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Completing the Square

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What if you had a quadratic equation like x^2 + 12x = 13 ? How could you solve it by taking the square root of both sides? After completing this Concept, you'll be able to complete the square to solve quadratic equations like this one.

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CK-12 Foundation: 1006S Solving Quadratic Equations by Completing the Square

Guidance

You saw in the last section that if you have a quadratic equation of the form (x - 2)^2 = 5 , you can easily solve it by taking the square root of each side:

x - 2 = \sqrt{5} \ \qquad \text{and} \qquad \ x - 2 = - \sqrt{5}

Simplify to get:

x = 2 + \sqrt{5} \approx 4.24 \ \qquad \text{and} \qquad \ x = 2 - \sqrt{5} \approx - 0.24

So what do you do with an equation that isn’t written in this nice form? In this section, you’ll learn how to rewrite any quadratic equation in this form by completing the square.

Complete the Square of a Quadratic Expression

Completing the square lets you rewrite a quadratic expression so that it contains a perfect square trinomial that you can factor as the square of a binomial.

Remember that the square of a binomial takes one of the following forms:

(x + a)^2 & = x^2 + 2ax + a^2\\(x - a)^2 & = x^2 - 2ax + a^2

So in order to have a perfect square trinomial, we need two terms that are perfect squares and one term that is twice the product of the square roots of the other terms.

Example A

Complete the square for the quadratic expression x^2 + 4x .

Solution

To complete the square we need a constant term that turns the expression into a perfect square trinomial. Since the middle term in a perfect square trinomial is always 2 times the product of the square roots of the other two terms, we re-write our expression as:

x^2 + 2(2)(x)

We see that the constant we are seeking must be 2^2:

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

Answer: By adding 4 to both sides, this can be factored as: (x + 2)^2

Notice, though, that we just changed the value of the whole expression by adding 4 to it. If it had been an equation, we would have needed to add 4 to the other side as well to make up for this.

Also, this was a relatively easy example because a , the coefficient of the x^2 term, was 1. When that coefficient doesn’t equal 1, we have to factor it out from the whole expression before completing the square.

Example B

Complete the square for the quadratic expression 4x^2 + 32x .

Solution

Factor the coefficient of the x^2 term:

4(x^2 + 8x)

Re-write the expression:

4 (x^2 + 2(4) (x))

We complete the square by adding the constant 4^2 :

4(x^2 + 2(4)(x) + 4^2)

Factor the perfect square trinomial inside the parenthesis:

4(x + 4)^2

The expression “completing the square” comes from a geometric interpretation of this situation. Let’s revisit the quadratic expression in Example 1: x^2 + 4x .

We can think of this expression as the sum of three areas. The first term represents the area of a square of side x . The second expression represents the areas of two rectangles with a length of 2 and a width of x :

We can combine these shapes as follows:

We obtain a square that is not quite complete. To complete the square, we need to add a smaller square of side length 2.

We end up with a square of side length (x + 2) ; its area is therefore (x + 2)^2 . Let’s demonstrate the method of completing the square with an example.

Example C

Solve the following quadratic equation: 3x^2 - 10x = -1

Solution

Divide all terms by the coefficient of the x^2 term:

x^2 - \frac{10}{3} x = - \frac{1}{3}

Rewrite: x^2 - 2 \left ( \frac{5}{3} \right ) (x) = - \frac{1}{3}

In order to have a perfect square trinomial on the right-hand-side we need to add the constant \left ( \frac{5}{3} \right )^2 . Add this constant to both sides of the equation:

x^2 - 2 \left ( \frac{5}{3} \right ) (x) + \left ( \frac{5}{3} \right )^2 = - \frac{1}{3} + \left ( \frac{5}{3} \right )^2

Factor the perfect square trinomial and simplify:

\left ( x - \frac{5}{3} \right )^2  & = - \frac{1}{3} + \frac{25}{9}\\\left (x - \frac{5}{3} \right )^2 & = \frac{22}{9}

Take the square root of both sides:

x - \frac{5}{3} &= \sqrt{\frac{22}{9}} && \text{and} && x - \frac{5}{3} = - \sqrt{ \frac{22}{9}}\\x &= \frac{5}{3} + \sqrt{\frac{22}{9}} \approx 3.23 && \text{and} && x = \frac{5}{3} - \sqrt{\frac{22}{9}} \approx 0.1

Answer: x = 3.23 and x = 0.1

Solving Quadratic Equations in Standard Form

If an equation is in standard form (ax^2 + bx + c = 0) , we can still solve it by the method of completing the square. All we have to do is start by moving the constant term to the right-hand-side of the equation.

Example D

Solve the following quadratic equation: x^2 + 15x + 12 = 0

Solution

Move the constant to the other side of the equation:

x^2 + 15x = -12

Rewrite: x^2 + 2 \left ( \frac{15}{2} \right )(x) = -12

Add the constant \left ( \frac{15}{2} \right )^2 to both sides of the equation:

x^2 + 2\left ( \frac{15}{2} \right )(x) + \left ( \frac{15}{2} \right )^2 = -12 + \left ( \frac{15}{2} \right )^2

Factor the perfect square trinomial and simplify:

\left (x + \frac{15}{2} \right )^2 & = -12 + \frac{225}{4}\\\left (x + \frac{15}{2} \right )^2 & = \frac{177}{4}

Take the square root of both sides:

x + \frac{15}{2} &= \sqrt{\frac{177}{4}} && \text{and} && x + \frac{15}{2} = - \sqrt{\frac{177}{4}}\\x &= - \frac{15}{2} + \sqrt{\frac{177}{4}} \approx - 0.85 && \text{and} && x = - \frac{15}{2} - \sqrt{\frac{177}{4}} \approx -14.15

Answer: x = -0.85 and x = -14.15

Watch this video for help with the Examples above.

CK-12 Foundation: 1006 Solving Quadratic Equations by Completing the Square

Vocabulary

  • A perfect square trinomial has the form a^2+2(ab)+b^2 , which factors into (a+b)^2 .

Guided Practice

Solve the following quadratic equation: -x^2 +22x = 5

Solution

Divide all terms by the coefficient of the x^2 term:

x^2 -22x = -6

Rewrite: x^2 - 2(11) (x) = - 6.

In order to have a perfect square trinomial on the right-hand-side we need to add the constant \left (11 \right )^2 . Add this constant to both sides of the equation:

x^2 - 2  ( 11 ) (x) + ( 11)^2 = - 6 + (11)^2

Factor the perfect square trinomial and simplify:

\left ( x - 11 \right )^2  & =- 6 + (11)^2 \\\left (x - \frac{5}{3} \right )^2 & = 16

Take the square root of both sides:

x - 11 &= \sqrt{16} && \text{and} && x - 11 = - \sqrt{ 16}\\x &= 11 + \sqrt{16} =15 && \text{and} && x =11 - \sqrt{4}= 7

Answer: x = 15 and x =7

Practice

Complete the square for each expression.

  1. x^2 + 5x
  2. x^2 - 2x
  3. x^2 + 3x
  4. x^2 - 4x
  5. 3x^2 + 18x
  6. 2x^2 - 22x
  7. 8x^2 - 10x
  8. 5x^2 + 12x

Solve each quadratic equation by completing the square.

  1. x^2 - 4x = 5
  2. x^2 - 5x = 10
  3. x^2 + 10x + 15 = 0
  4. x^2 + 15x + 20 = 0
  5. 2x^2 - 18x = 3
  6. 4x^2 + 5x = -1
  7. 10x^2 - 30x - 8 = 0
  8. 5x^2 + 15x - 40 = 0

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