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Intermediate Algebra
Quadratic Equations and Functions
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, 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(y-7)^2=12 because the left side was a perfect square.

(y−7)2=12y−7=±12y−7=±23y=7±23 \begin{array}{rcl} (y-7)^2 &=& 12 \\[4pt] 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 look at two examples to help us recognize the patterns.

(x+9)2(y−7)2(x+9)(x+9)(y−7)(y−7)x2+9x+9x+81y2−7y−7y+49x2+18x+81y2−14y+49 \begin{array}{cc} (x+9)^2 & (y-7)^2 \\ (x+9)(x+9) & (y-7)(y-7) \\ x^2+9x+9x+81 & y^2-7y-7y+49 \\ x^2+18x+81 & y^2-14y+49 \end{array}

We restate the patterns here for reference.

Binomial Squares Pattern. If aa and bb 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, a2+2ab+b2=(a+b)2a^2+2ab+b^2=(a+b)^2.

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 bb. But first we start with determining aa. Notice that the first term of x2+6xx^2+6x is a square, x2x^2. This tells us that a=xa=x.

What number, bb, when multiplied with 2x2x gives 6x6x? It would have to be 3, which is 12(6)\tfrac{1}{2}(6). So b=3b=3.

Now to complete the perfect square trinomial, we will find the last term by squaring bb, which is 32=93^2=9.

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

We can now factor.

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

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

Complete a 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 the (12b)2\left(\tfrac{1}{2}b\right)^2 to x2+bxx^2+bx.
  4. Factor the perfect square trinomial, writing it as a binomial squared.

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

(a) x2−26xx^2-26x   (b) y2−9yy^2-9y   (c) n2+12nn^2+\tfrac{1}{2}n

For (a), the coefficient of xx is −26-26.

Find (12b)2.(12(−26))2=(−13)2=169Add 169 to the binomial to complete the square.x2−26x+169Factor the perfect square trinomial.(x−13)2 \begin{array}{lrcl} \text{Find }\left(\tfrac{1}{2}b\right)^2. &&& \left(\tfrac{1}{2}(-26)\right)^2=(-13)^2=169 \\[10pt] \text{Add 169 to the binomial to complete the square.} &&& x^2-26x+169 \\[4pt] \text{Factor the perfect square trinomial.} &&& (x-13)^2 \end{array}

For (b), the coefficient of yy is −9-9.

Find (12b)2.(12(−9))2=(−92)2=814Add 814 to the binomial to complete the square.y2−9y+814Factor the perfect square trinomial.(y−92)2 \begin{array}{lrcl} \text{Find }\left(\tfrac{1}{2}b\right)^2. &&& \left(\tfrac{1}{2}(-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.} &&& y^2-9y+\tfrac{81}{4} \\[10pt] \text{Factor the perfect square trinomial.} &&& \left(y-\tfrac{9}{2}\right)^2 \end{array}

For (c), the coefficient of nn is 12\tfrac{1}{2}.

Find (12b)2.(12⋅12)2=(14)2=116Add 116 to the binomial to complete the square.n2+12n+116Rewrite as a binomial square.(n+14)2 \begin{array}{lrcl} \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.} &&& n^2+\tfrac{1}{2}n+\tfrac{1}{16} \\[10pt] \text{Rewrite as a binomial square.} &&& \left(n+\tfrac{1}{4}\right)^2 \end{array}

Complete the square for a2−20aa^2-20a. Enter the resulting binomial squared.

Complete the square for m2−5mm^2-5m. Enter the resulting binomial squared.

Complete the square for p2+14pp^2+\tfrac{1}{4}p. Enter the resulting binomial squared.

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 9 to both sides of the equation.

x2+6x=40x2+6x+__=40+__Add 9 to both sides to complete the square.x2+6x+9=40+9(x+3)2=49 \begin{array}{lrcl} &&& x^2+6x=40 \\[4pt] &&& x^2+6x+\_\_=40+\_\_ \\[4pt] \text{Add 9 to both sides to complete the square.} & x^2+6x+9 &=& 40+9 \\[4pt] &&& (x+3)^2=49 \end{array}

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. How to solve a quadratic equation of the form x2+bx+c=0x^2+bx+c=0 by completing the square. Solve by completing the square: x2+8x=48x^2+8x=48.

The variable terms are already isolated.x2+8x=48Take half of 8 and square it. Add 16 to both sides.x2+8x+16=48+16Factor the perfect square trinomial and add on the right.(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 already isolated.} & x^2+8x &=& 48 \\[4pt] \text{Take half of 8 and square it. Add 16 to both sides.} & x^2+8x+16 &=& 48+16 \\[4pt] \text{Factor the perfect square trinomial and add on the right.} & (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 in the original equation.

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

Solve by completing the square: x2+4x=5x^2+4x=5. Enter both solutions, separated by commas.

Solve by completing the square: y2−10y=−9y^2-10y=-9. Enter both solutions, separated by commas.

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

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 (12⋅b)2\left(\tfrac{1}{2}\cdot b\right)^2, the number needed to complete the square. Add it to both sides of the equation.
  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. Use the Square Root Property.
  5. Simplify the radical and then solve the two resulting equations.
  6. Check the solutions.

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

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

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

The two solutions are x=−2+17ix=-2+\sqrt{17}i and x=−2−17ix=-2-\sqrt{17}i. We leave the check to you.

Solve by completing the square: y2−10y=−35y^2-10y=-35. Enter both solutions, separated by commas.

Solve by completing the square: z2+8z=−19z^2+8z=-19. Enter both solutions, separated by commas.

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

Example. Solve by completing the square: y2−18y=−6y^2-18y=-6.

The variable terms are on the left. Take half of −18 and square it.y2−18y+(12(−18))2=−6(12(−18))2=81. Add 81 to both sides.y2−18y+81=−6+81Factor the perfect square trinomial.(y−9)2=75Use the Square Root Property.y−9=±75Simplify the radical.y−9=±53Solve for y.y=9±53 \begin{array}{lrcl} \text{The variable terms are on the left. Take half of }-18\text{ and square it.} & y^2-18y+\left(\tfrac{1}{2}(-18)\right)^2 &=& -6 \\[10pt] \left(\tfrac{1}{2}(-18)\right)^2=81.\ \text{Add 81 to both sides.} & y^2-18y+81 &=& -6+81 \\[4pt] \text{Factor the perfect square trinomial.} & (y-9)^2 &=& 75 \\[4pt] \text{Use the Square Root Property.} & y-9 &=& \pm\sqrt{75} \\[4pt] \text{Simplify the radical.} & y-9 &=& \pm5\sqrt{3} \\[4pt] \text{Solve for }y. & y &=& 9\pm5\sqrt{3} \end{array}

Check:

(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} (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}

Another way to check this would be to use a calculator. Evaluate y2−18yy^2-18y for both of the solutions. The answer should be −6-6.

Solve by completing the square: x2−16x=−16x^2-16x=-16. Enter both solutions, separated by commas.

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

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

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

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

Check:

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

Solve by completing the square: a2+4a+9=30a^2+4a+9=30. Enter both solutions, separated by commas.

Solve by completing the square: b2+8b−4=16b^2+8b-4=16. Enter both solutions, separated by commas.

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

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.(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±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}(-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.} & \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} \end{array}

The two solutions are n=32+532n=\tfrac{3}{2}+\tfrac{\sqrt{53}}{2} and n=32−532n=\tfrac{3}{2}-\tfrac{\sqrt{53}}{2}. We leave the check for you!

Solve by completing the square: p2=5p+9p^2=5p+9. Enter both solutions, separated by commas.

Solve by completing the square: q2=7q−3q^2=7q-3. Enter both solutions, separated by commas.

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=0a\cdot b=0, then a=0a=0 or b=0b=0.” Instead, we multiply the factors and then put the equation into standard form to continue by completing the square.

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

(x−3)(x+5)=9Multiply the binomials on the left.x2+2x−15=9Add 15 to isolate the constant terms on the right.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.(x+1)2=25Use the Square Root Property.x+1=±25Solve for x.x=−1±5Simplify.x=4, −6 \begin{array}{lrcl} &&& (x-3)(x+5)=9 \\[4pt] \text{Multiply the binomials on the left.} & x^2+2x-15 &=& 9 \\[4pt] \text{Add 15 to isolate the constant terms on the right.} & x^2+2x &=& 24 \\[4pt] \text{Take half of 2 and square it.} && \left(\tfrac{1}{2}(2)\right)^2=1 \\[10pt] \text{Add 1 to both sides.} & x^2+2x+1 &=& 24+1 \\[4pt] \text{Factor the perfect square trinomial.} & (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{Simplify.} & x &=& 4,\ -6 \end{array}

We leave the check for you!

Solve by completing the square: (c−2)(c+8)=11(c-2)(c+8)=11. Enter both solutions, separated by commas.

Solve by completing the square: (d−7)(d+3)=56(d-7)(d+3)=56. Enter both solutions, separated by commas.

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 coefficient of x2x^2 is 1, 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 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. Solve by completing the square: 3x2−12x−15=03x^2-12x-15=0.

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 3.3(x2−4x−5)3=03Simplify.x2−4x−5=0Add 5 to get the constant terms on the right side.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.(x−2)2=9Use the Square Root Property.x−2=±9Solve for x.x=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.} & \tfrac{3(x^2-4x-5)}{3} &=& \tfrac{0}{3} \\[10pt] \text{Simplify.} & x^2-4x-5 &=& 0 \\[4pt] \text{Add 5 to get the constant terms on the right side.} & x^2-4x &=& 5 \\[4pt] \text{Take half of }-4\text{ and square it.} && \left(\tfrac{1}{2}(-4)\right)^2=4 \\[10pt] \text{Add 4 to both sides.} & x^2-4x+4 &=& 5+4 \\[4pt] \text{Factor the perfect square trinomial.} & (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{Simplify.} & x &=& 5,\ -1 \end{array}

Check:

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

Solve by completing the square: 2m2+16m+14=02m^2+16m+14=0. Enter both solutions, separated by commas.

Solve by completing the square: 4n2−24n−56=84n^2-24n-56=8. Enter both solutions, separated by commas.

To complete the square, the coefficient of the x2x^2 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. Solve by completing the square: 2x2−3x=202x^2-3x=20.

To complete the square we need the coefficient of x2x^2 to be one. We will divide both sides of the equation by the coefficient of x2x^2. Then we can continue with solving the equation by completing the square.

2x2−3x=20Divide both sides by 2.2x2−3x2=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.(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±134Simplify.x=4, −52 \begin{array}{lrcl} &&& 2x^2-3x=20 \\[4pt] \text{Divide both sides by 2.} & \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}\left(-\tfrac{3}{2}\right)\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.} & \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{Simplify.} & x &=& 4,\ -\tfrac{5}{2} \end{array}

We leave the check for you!

Solve by completing the square: 3r2−2r=213r^2-2r=21. Enter both solutions, separated by commas.

Solve by completing the square: 4t2+2t=204t^2+2t=20. Enter both solutions, separated by commas.

Now that we have seen that the coefficient of x2x^2 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=0ax^2+bx+c=0.

Solve a quadratic equation of the form ax2+bx+c=0ax^2+bx+c=0 by completing the square.

  1. Divide by aa to make the coefficient of the x2x^2 term 1.
  2. Isolate the variable terms on one side and the constant terms on the other.
  3. Find (12⋅b)2\left(\tfrac{1}{2}\cdot b\right)^2, the number needed to complete the square. Add it to both sides of the equation.
  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. Use the Square Root Property.
  6. Simplify the radical and then solve the two resulting equations.
  7. Check the solutions.

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

Again, our first step will be to make the coefficient of x2x^2 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 3.3x2+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.(x+13)2=129+19Use the Square Root Property.x+13=±139Simplify the radical.x+13=±133Solve for x.x=−13±133 \begin{array}{lrcl} &&& 3x^2+2x=4 \\[4pt] \text{Divide both sides by 3.} & \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.} & \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} \end{array}

The two solutions are x=−13+133x=-\tfrac{1}{3}+\tfrac{\sqrt{13}}{3} and x=−13−133x=-\tfrac{1}{3}-\tfrac{\sqrt{13}}{3}. We leave the check for you!

Solve by completing the square: 4x2+3x=24x^2+3x=2. Enter both solutions, separated by commas.

Solve by completing the square: 3y2−10y=−53y^2-10y=-5. Enter both solutions, separated by commas.

Key terms

completing the square — a process used to make a perfect square trinomial, which transforms a quadratic equation into a form that can be solved using the Square Root Property.

This section is adapted from Intermediate Algebra 2e, Section 9.2: Solve Quadratic Equations by Completing the Square by Lynn Marecek and Andrea Honeycutt Mathis, © OpenStax, licensed under CC BY-NC-SA 4.0. Access the original for free at openstax.org. Changes: reformatted worked-example steps as accessible typeset mathematics; omitted the Be Prepared quiz, media links, and end-of-section exercises; and converted the practice problems (“Try Its”) into interactive exercises with instant feedback.

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