation shows '(x + 5)^2 = 36' centered on a white background.], alt: "A mathematical equation shows '(x + 5)^2 = 36' centered on a white background.", caption: none)],
  [Use the Square Root Property.], [#figure(figph[A mathematical equation is displayed, reading 'x + 5 = +/- square root 36', showing a step in solving for x in an algebraic problem involving a square root.], alt: "A mathematical equation is displayed, reading 'x + 5 = +/- square root 36', showing a step in solving for x in an algebraic problem involving a square root.", caption: none)],
  [Simplify the radical.], [#figure(figph[A mathematical equation is displayed, showing 'x + 5 =  6' with a plus-minus symbol before the 6.], alt: "A mathematical equation is displayed, showing 'x + 5 = 6' with a plus-minus symbol before the 6.", caption: none)],
  [Solve for #emph[x].], [#figure(figph[A mathematical equation is displayed on a white background: x = -5 ×1 6. This represents two possible values for x: x = -5 + 6 and x = -5 - 6.], alt: "A mathematical equation is displayed on a white background: x = -5 ×1 6. This represents two possible values for x: x = -5 + 6 and x = -5 - 6.", caption: none)],
  [Rewrite to show two solutions.], [#figure(figph[The image displays two algebraic expressions for the variable x: x = -5 + 6 and x = -5 - 6, indicating two different calculations.], alt: "The image displays two algebraic expressions for the variable x: x = -5 + 6 and x = -5 - 6, indicating two different calculations.", caption: none)],
  [Solve the equations.], [#figure(figph[Two mathematical equations are displayed: x=1 and x=-11, set against a plain white background.], alt: "Two mathematical equations are displayed: x=1 and x=-11, set against a plain white background.", caption: none)],
  [Check: #linebreak()  #linebreak() 
#figure(figph[Verification of the quadratic equation x^2 + 10x + 4 = 15 by substituting x=1 and x=-11, showing that both values yield 15=15 and are therefore correct solutions.], alt: "Verification of the quadratic equation x^2 + 10x + 4 = 15 by substituting x=1 and x=-11, showing that both values yield 15=15 and are therefore correct solutions.", caption: none)], [],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve by completing the square: #math.equation(block: false, alt: "a squared plus 4 a plus 9 equals 30 .")[$a^(2) + 4 a + 9 = 30 .$]

#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 by completing the square: #math.equation(block: false, alt: "b squared plus 8 b minus 4 equals 16 .")[$b^(2) + 8 b − 4 = 16 .$]

#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 on the left side of the equation. Then we proceed as we did in the previous examples.

#examplebox("Example 6")[][
Solve by completing the square: #math.equation(block: false, alt: "n squared equals 3 n plus 11 .")[$n^(2) = 3 n + 11 .$]

#solutionbox[
#figure(table(
  columns: 3, align: left, inset: 6pt,
  [], [], [#figure(figph[A mathematical equation is displayed on a white background: n squared equals three n plus eleven (n^2 = 3n + 11).], alt: "A mathematical equation is displayed on a white background: n squared equals three n plus eleven (n^2 = 3n + 11).", caption: none)],
  [Subtract #math.equation(block: false, alt: "3 n")[$3 n$] to get the variable terms on the left side.], [#figure(figph[A mathematical equation is displayed on a white background, reading 'n squared minus 3n equals 11'.], alt: "A mathematical equation is displayed on a white background, reading 'n squared minus 3n equals 11'.", 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")[$attach(( frac(1, 2) ( −3 ) ), t: 2) = frac(9, 4)$]], [#figure(figph[A mathematical equation illustrating the process of completing the square: n^2 - 3n + (empty numerator) / ((1/2)\*(-3))^2 = 11. The term to be added to complete the square is represented by the fraction.], alt: "A mathematical equation illustrating the process of completing the square: n^2 - 3n + (empty numerator) / ((1/2)*(-3))^2 = 11. The term to be added to complete the square is represented by the fraction.", caption: none)], [],
  [Add #math.equation(block: false, alt: "the fraction 9 over 4")[$frac(9, 4)$] to both sides.], [#figure(figph[An algebraic equation is displayed: n squared minus 3n plus 9 over 4 equals 11 plus 9 over 4. This likely represents a step in completing the square for a quadratic equation.], alt: "An algebraic equation is displayed: n squared minus 3n plus 9 over 4 equals 11 plus 9 over 4. This likely represents a step in completing the square for a quadratic equation.", caption: none)], [],
  [Factor the perfect square trinomial, writing it as #linebreak() a binomial squared.], [#figure(figph[A mathematical equation shows (n - 3/2)^2 = 44/4 + 9/4. The expression on the left is a binomial squared with variable 'n' and constant 3/2. The right side is a sum of two fractions, both with a denominator of 4.], alt: "A mathematical equation shows (n - 3/2)^2 = 44/4 + 9/4. The expression on the left is a binomial squared with variable 'n' and constant 3/2. The right side is a sum of two fractions, both with a denominator of 4.", caption: none)], [],
  [Add the fractions on the right side.], [#figure(figph[A mathematical equation shown on a white background, which reads '(n - 3/2)^2 = 53/4'], alt: "A mathematical equation shown on a white background, which reads '(n - 3/2)^2 = 53/4'", caption: none)], [],
  [Use the Square Root Property.], [#figure(figph[A mathematical equation is displayed, showing 'n - 3/2 = ±√(53/4)' in black text on a white background.], alt: "A mathematical equation is displayed, showing 'n - 3/2 = ±√(53/4)' in black text on a white background.", caption: none)], [],
  [Simplify the radical.], [#figure(figph[A mathematical equation shows 'n minus 3 over 2 equals plus or minus the square root of 53 over 2'.], alt: "A mathematical equation shows 'n minus 3 over 2 equals plus or minus the square root of 53 over 2'.", caption: none)], [],
  [Solve for #emph[n].], [#figure(figph[A mathematical equation displays 'n = 3/2 plus or minus the square root of 53, all divided by 2' in black text on a white background.], alt: "A mathematical equation displays 'n = 3/2 plus or minus the square root of 53, all divided by 2' in black text on a white background.", caption: none)], [],
  [Rewrite to show two solutions.], [#figure(figph[Two mathematical expressions for 'n' are shown: n = 3/2 + sqrt(53)/2, and n = 3/2 - sqrt(53)/2.], alt: "Two mathematical expressions for 'n' are shown: n = 3/2 + sqrt(53)/2, and n = 3/2 - sqrt(53)/2.", caption: none)], [],
  [Check: #linebreak() 
We leave the check for you!], [], [],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve by completing the square: #math.equation(block: false, alt: "p squared equals 5 p plus 9 .")[$p^(2) = 5 p + 9 .$]

#solutionbox[
#math.equation(block: true, alt: "p equals the fraction 5 over 2 plus the fraction the square root of 61 over 2 , p equals the fraction 5 over 2 minus the fraction the square root of 61 over 2")[$p = frac(5, 2) + frac(sqrt(61), 2) , #h(0.2em) #h(0.2em) p = frac(5, 2) − frac(sqrt(61), 2)$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve by completing the square: #math.equation(block: false, alt: "q squared equals 7 q minus 3 .")[$q^(2) = 7 q − 3 .$]

#solutionbox[
#math.equation(block: true, alt: "q equals the fraction 7 over 2 plus the fraction the square root of 37 over 2 , q equals the fraction 7 over 2 minus the fraction the square root of 37 over 2")[$q = frac(7, 2) + frac(sqrt(37), 2) , #h(0.2em) "q" = frac(7, 2) − frac(sqrt(37), 2)$]
]
]

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 #strong[Zero Product Property] since it says “If #math.equation(block: false, alt: "a times b equals 0 ,")[$a #h(0.2em) · #h(0.2em) b = 0 ,$] then #emph[a] = 0 or #emph[b] = 0.” Instead, we multiply the factors and then put the equation into standard form to solve by completing the square.

#examplebox("Example 7")[][
Solve by completing the square: #math.equation(block: false, alt: "open parenthesis x minus 3 close parenthesis open parenthesis x plus 5 close parenthesis equals 9 .")[$( x − 3 ) ( x + 5 ) = 9 .$]

#solutionbox[
#figure(table(
  columns: 4, align: left, inset: 6pt,
  [], [], [], [#figure(figph[A mathematical equation is displayed against a white background: (x - 3)(x + 5) = 9. The equation involves a variable 'x' in two binomials that are multiplied together and set equal to 9.], alt: "A mathematical equation is displayed against a white background: (x - 3)(x + 5) = 9. The equation involves a variable 'x' in two binomials that are multiplied together and set equal to 9.", caption: none)],
  [We multiply the binomials on the left.], [], [#figure(figph[A quadratic equation is displayed on a white background, reading 'x^2 + 2x - 15 = 9'.], alt: "A quadratic equation is displayed on a white background, reading 'x^2 + 2x - 15 = 9'.", caption: none)], [],
  [Add 15 to isolate the constant terms on the right.], [], [#figure(figph[A mathematical equation is displayed on a white background, reading 'x^2 + 2x = 24'. The equation features a quadratic term, a linear term, and a constant.], alt: "A mathematical equation is displayed on a white background, reading 'x^2 + 2x = 24'. The equation features a quadratic term, a linear term, and a constant.", caption: none)], [],
  [Take half of 2 and square it.], [], [], [],
  [#math.equation(block: false, alt: "open parenthesis the fraction 1 over 2 times open parenthesis 2 close parenthesis close parenthesis squared equals 1")[$attach(( frac(1, 2) · ( 2 ) ), t: 2) = 1$]], [], [#figure(figph[An algebraic equation showing a step to complete the square. The expression is x^2 + 2x + a fractional term with a blank numerator and (1/2 \* 2)^2 in the denominator, equaling 24.], alt: "An algebraic equation showing a step to complete the square. The expression is x^2 + 2x + a fractional term with a blank numerator and (1/2 * 2)^2 in the denominator, equaling 24.", caption: none)], [],
  [Add 1 to both sides.], [], [#figure(figph[The equation x^2 + 2x + 1 = 24 + 1, with the added '1' on each side emphasized in red.], alt: "The equation x^2 + 2x + 1 = 24 + 1, with the added '1' on each side emphasized in red.", caption: none)], [],
  [Factor the perfect square trinomial, writing it as #linebreak() a binomial squared.], [], [#figure(figph[A mathematical equation is displayed on a white background, reading '(x+1)^2 = 25' in black text.], alt: "A mathematical equation is displayed on a white background, reading '(x+1)^2 = 25' in black text.", caption: none)], [],
  [Use the Square Root Property.], [], [#figure(figph[A mathematical equation on a white background reads 'x + 1 = ±√25'.], alt: "A mathematical equation on a white background reads 'x + 1 = ±√25'.", caption: none)], [],
  [Solve for #emph[x].], [], [#figure(figph[A close-up shot of a math equation on a white background, which reads 'x = -1 ± 5'.], alt: "A close-up shot of a math equation on a white background, which reads 'x = -1 ± 5'.", caption: none)], [],
  [Rewrite to show two solutions.], [#figure(figph[Two mathematical equations are displayed on a white background: 'x = -1 + 5' and 'x = -1 - 5'. These represent two possible solutions for the variable 'x'.], alt: "Two mathematical equations are displayed on a white background: 'x = -1 + 5' and 'x = -1 - 5'. These represent two possible solutions for the variable 'x'.", caption: none)], [], [],
  [Simplify.], [#figure(figph[Two mathematical expressions are displayed on a white background: 'x = 4,' and 'x = -6'.], alt: "Two mathematical expressions are displayed on a white background: 'x = 4,' and 'x = -6'.", caption: none)], [], [],
  [Check: #linebreak() 
We leave the check for you!], [], [], [],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve by completing the square: #math.equation(block: false, alt: "open parenthesis c minus 2 close parenthesis open parenthesis c plus 8 close parenthesis equals 11 .")[$( c − 2 ) ( c + 8 ) = 11 .$]

#solutionbox[
#math.equation(block: true, alt: "c equals −9 , c equals 3")[$c = −9 , c = 3$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve by completing the square: #math.equation(block: false, alt: "open parenthesis d minus 7 close parenthesis open parenthesis d plus 3 close parenthesis equals 56 .")[$( d − 7 ) ( d + 3 ) = 56 .$]

#solutionbox[
#math.equation(block: true, alt: "d equals 11 , d equals −7")[$d = 11 , d = −7$]
]
]


=== Solve Quadratic Equations of the Form #emph[ax]#super[2] + #emph[bx] + #emph[c] = 0 by Completing the Square

The process of #strong[completing the square] works best when the coefficient of #emph[x]#super[2] is 1, so the left side of the equation is of the form #emph[x]#super[2] + #emph[bx] + #emph[c]. If the #emph[x]#super[2] term has a coefficient other than 1, we take some preliminary steps to make the coefficient equal to 1.

Sometimes the coefficient can be factored from all three terms of the trinomial. This will be our strategy in the next example.

#examplebox("Example 8")[][
Solve by completing the square: #math.equation(block: false, alt: "3 x squared minus 12 x minus 15 equals 0 .")[$3 x^(2) − 12 x − 15 = 0 .$]

#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[A mathematical equation reads 3x^2 - 12x - 15 = 0.], alt: "A mathematical equation reads 3x^2 - 12x - 15 = 0.", caption: none)]),
  [Factor out the greatest common factor.], [#figure(figph[A mathematical equation is displayed with the expression 3(x^2 - 4x - 5) = 0, indicating a quadratic equation to be solved.], alt: "A mathematical equation is displayed with the expression 3(x^2 - 4x - 5) = 0, indicating a quadratic equation to be solved.", caption: none)],
  [Divide both sides by 3 to isolate the trinomial #linebreak() with coefficient 1.], [#figure(figph[A mathematical equation shown as 3(x^2 - 4x - 5) / 3 = 0 / 3, demonstrating division by 3 on both sides of the equation.], alt: "A mathematical equation shown as 3(x^2 - 4x - 5) / 3 = 0 / 3, demonstrating division by 3 on both sides of the equation.", caption: none)],
  [Simplify.], [#figure(figph[A quadratic equation, x^2 - 4x - 5 = 0, is displayed in a grayscale image. The equation is centered against a white background, representing a mathematical problem to be solved.], alt: "A quadratic equation, x^2 - 4x - 5 = 0, is displayed in a grayscale image. The equation is centered against a white background, representing a mathematical problem to be solved.", caption: none)],
  [Add 5 to get the constant terms on the right side.], [#figure(figph[A mathematical equation is displayed, showing 'x squared minus 4x equals 5' in a white background.], alt: "A mathematical equation is displayed, showing 'x squared minus 4x equals 5' in a white background.", 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")[$attach(( frac(1, 2) ( −4 ) ), t: 2) = 4$]], [#figure(figph[An algebraic equation showing x^2 - 4x plus a blank numerator over a red (1/2 \* 4)^2 term, set equal to 5. This illustrates a step in completing the square for a quadratic expression.], alt: "An algebraic equation showing x^2 - 4x plus a blank numerator over a red (1/2 * 4)^2 term, set equal to 5. This illustrates a step in completing the square for a quadratic expression.", caption: none)],
  [Add 4 to both sides.], [#figure(figph[A step in solving x^2 - 4x = 5 by adding 4 to both sides to complete the square, resulting in x^2 - 4x + 4 = 5 + 4.], alt: "A step in solving x^2 - 4x = 5 by adding 4 to both sides to complete the square, resulting in x^2 - 4x + 4 = 5 + 4.", caption: none)],
  [Factor the perfect square trinomial, writing it #linebreak() as a binomial squared.], [#figure(figph[The image displays the algebraic equation (x-2)^2 = 9, centered on a white background.], alt: "The image displays the algebraic equation (x-2)^2 = 9, centered on a white background.", caption: none)],
  [Use the Square Root Property.], [#figure(figph[A mathematical equation is displayed on a white background: x - 2 = plus or minus the square root of 9. This equation involves a variable, arithmetic operations, and a radical expression.], alt: "A mathematical equation is displayed on a white background: x - 2 = plus or minus the square root of 9. This equation involves a variable, arithmetic operations, and a radical expression.", caption: none)],
  [Solve for #emph[x].], [#figure(figph[A mathematical equation 'x - 2 equals plus or minus 3'], alt: "A mathematical equation 'x - 2 equals plus or minus 3'", caption: none)],
  [Rewrite to show two solutions.], [#figure(figph[The image displays two simple mathematical equations: x = 2 + 3 and x = 2 - 3, separated by a comma. These equations show two possible values for x, one from addition and one from subtraction.], alt: "The image displays two simple mathematical equations: x = 2 + 3 and x = 2 - 3, separated by a comma. These equations show two possible values for x, one from addition and one from subtraction.", caption: none)],
  [Simplify.], [#figure(figph[Two mathematical equations are displayed against a white background: 'x = 5,' and 'x = -1'.], alt: "Two mathematical equations are displayed against a white background: 'x = 5,' and 'x = -1'.", caption: none)],
  [Check: #linebreak()  #linebreak() 
#figure(figph[Checking the solutions x=5 and x=-1 for the quadratic equation 3x² - 12x - 15 = 0. Both substitutions correctly result in 0=0, confirming they are valid roots.], alt: "Checking the solutions x=5 and x=-1 for the quadratic equation 3x² - 12x - 15 = 0. Both substitutions correctly result in 0=0, confirming they are valid roots.", caption: none)], [],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve by completing the square: #math.equation(block: false, alt: "2 m squared plus 16 m plus 14 equals 0 .")[$2 m^(2) + 16 m + 14 = 0 .$]

#solutionbox[
#math.equation(block: true, alt: "m equals −7 , m equals −1")[$m = −7 , m = −1$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve by completing the square: #math.equation(block: false, alt: "4 n squared minus 24 n minus 56 equals 8 .")[$4 n^(2) − 24 n − 56 = 8 .$]

#solutionbox[
#math.equation(block: true, alt: "n equals −2 , n equals 8")[$n = −2 , n = 8$]
]
]

To complete the square, the coefficient of the #emph[x]#super[2] must be 1. When the #strong[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 9")[][
Solve by completing the square: #math.equation(block: false, alt: "2 x squared minus 3 x equals 20 .")[$2 x^(2) − 3 x = 20 .$]

#solutionbox[
To complete the square we need the coefficient of #math.equation(block: false, alt: "x squared")[$x^(2)$] to be one. We will divide both sides of the equation by the coefficient of #emph[x]#super[2]. Then we can continue with solving the equation by completing the square.

#figure(table(
  columns: 3, align: left, inset: 6pt,
  [], [], [#figure(figph[A mathematical equation is displayed, reading '2x^2 - 3x = 20' in black text against a white background.], alt: "A mathematical equation is displayed, reading '2x^2 - 3x = 20' in black text against a white background.", caption: none)],
  [Divide both sides by 2 to get the #linebreak() coefficient of #math.equation(block: false, alt: "x squared")[$x^(2)$] to be 1.], [#figure(figph[A mathematical equation is shown with the expression (2x^2 - 3x) / 2 = 20 / 2, where both sides of the equation are divided by 2.], alt: "A mathematical equation is shown with the expression (2x^2 - 3x) / 2 = 20 / 2, where both sides of the equation are divided by 2.", caption: none)], [],
  [Simplify.], [#figure(figph[A mathematical equation is displayed, showing 'x squared minus three halves x equals ten' in black text on a white background.], alt: "A mathematical equation is displayed, showing 'x squared minus three halves x equals ten' in black text on a white background.", 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")[$attach(( frac(1, 2) ( − frac(3, 2) ) ), t: 2) = frac(9, 16)$]], [#figure(figph[A mathematical equation demonstrating the process of completing the square. It shows x squared minus three-halves x, plus a blank term, which is indicated by one-half times negative three-halves, all squared, equals 10.], alt: "A mathematical equation demonstrating the process of completing the square. It shows x squared minus three-halves x, plus a blank term, which is indicated by one-half times negative three-halves, all squared, equals 10.", caption: none)], [],
  [Add #math.equation(block: false, alt: "the fraction 9 over 16")[$frac(9, 16)$] to both sides.], [#figure(figph[A mathematical equation showing x squared minus three-halves x plus nine-sixteenths equals ten plus nine-sixteenths, with the fractions and plus signs in red.], alt: "A mathematical equation showing x squared minus three-halves x plus nine-sixteenths equals ten plus nine-sixteenths, with the fractions and plus signs in red.", caption: none)], [],
  [Factor the perfect square trinomial, #linebreak() writing it as a binomial squared.], [#figure(figph[A mathematical equation shows '(x - 3/4)^2 = 160/16 + 9/16' written in black characters on a white background.], alt: "A mathematical equation shows '(x - 3/4)^2 = 160/16 + 9/16' written in black characters on a white background.", caption: none)], [],
  [Add the fractions on the right side.], [#figure(figph[A mathematical equation shows (x - 3/4)^2 = 169/16.], alt: "A mathematical equation shows (x - 3/4)^2 = 169/16.", caption: none)], [],
  [Use the Square Root Property.], [#figure(figph[A mathematical equation is displayed, showing x minus three-fourths equals plus or minus the square root of 169 over 16.], alt: "A mathematical equation is displayed, showing x minus three-fourths equals plus or minus the square root of 169 over 16.", caption: none)], [],
  [Simplify the radical.], [#figure(figph[A mathematical equation shows 'x minus 3/4 equals plus or minus 13/4' on a white background. This equation represents a step in solving for x, likely from a quadratic equation.], alt: "A mathematical equation shows 'x minus 3/4 equals plus or minus 13/4' on a white background. This equation represents a step in solving for x, likely from a quadratic equation.", caption: none)], [],
  [Solve for #emph[x].], [#figure(figph[A mathematical equation shows x equals three-fourths plus or minus thirteen-fourths, represented as x = 3/4 ×1 13/4.], alt: "A mathematical equation shows x equals three-fourths plus or minus thirteen-fourths, represented as x = 3/4 ×1 13/4.", caption: none)], [],
  [Rewrite to show two solutions.], [#figure(figph[Two mathematical equations are displayed horizontally. The first reads x equals 3 over 4 plus 13 over 4. The second reads x equals 3 over 4 minus 13 over 4.], alt: "Two mathematical equations are displayed horizontally. The first reads x equals 3 over 4 plus 13 over 4. The second reads x equals 3 over 4 minus 13 over 4.", caption: none)], [],
  [Simplify.], [#figure(figph[The image displays two values for x: x=4 and x=-5/2.], alt: "The image displays two values for x: x=4 and x=-5/2.", caption: none)], [],
  [Check: #linebreak() 
We leave the check for you!], [], [],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve by completing the square: #math.equation(block: false, alt: "3 r squared minus 2 r equals 21 .")[$3 r^(2) − 2 r = 21 .$]

#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 by completing the square: #math.equation(block: false, alt: "4 t squared plus 2 t equals 20 .")[$4 t^(2) + 2 t = 20 .$]

#solutionbox[
#math.equation(block: true, alt: "t equals minus the fraction 5 over 2 , t equals 2")[$t = − frac(5, 2) , t = 2$]
]
]

Now that we have seen that the coefficient of #emph[x]#super[2] must be 1 for us to complete the square, we update our procedure for solving a #strong[quadratic equation] by completing the square to include equations of the form #emph[ax]#super[2] + #emph[bx] + #emph[c] = 0.

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Solve a quadratic equation of the form #math.equation(block: false, alt: "a x squared plus b x plus c equals 0")[$a x^(2) + b x + c = 0$] by completing the square.]

+ Divide by #math.equation(block: false, alt: "a")[$a$] to make the coefficient of #emph[x]#super[2] term 1.
+ 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 ,")[$attach(( frac(1, 2) #h(0.2em) · #h(0.2em) b ), t: 2) ,$] the number needed to complete the square. Add it to both sides of the equation.
+ Factor the perfect square trinomial, writing it as a binomial squared on the left and simplify by adding the terms on the right
+ Use the Square Root Property.
+ Simplify the radical and then solve the two resulting equations.
+ Check the solutions.
]

#examplebox("Example 10")[][
Solve by completing the square: #math.equation(block: false, alt: "3 x squared plus 2 x equals 4 .")[$3 x^(2) + 2 x = 4 .$]

#solutionbox[
Again, our first step will be to make the coefficient of #emph[x]#super[2] one. By dividing both sides of the equation by the coefficient of #emph[x]#super[2], we can then continue with solving the equation by completing the square.

#figure(table(
  columns: 3, align: left, inset: 6pt,
  [], [], [#figure(figph[A mathematical equation reads '3x² + 2x = 4' against a plain white background.], alt: "A mathematical equation reads '3x² + 2x = 4' against a plain white background.", caption: none)],
  [Divide both sides by 3 to make the #linebreak() coefficient of #math.equation(block: false, alt: "x squared")[$x^(2)$] equal 1.], [#figure(figph[A mathematical equation shows (3x^2 + 2x) divided by 3, equaling 4/3. The equation is presented in black text on a white background.], alt: "A mathematical equation shows (3x^2 + 2x) divided by 3, equaling 4/3. The equation is presented in black text on a white background.", caption: none)], [],
  [Simplify.], [#figure(figph[A quadratic equation displayed on a white background, reading x squared plus two-thirds x equals four-thirds.], alt: "A quadratic equation displayed on a white background, reading x squared plus two-thirds x equals four-thirds.", 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")[$attach(( frac(1, 2) · frac(2, 3) ), t: 2) = frac(1, 9)$]], [#figure(figph[A quadratic equation, x squared plus two-thirds x plus a blank equals four-thirds. A red box highlights the term one-half times two-thirds, all squared, which is being added to complete the square.], alt: "A quadratic equation, x squared plus two-thirds x plus a blank equals four-thirds. A red box highlights the term one-half times two-thirds, all squared, which is being added to complete the square.", caption: none)], [],
  [Add #math.equation(block: false, alt: "the fraction 1 over 9")[$frac(1, 9)$] to both sides.], [#figure(figph[A quadratic equation displayed as x squared plus two-thirds x plus one-ninth equals four-thirds plus one-ninth, with the one-ninth terms highlighted in red on both sides.], alt: "A quadratic equation displayed as x squared plus two-thirds x plus one-ninth equals four-thirds plus one-ninth, with the one-ninth terms highlighted in red on both sides.", caption: none)], [],
  [Factor the perfect square trinomial, writing it as #linebreak() a binomial squared.], [#figure(figph[A mathematical equation shows (x + 1/3) squared equals 12/9 plus 1/9, demonstrating steps in solving a quadratic equation or simplifying expressions.], alt: "A mathematical equation shows (x + 1/3) squared equals 12/9 plus 1/9, demonstrating steps in solving a quadratic equation or simplifying expressions.", caption: none)], [],
  [Use the Square Root Property.], [#figure(figph[A mathematical equation is displayed, showing x plus one-third equals plus or minus the square root of thirteen over nine.], alt: "A mathematical equation is displayed, showing x plus one-third equals plus or minus the square root of thirteen over nine.", caption: none)], [],
  [Simplify the radical.], [#figure(figph[A mathematical equation is displayed, reading 'x + 1/3 = +/- sqrt(13)/3' on a white background. The equation involves a variable 'x', fractions, a square root, and the plus-minus symbol.], alt: "A mathematical equation is displayed, reading 'x + 1/3 = +/- sqrt(13)/3' on a white background. The equation involves a variable 'x', fractions, a square root, and the plus-minus symbol.", caption: none)], [],
  [Solve for #emph[x] .], [#figure(figph[A mathematical equation shows 'X = -1/3 ×1 (sqrt)13 / 3' written in the center of a plain white background.], alt: "A mathematical equation shows 'X = -1/3 ×1 (sqrt)13 / 3' written in the center of a plain white background.", caption: none)], [],
  [Rewrite to show two solutions.], [#figure(figph[The image displays two mathematical solutions for x: X = -1/3 + sqrt(13)/3 and X = -1/3 - sqrt(13)/3, representing the roots of a quadratic equation.], alt: "The image displays two mathematical solutions for x: X = -1/3 + sqrt(13)/3 and X = -1/3 - sqrt(13)/3, representing the roots of a quadratic equation.", caption: none)], [],
  [Check: #linebreak() 
We leave the check for you!], [], [],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve by completing the square: #math.equation(block: false, alt: "4 x squared plus 3 x equals 2 .")[$4 x^(2) + 3 x = 2 .$]

#solutionbox[
#math.equation(block: true, alt: "x equals minus the fraction 3 over 8 plus the fraction the square root of 41 over 8 , x equals minus the fraction 3 over 8 minus the fraction the square root of 41 over 8")[$x = − frac(3, 8) + frac(sqrt(41), 8) , #h(0.2em) #h(0.2em) x = − frac(3, 8) − frac(sqrt(41), 8)$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve by completing the square: #math.equation(block: false, alt: "3 y squared minus 10 y equals −5 .")[$3 y^(2) − 10 y = −5 .$]

#solutionbox[
#math.equation(block: true, alt: "y equals the fraction 5 over 3 plus the fraction the square root of 10 over 3 , y equals the fraction 5 over 3 minus the fraction the square root of 10 over 3")[$y = frac(5, 3) + frac(sqrt(10), 3) , #h(0.2em) #h(0.2em) y = frac(5, 3) − frac(sqrt(10), 3)$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Access these online resources for additional instruction and practice with completing the square.

- #link("https://openstax.org/l/37CompTheSq1")[Completing Perfect Square Trinomials]
- #link("https://openstax.org/l/37CompTheSq2")[Completing the Square 1]
- #link("https://openstax.org/l/37CompTheSq3")[Completing the Square to Solve Quadratic Equations]
- #link("https://openstax.org/l/37CompTheSq4")[Completing the Square to Solve Quadratic Equations: More Examples]
- #link("https://openstax.org/l/37CompTheSq5")[Completing the Square 4]
]


=== Key Concepts

- Binomial Squares Pattern #linebreak() 
If #emph[a] and #emph[b] are real numbers, #linebreak() 
#figure(figph[Quantity a plus b squared equals a squared plus 2 a b plus b2 where the binomial squared equals the first term squared plus 2 times the product of terms plus the second term squared. Quantity a minus b squared equals a squared minus 2 a b plus b2 where the binomial squared equals the first term squared minus 2 times the product of terms plus the second term squared.], alt: "Quantity a plus b squared equals a squared plus 2 a b plus b2 where the binomial squared equals the first term squared plus 2 times the product of terms plus the second term squared. Quantity a minus b squared equals a squared minus 2 a b plus b2 where the binomial squared equals the first term squared minus 2 times the product of terms plus the second term squared.", caption: none)
- How to Complete a Square
+ Identify #emph[b], the coefficient of #emph[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 #emph[x]#super[2] + #emph[bx]
+ Rewrite the trinomial as a binomial square
- How to solve a quadratic equation of the form #emph[ax]#super[2] + #emph[bx] + #emph[c] = 0 by completing the square.
+ Divide by #emph[a] to make the coefficient of #emph[x]#super[2] term 1.
+ 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 ,")[$attach(( frac(1, 2) #h(0.2em) · #h(0.2em) b ), t: 2) ,$] the number needed to complete the square. Add it to both sides of the equation.
+ Factor the perfect square trinomial, writing it as a binomial squared on the left and simplify by adding the terms on the right.
+ Use the Square Root Property.
+ Simplify the radical and then solve the two resulting equations.
+ Check the solutions.


==== 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: false, alt: "m squared minus 24 m")[$m^(2) − 24 m$] ⓑ #math.equation(block: false, alt: "x squared minus 11 x")[$x^(2) − 11 x$] ⓒ #math.equation(block: false, alt: "p squared minus the fraction 1 over 3 p")[$p^(2) − frac(1, 3) p$]

#solutionbox[
ⓐ #math.equation(block: false, alt: "open parenthesis m minus 12 close parenthesis squared")[$attach(( m − 12 ), t: 2)$] ⓑ #math.equation(block: false, alt: "open parenthesis x minus the fraction 11 over 2 close parenthesis squared")[$attach(( x − frac(11, 2) ), t: 2)$] #linebreak() ⓒ #math.equation(block: false, alt: "open parenthesis p minus the fraction 1 over 6 close parenthesis squared")[$attach(( p − frac(1, 6) ), t: 2)$]
]

ⓐ #math.equation(block: false, alt: "n squared minus 16 n")[$n^(2) − 16 n$] ⓑ #math.equation(block: false, alt: "y squared plus 15 y")[$y^(2) + 15 y$] ⓒ #math.equation(block: false, alt: "q squared plus the fraction 3 over 4 q")[$q^(2) + frac(3, 4) q$]

ⓐ #math.equation(block: false, alt: "p squared minus 22 p")[$p^(2) − 22 p$] ⓑ #math.equation(block: false, alt: "y squared plus 5 y")[$y^(2) + 5 y$] ⓒ #math.equation(block: false, alt: "m squared plus the fraction 2 over 5 m")[$m^(2) + frac(2, 5) m$]

#solutionbox[
ⓐ #math.equation(block: false, alt: "open parenthesis p minus 11 close parenthesis squared")[$attach(( p − 11 ), t: 2)$] ⓑ #math.equation(block: false, alt: "open parenthesis y plus the fraction 5 over 2 close parenthesis squared")[$attach(( y + frac(5, 2) ), t: 2)$] #linebreak() ⓒ #math.equation(block: false, alt: "open parenthesis m plus the fraction 1 over 5 close parenthesis squared")[$attach(( m + frac(1, 5) ), t: 2)$]
]

ⓐ #math.equation(block: false, alt: "q squared minus 6 q")[$q^(2) − 6 q$] ⓑ #math.equation(block: false, alt: "x squared minus 7 x")[$x^(2) − 7 x$] ⓒ #math.equation(block: false, alt: "n squared minus the fraction 2 over 3 n")[$n^(2) − frac(2, 3) n$]

#strong[Solve Quadratic Equations of the form #emph[x]#super[2] + #emph[bx] + #emph[c] = 0 by Completing the Square]

In the following exercises, solve by completing the square.

#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: "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[
#math.equation(block: true, alt: "m equals −2 plus or minus 2 the square root of 10 i")[$m = −2 ± 2 sqrt(10) i$]
]

#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[
#math.equation(block: true, alt: "r equals −3 plus or minus the square root of 2 i")[$r = −3 ± sqrt(2) i$]
]

#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: "x squared plus 5 x equals 2")[$x^(2) + 5 x = 2$]

#solutionbox[
#math.equation(block: true, alt: "x equals minus the fraction 5 over 2 plus or minus the fraction the square root of 33 over 2")[$x = − frac(5, 2) ± frac(sqrt(33), 2)$]
]

#math.equation(block: true, alt: "y squared minus 3 y equals 2")[$y^(2) − 3 y = 2$]

#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: "r squared minus 4 r minus 3 equals 9")[$r^(2) − 4 r − 3 = 9$]

#solutionbox[
#math.equation(block: true, alt: "r equals −2 , r equals 6")[$r = −2 , r = 6$]
]

#math.equation(block: true, alt: "t squared minus 10 t minus 6 equals 5")[$t^(2) − 10 t − 6 = 5$]

#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: "x squared minus 5 equals 10 x")[$x^(2) − 5 = 10 x$]

#solutionbox[
#math.equation(block: true, alt: "x equals 5 plus or minus the square root of 30")[$x = 5 ± sqrt(30)$]
]

#math.equation(block: true, alt: "y squared minus 14 equals 6 y")[$y^(2) − 14 = 6 y$]

#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 80")[$( y + 9 ) ( y + 7 ) = 80$]

#math.equation(block: true, alt: "open parenthesis x plus 2 close parenthesis open parenthesis x plus 4 close parenthesis equals 3")[$( x + 2 ) ( x + 4 ) = 3$]

#solutionbox[
#math.equation(block: true, alt: "x equals −5 , x equals −1")[$x = −5 , x = −1$]
]

#math.equation(block: true, alt: "open parenthesis x minus 2 close parenthesis open parenthesis x minus 6 close parenthesis equals 5")[$( x − 2 ) ( x − 6 ) = 5$]

#strong[Solve Quadratic Equations of the form #emph[ax]#super[2] + #emph[bx] + #emph[c] = 0 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 x squared minus 14 x plus 12 equals 0")[$2 x^(2) − 14 x + 12 = 0$]

#math.equation(block: true, alt: "2 n squared plus 4 n equals 26")[$2 n^(2) + 4 n = 26$]

#solutionbox[
#math.equation(block: true, alt: "n equals minus 1 plus or minus the square root of 14")[$n = − 1 ± sqrt(14)$]
]

#math.equation(block: true, alt: "5 x squared plus 20 x equals 15")[$5 x^(2) + 20 x = 15$]

#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 x squared plus 7 x minus 15 equals 0")[$2 x^(2) + 7 x − 15 = 0$]

#solutionbox[
#math.equation(block: true, alt: "x equals −5 , x equals the fraction 3 over 2")[$x = −5 , x = frac(3, 2)$]
]

#math.equation(block: true, alt: "3 x squared minus 14 x plus 8 equals 0")[$3 x^(2) − 14 x + 8 = 0$]

#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")[$3 q^(2) − 5 q = 9$]

#math.equation(block: true, alt: "5 x squared minus 3 x equals −10")[$5 x^(2) − 3 x = −10$]

#solutionbox[
#math.equation(block: true, alt: "x equals the fraction 3 over 10 plus or minus the fraction the square root of 191 over 10 i")[$x = frac(3, 10) ± frac(sqrt(191), 10) i$]
]

#math.equation(block: true, alt: "7 x squared plus 4 x equals −3")[$7 x^(2) + 4 x = −3$]


==== 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

ⓑ by Completing the Square

ⓒ Which method do you prefer? Why?

#solutionbox[
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 provides a checklist to evaluate mastery of the objectives of this section. Choose how would you respond to the statement “I can complete the square of a binomial expression.” “Confidently,” “with some help,” or “No, I don’t get it.” Choose how would you respond to the statement “I can solve quadratic equations of the form x squared plus b times x plus c equals 0 by completing the square.” “Confidently,” “with some help,” or “No, I don’t get it.” Choose how would you respond to the statement “I can solve quadratic equations of the form a times x squared plus b times x plus c equals 0 by completing the square.” “Confidently,” “with some help,” or “No, I don’t get it.”], alt: "This table provides a checklist to evaluate mastery of the objectives of this section. Choose how would you respond to the statement “I can complete the square of a binomial expression.” “Confidently,” “with some help,” or “No, I don’t get it.” Choose how would you respond to the statement “I can solve quadratic equations of the form x squared plus b times x plus c equals 0 by completing the square.” “Confidently,” “with some help,” or “No, I don’t get it.” Choose how would you respond to the statement “I can solve quadratic equations of the form a times x squared plus b times x plus c equals 0 by completing the square.” “Confidently,” “with some help,” or “No, I don’t get it.”", caption: none)

ⓑ After reviewing this checklist, what will you do to become confident for all objectives?