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Quadratic Equations
Solve Quadratic Equations by Completing the Square

Solve Quadratic Equations by Completing the Square

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=0x^2+bx+c=0 by completing the square, and solve quadratic equations of the form ax2+bx+c=0ax^2+bx+c=0 by completing the square.

So far, we have solved quadratic equations by factoring and using the Square Root Property. In this section, we will solve quadratic equations by a process called “completing the square.”

Complete the Square of a Binomial Expression

In the last section, we were able to use the Square Root Property to solve the equation (y−7)2=12(y-7)^2=12 because the left side was a perfect square.

y−7=±12y−7=±23y=7±23 \begin{array}{rcl} y-7 &=& \pm\sqrt{12} \\[4pt] y-7 &=& \pm2\sqrt{3} \\[4pt] y &=& 7\pm2\sqrt{3} \end{array}

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

x2−10x+25=18(x−5)2=18 \begin{array}{rcl} x^2-10x+25 &=& 18 \\[4pt] (x-5)^2 &=& 18 \end{array}

What happens if the variable is not part of a perfect square? Can we use algebra to make a perfect square?

Let’s study the binomial square pattern we have used many times. We will look at two examples.

(x+9)2=(x+9)(x+9)=x2+9x+9x+81=x2+18x+81 \begin{array}{rcl} (x+9)^2 &=& (x+9)(x+9) \\[4pt] &=& x^2+9x+9x+81 \\[4pt] &=& x^2+18x+81 \end{array} (y−7)2=(y−7)(y−7)=y2−7y−7y+49=y2−14y+49 \begin{array}{rcl} (y-7)^2 &=& (y-7)(y-7) \\[4pt] &=& y^2-7y-7y+49 \\[4pt] &=& y^2-14y+49 \end{array}

Binomial Squares Pattern. If a,ba,b are real numbers,

(a+b)2=a2+2ab+b2(a+b)^2=a^2+2ab+b^2(a−b)2=a2−2ab+b2(a-b)^2=a^2-2ab+b^2

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

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

Notice that the first term of x2+6xx^2+6x is a square, x2x^2. We now know a=xa=x.

What number can we add to x2+6xx^2+6x to make a perfect square trinomial?

The middle term of the Binomial Squares Pattern, 2ab2ab, is twice the product of the two terms of the binomial. This means twice the product of xx and some number is 6x6x. So, two times some number must be six. The number we need is 12⋅6=3\tfrac{1}{2}\cdot6=3. The second term in the binomial, bb, must be 33.

Now, we just square the second term of the binomial to get the last term of the perfect square trinomial, so we square three to get the last term, nine.

x2+6x+9x^2+6x+9

We can now factor to

(x+3)2.(x+3)^2.

So, we found that adding nine to x2+6xx^2+6x “completes the square,” and we write it as (x+3)2(x+3)^2.

Complete a square. To complete the square of x2+bxx^2+bx:

  1. Identify bb, the coefficient of xx.
  2. Find (12b)2\left(\tfrac{1}{2}b\right)^2, the number to complete the square.
  3. Add (12b)2\left(\tfrac{1}{2}b\right)^2 to x2+bxx^2+bx.

Example 10.14. Complete the square to make a perfect square trinomial. Then, write the result as a binomial square: x2+14xx^2+14x.

The coefficient of x is 14.x2+14xFind (12b)2.(12⋅14)2=72=49Add 49 to the binomial to complete the square.x2+14x+49Rewrite as a binomial square.(x+7)2 \begin{array}{lrcl} \text{The coefficient of }x\text{ is }14. &&& x^2+14x \\[4pt] \text{Find }\left(\tfrac{1}{2}b\right)^2. &&& \left(\tfrac{1}{2}\cdot14\right)^2=7^2=49 \\[10pt] \text{Add }49\text{ to the binomial to complete the square.} &&& x^2+14x+49 \\[4pt] \text{Rewrite as a binomial square.} &&& (x+7)^2 \end{array}

Example 10.15. Complete the square to make a perfect square trinomial. Then, write the result as a binomial squared: m2−26mm^2-26m.

The coefficient of m is −26.m2−26mFind (12b)2.(12⋅(−26))2=(−13)2=169Add 169 to the binomial to complete the square.m2−26m+169Rewrite as a binomial square.(m−13)2 \begin{array}{lrcl} \text{The coefficient of }m\text{ is }-26. &&& m^2-26m \\[4pt] \text{Find }\left(\tfrac{1}{2}b\right)^2. &&& \left(\tfrac{1}{2}\cdot(-26)\right)^2=(-13)^2=169 \\[10pt] \text{Add }169\text{ to the binomial to complete the square.} &&& m^2-26m+169 \\[4pt] \text{Rewrite as a binomial square.} &&& (m-13)^2 \end{array}

Example 10.16. Complete the square to make a perfect square trinomial. Then, write the result as a binomial squared: u2−9uu^2-9u.

The coefficient of u is −9.u2−9uFind (12b)2.(12⋅(−9))2=(−92)2=814Add 814 to the binomial to complete the square.u2−9u+814Rewrite as a binomial square.(u−92)2 \begin{array}{lrcl} \text{The coefficient of }u\text{ is }-9. &&& u^2-9u \\[4pt] \text{Find }\left(\tfrac{1}{2}b\right)^2. &&& \left(\tfrac{1}{2}\cdot(-9)\right)^2=\left(-\tfrac{9}{2}\right)^2=\tfrac{81}{4} \\[10pt] \text{Add }\tfrac{81}{4}\text{ to the binomial to complete the square.} &&& u^2-9u+\tfrac{81}{4} \\[10pt] \text{Rewrite as a binomial square.} &&& \left(u-\tfrac{9}{2}\right)^2 \end{array}

Example 10.17. Complete the square to make a perfect square trinomial. Then, write the result as a binomial squared: p2+12pp^2+\tfrac{1}{2}p.

The coefficient of p is 12.p2+12pFind (12b)2.(12⋅12)2=(14)2=116Add 116 to the binomial to complete the square.p2+12p+116Rewrite as a binomial square.(p+14)2 \begin{array}{lrcl} \text{The coefficient of }p\text{ is }\tfrac{1}{2}. &&& p^2+\tfrac{1}{2}p \\[10pt] \text{Find }\left(\tfrac{1}{2}b\right)^2. &&& \left(\tfrac{1}{2}\cdot\tfrac{1}{2}\right)^2= \left(\tfrac{1}{4}\right)^2=\tfrac{1}{16} \\[10pt] \text{Add }\tfrac{1}{16}\text{ to the binomial to complete the square.} &&& p^2+\tfrac{1}{2}p+\tfrac{1}{16} \\[10pt] \text{Rewrite as a binomial square.} &&& \left(p+\tfrac{1}{4}\right)^2 \end{array}

Complete the square to make a perfect square trinomial. Enter the result as a binomial square: a2−20aa^2-20a.

Solve Quadratic Equations of the Form x2+bx+c=0x^2+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=40x^2+6x=40 and we want to complete the square on the left, we will add nine to both sides of the equation.

x2+6x=40x2+6x+9=40+9 \begin{array}{rcl} x^2+6x &=& 40 \\[4pt] x^2+6x+9 &=& 40+9 \end{array}

Then, we factor on the left and simplify on the right.

(x+3)2=49(x+3)^2=49

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

Example 10.18. How to Solve a Quadratic Equation of the Form x2+bx+c=0x^2+bx+c=0 by Completing the Square. Solve x2+8x=48x^2+8x=48 by completing the square.

The variable terms are on the left side.x2+8x=48Take half of 8 and square it.(12⋅8)2=16Add 16 to both sides.x2+8x+16=48+16Factor the perfect square trinomial as a binomial square.(x+4)2=64Use the Square Root Property.x+4=±64Simplify the radical.x+4=±8Solve the two resulting equations.x=4, −12 \begin{array}{lrcl} \text{The variable terms are on the left side.} & x^2+8x &=& 48 \\[4pt] \text{Take half of }8\text{ and square it.} & \left(\tfrac{1}{2}\cdot8\right)^2 &=& 16 \\[10pt] \text{Add }16\text{ to both sides.} & x^2+8x+16 &=& 48+16 \\[4pt] \text{Factor the perfect square trinomial as a binomial square.} & (x+4)^2 &=& 64 \\[4pt] \text{Use the Square Root Property.} & x+4 &=& \pm\sqrt{64} \\[4pt] \text{Simplify the radical.} & x+4 &=& \pm8 \\[4pt] \text{Solve the two resulting equations.} & x &=& 4,\ -12 \end{array}

Check the solutions.

(4)2+8(4)=?4816+32=?4848=48 ✓(−12)2+8(−12)=?48144−96=?4848=48 ✓ \begin{array}{rcl} (4)^2+8(4) &\overset{?}{=}& 48 \\[4pt] 16+32 &\overset{?}{=}& 48 \\[4pt] 48 &=& 48\ \checkmark \end{array} \qquad \begin{array}{rcl} (-12)^2+8(-12) &\overset{?}{=}& 48 \\[4pt] 144-96 &\overset{?}{=}& 48 \\[4pt] 48 &=& 48\ \checkmark \end{array}

Solve a quadratic equation of the form x2+bx+c=0x^2+bx+c=0 by completing the square.

  1. Isolate the variable terms on one side and the constant terms on the other.
  2. Find (12b)2\left(\tfrac{1}{2}b\right)^2, the number to complete the square. Add it to both sides of the equation.
  3. Factor the perfect square trinomial as a binomial square.
  4. Use the Square Root Property.
  5. Simplify the radical and then solve the two resulting equations.
  6. Check the solutions.

Example 10.19. Solve y2−6y=16y^2-6y=16 by completing the square.

The variable terms are on the left side.y2−6y=16Take half of −6 and square it.(12⋅(−6))2=9Add 9 to both sides.y2−6y+9=16+9Factor the perfect square trinomial as a binomial square.(y−3)2=25Use the Square Root Property.y−3=±25Simplify the radical.y−3=±5Solve for y.y=3±5Rewrite to show two solutions.y=3+5, 3−5Solve the equations.y=8, −2 \begin{array}{lrcl} \text{The variable terms are on the left side.} & y^2-6y &=& 16 \\[4pt] \text{Take half of }-6\text{ and square it.} & \left(\tfrac{1}{2}\cdot(-6)\right)^2 &=& 9 \\[10pt] \text{Add }9\text{ to both sides.} & y^2-6y+9 &=& 16+9 \\[4pt] \text{Factor the perfect square trinomial as a binomial square.} & (y-3)^2 &=& 25 \\[4pt] \text{Use the Square Root Property.} & y-3 &=& \pm\sqrt{25} \\[4pt] \text{Simplify the radical.} & y-3 &=& \pm5 \\[4pt] \text{Solve for }y. & y &=& 3\pm5 \\[4pt] \text{Rewrite to show two solutions.} & y &=& 3+5,\ 3-5 \\[4pt] \text{Solve the equations.} & y &=& 8,\ -2 \end{array}

Check.

82−6⋅8=?1664−48=?1616=16 ✓(−2)2−6(−2)=?164+12=?1616=16 ✓ \begin{array}{rcl} 8^2-6\cdot8 &\overset{?}{=}& 16 \\[4pt] 64-48 &\overset{?}{=}& 16 \\[4pt] 16 &=& 16\ \checkmark \end{array} \qquad \begin{array}{rcl} (-2)^2-6(-2) &\overset{?}{=}& 16 \\[4pt] 4+12 &\overset{?}{=}& 16 \\[4pt] 16 &=& 16\ \checkmark \end{array}

Example 10.20. Solve x2+4x=−21x^2+4x=-21 by completing the square.

The variable terms are on the left side.x2+4x=−21Take half of 4 and square it.(12⋅4)2=4Add 4 to both sides.x2+4x+4=−21+4Factor the perfect square trinomial as a binomial square.(x+2)2=−17Use the Square Root Property.x+2=±−17 \begin{array}{lrcl} \text{The variable terms are on the left side.} & x^2+4x &=& -21 \\[4pt] \text{Take half of }4\text{ and square it.} & \left(\tfrac{1}{2}\cdot4\right)^2 &=& 4 \\[10pt] \text{Add }4\text{ to both sides.} & x^2+4x+4 &=& -21+4 \\[4pt] \text{Factor the perfect square trinomial as a binomial square.} & (x+2)^2 &=& -17 \\[4pt] \text{Use the Square Root Property.} & x+2 &=& \pm\sqrt{-17} \end{array}

We cannot take the square root of a negative number. There is no real solution.

In the previous example, there was no real solution because (x+k)2(x+k)^2 was equal to a negative number.

Example 10.21. Solve p2−18p=−6p^2-18p=-6 by completing the square.

The variable terms are on the left side.p2−18p=−6Take half of −18 and square it.(12⋅(−18))2=81Add 81 to both sides.p2−18p+81=−6+81Factor the perfect square trinomial as a binomial square.(p−9)2=75Use the Square Root Property.p−9=±75Simplify the radical.p−9=±53Solve for p.p=9±53Rewrite to show two solutions.p=9+53, 9−53 \begin{array}{lrcl} \text{The variable terms are on the left side.} & p^2-18p &=& -6 \\[4pt] \text{Take half of }-18\text{ and square it.} & \left(\tfrac{1}{2}\cdot(-18)\right)^2 &=& 81 \\[10pt] \text{Add }81\text{ to both sides.} & p^2-18p+81 &=& -6+81 \\[4pt] \text{Factor the perfect square trinomial as a binomial square.} & (p-9)^2 &=& 75 \\[4pt] \text{Use the Square Root Property.} & p-9 &=& \pm\sqrt{75} \\[4pt] \text{Simplify the radical.} & p-9 &=& \pm5\sqrt{3} \\[4pt] \text{Solve for }p. & p &=& 9\pm5\sqrt{3} \\[4pt] \text{Rewrite to show two solutions.} & p &=& 9+5\sqrt{3},\ 9-5\sqrt{3} \end{array}

Check.

(9+53)2−18(9+53)=?−681+903+75−162−903=?−6−6=−6 ✓(9−53)2−18(9−53)=?−681−903+75−162+903=?−6−6=−6 ✓ \begin{array}{rcl} (9+5\sqrt{3})^2-18(9+5\sqrt{3}) &\overset{?}{=}& -6 \\[4pt] 81+90\sqrt{3}+75-162-90\sqrt{3} &\overset{?}{=}& -6 \\[4pt] -6 &=& -6\ \checkmark \end{array} \qquad \begin{array}{rcl} (9-5\sqrt{3})^2-18(9-5\sqrt{3}) &\overset{?}{=}& -6 \\[4pt] 81-90\sqrt{3}+75-162+90\sqrt{3} &\overset{?}{=}& -6 \\[4pt] -6 &=& -6\ \checkmark \end{array}

Another way to check this would be to use a calculator. Evaluate p2−18pp^2-18p for both of the solutions. The answer should be −6-6.

Solve y2+8y=11y^2+8y=11 by completing the square. Enter both solutions separated by commas, least to greatest.

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

Example 10.22. Solve x2+10x+4=15x^2+10x+4=15 by completing the square.

The variable terms are on the left side.x2+10x+4=15Subtract 4 to get the constant terms on the right side.x2+10x=11Take half of 10 and square it.(12⋅10)2=25Add 25 to both sides.x2+10x+25=11+25Factor the perfect square trinomial as a binomial square.(x+5)2=36Use the Square Root Property.x+5=±36Simplify the radical.x+5=±6Solve for x.x=−5±6Rewrite to show two equations.x=−5+6, −5−6Solve the equations.x=1, −11 \begin{array}{lrcl} \text{The variable terms are on the left side.} & x^2+10x+4 &=& 15 \\[4pt] \text{Subtract }4\text{ to get the constant terms on the right side.} & x^2+10x &=& 11 \\[4pt] \text{Take half of }10\text{ and square it.} & \left(\tfrac{1}{2}\cdot10\right)^2 &=& 25 \\[10pt] \text{Add }25\text{ to both sides.} & x^2+10x+25 &=& 11+25 \\[4pt] \text{Factor the perfect square trinomial as a binomial square.} & (x+5)^2 &=& 36 \\[4pt] \text{Use the Square Root Property.} & x+5 &=& \pm\sqrt{36} \\[4pt] \text{Simplify the radical.} & x+5 &=& \pm6 \\[4pt] \text{Solve for }x. & x &=& -5\pm6 \\[4pt] \text{Rewrite to show two equations.} & x &=& -5+6,\ -5-6 \\[4pt] \text{Solve the equations.} & x &=& 1,\ -11 \end{array}

Check.

(1)2+10(1)+4=?151+10+4=?1515=15 ✓(−11)2+10(−11)+4=?15121−110+4=?1515=15 ✓ \begin{array}{rcl} (1)^2+10(1)+4 &\overset{?}{=}& 15 \\[4pt] 1+10+4 &\overset{?}{=}& 15 \\[4pt] 15 &=& 15\ \checkmark \end{array} \qquad \begin{array}{rcl} (-11)^2+10(-11)+4 &\overset{?}{=}& 15 \\[4pt] 121-110+4 &\overset{?}{=}& 15 \\[4pt] 15 &=& 15\ \checkmark \end{array}

To solve the next equation, we must first collect all the variable terms to the left side of the equation. Then, we proceed as we did in the previous examples.

Example 10.23. Solve n2=3n+11n^2=3n+11 by completing the square.

n2=3n+11Subtract 3n to get the variable terms on the left side.n2−3n=11Take half of −3 and square it.(12⋅(−3))2=94Add 94 to both sides.n2−3n+94=11+94Factor the perfect square trinomial as a binomial square.(n−32)2=444+94Add the fractions on the right side.(n−32)2=534Use the Square Root Property.n−32=±534Simplify the radical.n−32=±532Solve for n.n=32±532Rewrite to show two equations.n=32+532, 32−532 \begin{array}{lrcl} & n^2 &=& 3n+11 \\[4pt] \text{Subtract }3n\text{ to get the variable terms on the left side.} & n^2-3n &=& 11 \\[4pt] \text{Take half of }-3\text{ and square it.} & \left(\tfrac{1}{2}\cdot(-3)\right)^2 &=& \tfrac{9}{4} \\[10pt] \text{Add }\tfrac{9}{4}\text{ to both sides.} & n^2-3n+\tfrac{9}{4} &=& 11+\tfrac{9}{4} \\[10pt] \text{Factor the perfect square trinomial as a binomial square.} & \left(n-\tfrac{3}{2}\right)^2 &=& \tfrac{44}{4}+\tfrac{9}{4} \\[10pt] \text{Add the fractions on the right side.} & \left(n-\tfrac{3}{2}\right)^2 &=& \tfrac{53}{4} \\[10pt] \text{Use the Square Root Property.} & n-\tfrac{3}{2} &=& \pm\sqrt{\tfrac{53}{4}} \\[10pt] \text{Simplify the radical.} & n-\tfrac{3}{2} &=& \pm\tfrac{\sqrt{53}}{2} \\[10pt] \text{Solve for }n. & n &=& \tfrac{3}{2}\pm\tfrac{\sqrt{53}}{2} \\[10pt] \text{Rewrite to show two equations.} & n &=& \tfrac{3}{2}+\tfrac{\sqrt{53}}{2},\ \tfrac{3}{2}-\tfrac{\sqrt{53}}{2} \end{array}

Check. We leave the check for you!

Notice that the left side of the next equation is in factored form. But the right side is not zero, so we cannot use the Zero Product Property. Instead, we multiply the factors and then put the equation into the standard form to solve by completing the square.

Example 10.24. Solve (x−3)(x+5)=9(x-3)(x+5)=9 by completing the square.

(x−3)(x+5)=9We multiply binomials on the left.x2+2x−15=9Add 15 to get the variable terms on the left side.x2+2x=24Take half of 2 and square it.(12⋅2)2=1Add 1 to both sides.x2+2x+1=24+1Factor the perfect square trinomial as a binomial square.(x+1)2=25Use the Square Root Property.x+1=±25Solve for x.x=−1±5Rewrite to show two solutions.x=−1+5, −1−5Simplify.x=4, −6 \begin{array}{lrcl} & (x-3)(x+5) &=& 9 \\[4pt] \text{We multiply binomials on the left.} & x^2+2x-15 &=& 9 \\[4pt] \text{Add }15\text{ to get the variable terms on the left side.} & x^2+2x &=& 24 \\[4pt] \text{Take half of }2\text{ and square it.} & \left(\tfrac{1}{2}\cdot2\right)^2 &=& 1 \\[10pt] \text{Add }1\text{ to both sides.} & x^2+2x+1 &=& 24+1 \\[4pt] \text{Factor the perfect square trinomial as a binomial square.} & (x+1)^2 &=& 25 \\[4pt] \text{Use the Square Root Property.} & x+1 &=& \pm\sqrt{25} \\[4pt] \text{Solve for }x. & x &=& -1\pm5 \\[4pt] \text{Rewrite to show two solutions.} & x &=& -1+5,\ -1-5 \\[4pt] \text{Simplify.} & x &=& 4,\ -6 \end{array}

Check. We leave the check for you!

Solve Quadratic Equations of the Form ax2+bx+c=0ax^2+bx+c=0 by Completing the Square

The process of completing the square works best when the leading coefficient is one, so the left side of the equation is of the form x2+bx+cx^2+bx+c. If the x2x^2 term has a coefficient, we take some preliminary steps to make the coefficient equal to one.

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

Example 10.25. Solve 3x2−12x−15=03x^2-12x-15=0 by completing the square.

To complete the square, we need the coefficient of x2x^2 to be one. If we factor out the coefficient of x2x^2 as a common factor, we can continue with solving the equation by completing the square.

3x2−12x−15=0Factor out the greatest common factor.3(x2−4x−5)=0Divide both sides by 3 to isolate the trinomial.3(x2−4x−5)3=03Simplify.x2−4x−5=0Subtract 5 to get the constant terms on the right.x2−4x=5Take half of −4 and square it.(12⋅(−4))2=4Add 4 to both sides.x2−4x+4=5+4Factor the perfect square trinomial as a binomial square.(x−2)2=9Use the Square Root Property.x−2=±9Solve for x.x−2=±3Rewrite to show two solutions.x=2+3, 2−3Simplify.x=5, −1 \begin{array}{lrcl} & 3x^2-12x-15 &=& 0 \\[4pt] \text{Factor out the greatest common factor.} & 3(x^2-4x-5) &=& 0 \\[4pt] \text{Divide both sides by }3\text{ to isolate the trinomial.} & \tfrac{3(x^2-4x-5)}{3} &=& \tfrac{0}{3} \\[10pt] \text{Simplify.} & x^2-4x-5 &=& 0 \\[4pt] \text{Subtract }5\text{ to get the constant terms on the right.} & x^2-4x &=& 5 \\[4pt] \text{Take half of }-4\text{ and square it.} & \left(\tfrac{1}{2}\cdot(-4)\right)^2 &=& 4 \\[10pt] \text{Add }4\text{ to both sides.} & x^2-4x+4 &=& 5+4 \\[4pt] \text{Factor the perfect square trinomial as a binomial square.} & (x-2)^2 &=& 9 \\[4pt] \text{Use the Square Root Property.} & x-2 &=& \pm\sqrt{9} \\[4pt] \text{Solve for }x. & x-2 &=& \pm3 \\[4pt] \text{Rewrite to show two solutions.} & x &=& 2+3,\ 2-3 \\[4pt] \text{Simplify.} & x &=& 5,\ -1 \end{array}

Check.

3(5)2−12(5)−15=?075−60−15=?00=0 ✓3(−1)2−12(−1)−15=?03+12−15=?00=0 ✓ \begin{array}{rcl} 3(5)^2-12(5)-15 &\overset{?}{=}& 0 \\[4pt] 75-60-15 &\overset{?}{=}& 0 \\[4pt] 0 &=& 0\ \checkmark \end{array} \qquad \begin{array}{rcl} 3(-1)^2-12(-1)-15 &\overset{?}{=}& 0 \\[4pt] 3+12-15 &\overset{?}{=}& 0 \\[4pt] 0 &=& 0\ \checkmark \end{array}

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

Example 10.26. Solve 2x2−3x=202x^2-3x=20 by completing the square.

Again, our first step will be to make the coefficient of x2x^2 be one. By dividing both sides of the equation by the coefficient of x2x^2, we can then continue with solving the equation by completing the square.

2x2−3x=20Divide both sides by 2 to get the coefficient of x2 to be 1.2x2−3x2=202Simplify.x2−32x=10Take half of −32 and square it.(12⋅−32)2=916Add 916 to both sides.x2−32x+916=10+916Factor the perfect square trinomial as a binomial square.(x−34)2=16016+916Add the fractions on the right side.(x−34)2=16916Use the Square Root Property.x−34=±16916Simplify the radical.x−34=±134Solve for x.x=34±134Rewrite to show two solutions.x=34+134, 34−134Simplify.x=4, −52 \begin{array}{lrcl} & 2x^2-3x &=& 20 \\[4pt] \text{Divide both sides by }2\text{ to get the coefficient of }x^2\text{ to be }1. & \tfrac{2x^2-3x}{2} &=& \tfrac{20}{2} \\[10pt] \text{Simplify.} & x^2-\tfrac{3}{2}x &=& 10 \\[10pt] \text{Take half of }-\tfrac{3}{2}\text{ and square it.} & \left(\tfrac{1}{2}\cdot-\tfrac{3}{2}\right)^2 &=& \tfrac{9}{16} \\[10pt] \text{Add }\tfrac{9}{16}\text{ to both sides.} & x^2-\tfrac{3}{2}x+\tfrac{9}{16} &=& 10+\tfrac{9}{16} \\[10pt] \text{Factor the perfect square trinomial as a binomial square.} & \left(x-\tfrac{3}{4}\right)^2 &=& \tfrac{160}{16}+\tfrac{9}{16} \\[10pt] \text{Add the fractions on the right side.} & \left(x-\tfrac{3}{4}\right)^2 &=& \tfrac{169}{16} \\[10pt] \text{Use the Square Root Property.} & x-\tfrac{3}{4} &=& \pm\sqrt{\tfrac{169}{16}} \\[10pt] \text{Simplify the radical.} & x-\tfrac{3}{4} &=& \pm\tfrac{13}{4} \\[10pt] \text{Solve for }x. & x &=& \tfrac{3}{4}\pm\tfrac{13}{4} \\[10pt] \text{Rewrite to show two solutions.} & x &=& \tfrac{3}{4}+\tfrac{13}{4},\ \tfrac{3}{4}-\tfrac{13}{4} \\[10pt] \text{Simplify.} & x &=& 4,\ -\tfrac{5}{2} \end{array}

Check. We leave the check for you.

Example 10.27. Solve 3x2+2x=43x^2+2x=4 by completing the square.

Again, our first step will be to make the coefficient of x2x^2 be one. By dividing both sides of the equation by the coefficient of x2x^2, we can then continue with solving the equation by completing the square.

3x2+2x=4Divide both sides by 3 to make the coefficient of x2 equal 1.3x2+2x3=43Simplify.x2+23x=43Take half of 23 and square it.(12⋅23)2=19Add 19 to both sides.x2+23x+19=43+19Factor the perfect square trinomial as a binomial square.(x+13)2=129+19Use the Square Root Property.x+13=±139Simplify the radical.x+13=±133Solve for x.x=−13±133Rewrite to show two solutions.x=−13+133, −13−133 \begin{array}{lrcl} & 3x^2+2x &=& 4 \\[4pt] \text{Divide both sides by }3\text{ to make the coefficient of }x^2\text{ equal }1. & \tfrac{3x^2+2x}{3} &=& \tfrac{4}{3} \\[10pt] \text{Simplify.} & x^2+\tfrac{2}{3}x &=& \tfrac{4}{3} \\[10pt] \text{Take half of }\tfrac{2}{3}\text{ and square it.} & \left(\tfrac{1}{2}\cdot\tfrac{2}{3}\right)^2 &=& \tfrac{1}{9} \\[10pt] \text{Add }\tfrac{1}{9}\text{ to both sides.} & x^2+\tfrac{2}{3}x+\tfrac{1}{9} &=& \tfrac{4}{3}+\tfrac{1}{9} \\[10pt] \text{Factor the perfect square trinomial as a binomial square.} & \left(x+\tfrac{1}{3}\right)^2 &=& \tfrac{12}{9}+\tfrac{1}{9} \\[10pt] \text{Use the Square Root Property.} & x+\tfrac{1}{3} &=& \pm\sqrt{\tfrac{13}{9}} \\[10pt] \text{Simplify the radical.} & x+\tfrac{1}{3} &=& \pm\tfrac{\sqrt{13}}{3} \\[10pt] \text{Solve for }x. & x &=& -\tfrac{1}{3}\pm\tfrac{\sqrt{13}}{3} \\[10pt] \text{Rewrite to show two solutions.} & x &=& -\tfrac{1}{3}+\tfrac{\sqrt{13}}{3},\ -\tfrac{1}{3}-\tfrac{\sqrt{13}}{3} \end{array}

Check. We leave the check for you.

Solve 4x2+3x=124x^2+3x=12 by completing the square. Enter both solutions separated by commas, least to greatest.

Key terms

Binomial Squares Pattern — (a+b)2=a2+2ab+b2(a+b)^2=a^2+2ab+b^2 and (a−b)2=a2−2ab+b2(a-b)^2=a^2-2ab+b^2. complete the square — for an expression x2+bxx^2+bx, add (12b)2\left(\tfrac{1}{2}b\right)^2 to make a perfect square trinomial. perfect square trinomial — a trinomial of the form a2+2ab+b2a^2+2ab+b^2 or a2−2ab+b2a^2-2ab+b^2; it factors to (a+b)2(a+b)^2 or (a−b)2(a-b)^2. Square Root Property — if x2=kx^2=k and k≥0k\geq0, then x=kx=\sqrt{k} or x=−kx=-\sqrt{k}.

This page is adapted from Elementary Algebra 2e, Section 10.2 by Lynn Marecek, MaryAnne Anthony-Smith, and Andrea Honeycutt Mathis, © OpenStax, licensed under CC BY-NC-SA 4.0. Access the original for free at openstax.org. Changes: reformatted the source for accessible web presentation and converted selected Try It problems into interactive exercises; the source exercise set and media links are omitted.

Solve Quadratic Equations Using the Square Root PropertySolve Quadratic Equations Using the Quadratic Formula