Skip to main content
Mathematics LibreTexts

9.3: Solve Quadratic Equations by Completing the Square

  • Page ID
    114238
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    \( \newcommand{\dsum}{\displaystyle\sum\limits} \)

    \( \newcommand{\dint}{\displaystyle\int\limits} \)

    \( \newcommand{\dlim}{\displaystyle\lim\limits} \)

    \( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)

    ( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\id}{\mathrm{id}}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\kernel}{\mathrm{null}\,}\)

    \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\)

    \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\)

    \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    \( \newcommand{\vectorA}[1]{\vec{#1}}      % arrow\)

    \( \newcommand{\vectorAt}[1]{\vec{\text{#1}}}      % arrow\)

    \( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vectorC}[1]{\textbf{#1}} \)

    \( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)

    \( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)

    \( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)

    \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \(\newcommand{\longvect}{\overrightarrow}\)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    \(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)
    Learning Objectives

    By the end of this section, you will be able to:

    • Complete the square of a binomial expression
    • Solve quadratic equations of the form x2+bx+c=0x2+bx+c=0 by completing the square
    • Solve quadratic equations of the form ax2+bx+c=0ax2+bx+c=0 by completing the square
    Be Prepared 9.4

    Before you get started, take this readiness quiz.

    Expand: (x+9)2.(x+9)2.
    If you missed this problem, review Example 5.32.

    Be Prepared 9.5

    Factor y214y+49.y214y+49.
    If you missed this problem, review Example 6.9.

    Be Prepared 9.6

    Factor 5n2+40n+80.5n2+40n+80.
    If you missed this problem, review Example 6.14.

    So far we have solved quadratic equations by factoring and using the Square Root Property. In this section, we will solve quadratic equations by a process called completing the square, 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 Square Root Property to solve the equation (y − 7)2 = 12 because the left side was a perfect square.

    (y7)2=12y7=±12y7=±23y=7±23(y7)2=12y7=±12y7=±23y=7±23

    We also solved an equation in which the left side was a perfect square trinomial, but we had to rewrite it the form (xk)2(xk)2 in order to use the Square Root Property.

    x210x+25=18(x5)2=18x210x+25=18(x5)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.

    (x+9)2(y7)2 (x+9)(x+9)(y7)(y7) x2+9x+9x+81y27y7y+49 x2+18x+81y214y+49(x+9)2(y7)2 (x+9)(x+9)(y7)(y7) x2+9x+9x+81y27y7y+49 x2+18x+81y214y+49

    We restate the patterns here for reference.

    Binomial Squares Pattern

    If a and b are real numbers,

    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.

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

    We will start with the expression x2 + 6x. Since there is a plus sign between the two terms, we will use the (a + b)2 pattern, a2 + 2ab + b2 = (a + b)2.

    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.

    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 b. But first we start with determining a. Notice that the first term of x2 + 6x is a square, x2. This tells us that a = x.

    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.

    What number, b, when multiplied with 2x gives 6x? It would have to be 3, which is 12(6).12(6). So b = 3.

    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.

    Now to complete the perfect square trinomial, we will find the last term by squaring b, which is 32 = 9.

    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.

    We can now factor.

    The factored expression, the square of a plus b, is shown over the square of the expression x + 3.

    So we found that adding 9 to x2 + 6x ‘completes the square’, and we write it as (x + 3)2.

    How To

    Complete a square of x2+bx.x2+bx.

    1. Step 1. Identify b, the coefficient of x.
    2. Step 2. Find (12b)2,(12b)2, the number to complete the square.
    3. Step 3. Add the (12b)2(12b)2 to x2 + bx.
    4. Step 4. Factor the perfect square trinomial, writing it as a binomial squared.
    Example 9.11

    Complete the square to make a perfect square trinomial. Then write the result as a binomial squared.

    x226xx226x y29yy29y n2+12nn2+12n

    Answer

      .
    The coefficient of xx is −26.  
    Find(12b)2.(12·(−26))2(13)2169Find(12b)2.(12·(−26))2(13)2169  
    Add 169 to the binomial to complete the square. .
    Factor the perfect square trinomial, writing it as
    a binomial squared.
    .

      .
    The coefficient of yy is −9−9.  
    Find(12b)2.(12·(−9))2(92)2814Find(12b)2.(12·(−9))2(92)2814  
    Add 814814 to the binomial to complete the square. .
    Factor the perfect square trinomial, writing it as
    a binomial squared.
    .

      .
    The coefficient of nn is 12.12.  
    Find(12b)2.(12·12)2(14)2116Find(12b)2.(12·12)2(14)2116  
    Add 116116 to the binomial to complete the square. .
    Rewrite as a binomial square. .
    Try It 9.21

    Complete the square to make a perfect square trinomial. Then write the result as a binomial squared.

    a220aa220a m25mm25m p2+14pp2+14p

    Try It 9.22

    Complete the square to make a perfect square trinomial. Then write the result as a binomial squared.

    b24bb24b n2+13nn2+13n q223qq223q

    Solve Quadratic Equations of the Form x2 + bx + c = 0 by Completing the Square

    In solving equations, we must always do the same thing to both sides of the equation. This is true, of course, when we solve a quadratic equation by 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 x2 + 6x = 40, and we want to complete the square on the left, we will add 9 to both sides of the equation.

      .
      .
      .
    Add 9 to both sides to complete the square. .

    Now the equation is in the form to solve using the Square Root Property! Completing the square is a way to transform an equation into the form we need to be able to use the Square Root Property.

    Example 9.12

    How to Solve a Quadratic Equation of the Form x2+bx+c=0x2+bx+c=0 by Completing the Square

    Solve by completing the square: x2+8x=48.x2+8x=48.

    Answer
    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. 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. 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. 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. 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. 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.
    Try It 9.23

    Solve by completing the square: x2+4x=5.x2+4x=5.

    Try It 9.24

    Solve by completing the square: y210y=−9.y210y=−9.

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

    How To

    Solve a quadratic equation of the form x2+bx+c=0x2+bx+c=0 by completing the square.

    1. Step 1. Isolate the variable terms on one side and the constant terms on the other.
    2. Step 2. Find (12·b)2,(12·b)2, the number needed to complete the square. Add it to both sides of the equation.
    3. 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
    4. Step 4. Use the Square Root Property.
    5. Step 5. Simplify the radical and then solve the two resulting equations.
    6. Step 6. Check the solutions.

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

    Example 9.13

    Solve by completing the square: x2+4x=−21.x2+4x=−21.

    Answer
        .
    The variable terms are on the left side.
    Take half of 4 and square it.
    .
    (12(4))2=4(12(4))2=4  
    Add 4 to both sides. .
    Factor the perfect square trinomial,
    writing it as a binomial squared.
    .
    Use the Square Root Property. .
    Simplify using complex numbers. .
    Subtract 2 from each side. .
    Rewrite to show two solutions. .
    We leave the check to you.  
    Try It 9.25

    Solve by completing the square: y210y=−35.y210y=−35.

    Try It 9.26

    Solve by completing the square: z2+8z=−19.z2+8z=−19.

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

    Example 9.14

    Solve by completing the square: y218y=−6.y218y=−6.

    Answer
      .
    The variable terms are on the left side.
    Take half of −18−18 and square it.
     
    (12(−18))2=81(12(−18))2=81 .
    Add 81 to both sides. .
    Factor the perfect square trinomial,
    writing it as a binomial squared.
    .
    Use the Square Root Property. .
    Simplify the radical. .
    Solve for yy. .
    Check.
    .

    Another way to check this would be to use a calculator. Evaluate y218yy218yfor both of the solutions. The answer should be −6.−6.

    Try It 9.27

    Solve by completing the square: x216x=−16.x216x=−16.

    Try It 9.28

    Solve by completing the square: y2+8y=11.y2+8y=11.

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

    Example 9.15

    Solve by completing the square: x2+10x+4=15.x2+10x+4=15.

    Answer
      .
    Isolate the variable terms on the left side.
    Subtract 4 to get the constant terms on the right side.
    .
    Take half of 10 and square it.  
    (12(10))2=25(12(10))2=25 .
    Add 25 to both sides. .
    Factor the perfect square trinomial, writing it as
    a binomial squared.
    .
    Use the Square Root Property. .
    Simplify the radical. .
    Solve for x. .
    Rewrite to show two solutions. .
    Solve the equations. .
    Check:

    .
    Try It 9.29

    Solve by completing the square: a2+4a+9=30.a2+4a+9=30.

    Try It 9.30

    Solve by completing the square: b2+8b4=16.b2+8b4=16.

    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.

    Example 9.16

    Solve by completing the square: n2=3n+11.n2=3n+11.

    Answer
        .
    Subtract 3n3n to get the variable terms on the left side. .
    Take half of −3−3 and square it.  
    (12(−3))2=94(12(−3))2=94 .
    Add 9494 to both sides. .
    Factor the perfect square trinomial, writing it as
    a binomial squared.
    .
    Add the fractions on the right side. .
    Use the Square Root Property. .
    Simplify the radical. .
    Solve for n. .
    Rewrite to show two solutions. .
    Check:
    We leave the check for you!
     
    Try It 9.31

    Solve by completing the square: p2=5p+9.p2=5p+9.

    Try It 9.32

    Solve by completing the square: q2=7q3.q2=7q3.

    Notice that the left side of the next equation is in factored form. But the right side is not zero. So, we cannot use the Zero Product Property since it says “If a·b=0,a·b=0, then a = 0 or b = 0.” Instead, we multiply the factors and then put the equation into standard form to solve by completing the square.

    Example 9.17

    Solve by completing the square: (x3)(x+5)=9.(x3)(x+5)=9.

    Answer
          .
    We multiply the binomials on the left.   .
    Add 15 to isolate the constant terms on the right.   .
    Take half of 2 and square it.    
    (12·(2))2=1(12·(2))2=1   .
    Add 1 to both sides.   .
    Factor the perfect square trinomial, writing it as
    a binomial squared.
      .
    Use the Square Root Property.   .
    Solve for x.   .
    Rewrite to show two solutions. .
    Simplify. .
    Check:
    We leave the check for you!
       
    Try It 9.33

    Solve by completing the square: (c2)(c+8)=11.(c2)(c+8)=11.

    Try It 9.34

    Solve by completing the square: (d7)(d+3)=56.(d7)(d+3)=56.

    Solve Quadratic Equations of the Form ax2 + bx + c = 0 by Completing the Square

    The process of completing the square works best when the coefficient of x2 is 1, so the left side of the equation is of the form x2 + bx + c. If the x2 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.

    Example 9.18

    Solve by completing the square: 3x212x15=0.3x212x15=0.

    Answer

    To complete the square, we need the coefficient of x2x2 to be one. If we factor out the coefficient of x2x2 as a common factor, we can continue with solving the equation by completing the square.

      .
    Factor out the greatest common factor. .
    Divide both sides by 3 to isolate the trinomial
    with coefficient 1.
    .
    Simplify. .
    Add 5 to get the constant terms on the right side. .
    Take half of 4 and square it.  
    (12(−4))2=4(12(−4))2=4 .
    Add 4 to both sides. .
    Factor the perfect square trinomial, writing it
    as a binomial squared.
    .
    Use the Square Root Property. .
    Solve for x. .
    Rewrite to show two solutions. .
    Simplify. .
    Check:

    .
     
    Try It 9.35

    Solve by completing the square: 2m2+16m+14=0.2m2+16m+14=0.

    Try It 9.36

    Solve by completing the square: 4n224n56=8.4n224n56=8.

    To complete the square, the coefficient of the x2 must be 1. When the leading coefficient is not a factor of all the terms, we will divide both sides of the equation by the leading coefficient! This will give us a fraction for the second coefficient. We have already seen how to complete the square with fractions in this section.

    Example 9.19

    Solve by completing the square: 2x23x=20.2x23x=20.

    Answer

    To complete the square we need the coefficient of x2x2 to be one. We will divide both sides of the equation by the coefficient of x2. Then we can continue with solving the equation by completing the square.

        .
    Divide both sides by 2 to get the
    coefficient of x2x2 to be 1.
    .
    Simplify. .
    Take half of 3232 and square it.  
    (12(32))2=916(12(32))2=916 .
    Add 916916 to both sides. .
    Factor the perfect square trinomial,
    writing it as a binomial squared.
    .
    Add the fractions on the right side. .
    Use the Square Root Property. .
    Simplify the radical. .
    Solve for x. .
    Rewrite to show two solutions. .
    Simplify. .
    Check:
    We leave the check for you!
     
    Try It 9.37

    Solve by completing the square: 3r22r=21.3r22r=21.

    Try It 9.38

    Solve by completing the square: 4t2+2t=20.4t2+2t=20.

    Now that we have seen that the coefficient of x2 must be 1 for us to complete the square, we update our procedure for solving a quadratic equation by completing the square to include equations of the form ax2 + bx + c = 0.

    How To

    Solve a quadratic equation of the form ax2+bx+c=0ax2+bx+c=0 by completing the square.

    1. Step 1. Divide by aa to make the coefficient of x2 term 1.
    2. Step 2. Isolate the variable terms on one side and the constant terms on the other.
    3. Step 3. Find (12·b)2,(12·b)2, the number needed to complete the square. Add it to both sides of the equation.
    4. Step 4. Factor the perfect square trinomial, writing it as a binomial squared on the left and simplify by adding the terms on the right
    5. Step 5. Use the Square Root Property.
    6. Step 6. Simplify the radical and then solve the two resulting equations.
    7. Step 7. Check the solutions.
    Example 9.20

    Solve by completing the square: 3x2+2x=4.3x2+2x=4.

    Answer

    Again, our first step will be to make the coefficient of x2 one. By dividing both sides of the equation by the coefficient of x2, we can then continue with solving the equation by completing the square.

        .
    Divide both sides by 3 to make the
    coefficient of x2x2 equal 1.
    .
    Simplify. .
    Take half of 2323 and square it.  
    (12·23)2=19(12·23)2=19 .
    Add 1919 to both sides. .
    Factor the perfect square trinomial, writing it as
    a binomial squared.
    .
    Use the Square Root Property. .
    Simplify the radical. .
    Solve for x . .
    Rewrite to show two solutions. .
    Check:
    We leave the check for you!
     
    Try It 9.39

    Solve by completing the square: 4x2+3x=2.4x2+3x=2.

    Try It 9.40

    Solve by completing the square: 3y210y=−5.3y210y=−5.

    Section 9.2 Exercises

    Practice Makes Perfect

    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.

    71.

    m224mm224m x211xx211x p213pp213p

    72.

    n216nn216n y2+15yy2+15y q2+34qq2+34q

    73.

    p222pp222p y2+5yy2+5y m2+25mm2+25m

    74.

    q26qq26q x27xx27x n223nn223n

    Solve Quadratic Equations of the form x2 + bx + c = 0 by Completing the Square

    In the following exercises, solve by completing the square.

    75.

    u 2 + 2 u = 3 u 2 + 2 u = 3

    76.

    z 2 + 12 z = −11 z 2 + 12 z = −11

    77.

    x 2 20 x = 21 x 2 20 x = 21

    78.

    y 2 2 y = 8 y 2 2 y = 8

    79.

    m 2 + 4 m = −44 m 2 + 4 m = −44

    80.

    n 2 2 n = −3 n 2 2 n = −3

    81.

    r 2 + 6 r = −11 r 2 + 6 r = −11

    82.

    t 2 14 t = −50 t 2 14 t = −50

    83.

    a 2 10 a = −5 a 2 10 a = −5

    84.

    b 2 + 6 b = 41 b 2 + 6 b = 41

    85.

    x 2 + 5 x = 2 x 2 + 5 x = 2

    86.

    y 2 3 y = 2 y 2 3 y = 2

    87.

    u 2 14 u + 12 = −1 u 2 14 u + 12 = −1

    88.

    z 2 + 2 z 5 = 2 z 2 + 2 z 5 = 2

    89.

    r 2 4 r 3 = 9 r 2 4 r 3 = 9

    90.

    t 2 10 t 6 = 5 t 2 10 t 6 = 5

    91.

    v 2 = 9 v + 2 v 2 = 9 v + 2

    92.

    w 2 = 5 w 1 w 2 = 5 w 1

    93.

    x 2 5 = 10 x x 2 5 = 10 x

    94.

    y 2 14 = 6 y y 2 14 = 6 y

    95.

    ( x + 6 ) ( x 2 ) = 9 ( x + 6 ) ( x 2 ) = 9

    96.

    ( y + 9 ) ( y + 7 ) = 80 ( y + 9 ) ( y + 7 ) = 80

    97.

    ( x + 2 ) ( x + 4 ) = 3 ( x + 2 ) ( x + 4 ) = 3

    98.

    ( x 2 ) ( x 6 ) = 5 ( x 2 ) ( x 6 ) = 5

    Solve Quadratic Equations of the form ax2 + bx + c = 0 by Completing the Square

    In the following exercises, solve by completing the square.

    99.

    3 m 2 + 30 m 27 = 6 3 m 2 + 30 m 27 = 6

    100.

    2 x 2 14 x + 12 = 0 2 x 2 14 x + 12 = 0

    101.

    2 n 2 + 4 n = 26 2 n 2 + 4 n = 26

    102.

    5 x 2 + 20 x = 15 5 x 2 + 20 x = 15

    103.

    2 c 2 + c = 6 2 c 2 + c = 6

    104.

    3 d 2 4 d = 15 3 d 2 4 d = 15

    105.

    2 x 2 + 7 x 15 = 0 2 x 2 + 7 x 15 = 0

    106.

    3 x 2 14 x + 8 = 0 3 x 2 14 x + 8 = 0

    107.

    2 p 2 + 7 p = 14 2 p 2 + 7 p = 14

    108.

    3 q 2 5 q = 9 3 q 2 5 q = 9

    109.

    5 x 2 3 x = −10 5 x 2 3 x = −10

    110.

    7 x 2 + 4 x = −3 7 x 2 + 4 x = −3

    Writing Exercises

    111.

    Solve the equation x2+10x=−25x2+10x=−25

    by using the Square Root Property

    by Completing the Square

    Which method do you prefer? Why?

    112.

    Solve the equation y2+8y=48y2+8y=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.

    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.”

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


    This page titled 9.3: Solve Quadratic Equations by Completing the Square is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to the style and standards of the LibreTexts platform.