Lesson Objectives
• Demonstrate an understanding of the square root property
• Learn how to solve a quadratic equation by completing the square

## How to Solve Quadratic Equations by Completing the Square

In our last lesson, we learned how to use the square root property to solve equations such as: $$x^2 = k$$ $$(ax + b)^2 = k$$ Recall the square root property tells us: $$x^2 = k$$ $$x = \pm \sqrt{k}$$ In most cases, we cannot use the square root property right away. Let's suppose we saw an equation such as: $$x^2 + 20x + 75 = 0$$ In its current state, we cannot use the square root property. What we need to do is perform a procedure known as completing the square. When we complete the square, we are transforming one side of the equation into a perfect square trinomial. This perfect square trinomial can then be factored into a binomial squared. Once this is done, we can use the square root property to solve our equation.

### Perfect Square Trinomial

$$x^2 + 2xy + y^2 = (x + y)^2$$ $$x^2 - 2xy + y^2 = (x - y)^2$$

### Completing the Square

• Write the quadratic equation where the terms ax2 and bx are on one side, the constant will be on the other
• Make sure the coefficient (a) for the squared term (ax2) is 1
• If the coefficient (a) is not 1, we will just divide both sides of the equation by (a) the coefficient of the squared term
• Complete the square by adding one-half of the coefficient (b) of the first-degree term (bx) squared to both sides of the equation
• Solve the equation using the square root property
Let's revisit our earlier example: $$x^2 + 20x + 75 = 0$$ Step 1) We want our ax2 and bx terms on one side and the constant on the other.
We will subtract 75 away from each side of the equation: $$x^2 + 20x = -75$$ Step 2) Make sure the coefficient for the squared term is 1.
In this case, we have x2, which is the same as 1x2 $$x^2 + 20x = -75$$ Step 3) Complete the square » add one-half of the coefficient of the first-degree term squared to both sides.
Our first-degree term is 20x, therefore, the coefficient is 20. We want to multiply this number by 1/2 or divide by 2: $$20 \cdot \frac{1}{2} = 10$$ Now square the result (10): $$10^2 = 100$$ Note we can do this in one step: $$\left(20 \cdot \frac{1}{2} \right)^2 = 10^2 = 100$$ Now we add this value of 100 to both sides of the equation: $$x^2 + 20x + 100 = -75 + 100$$ Simplify: $$x^2 + 20x + 100 = 25$$ Step 4) Solve the equation.
The left side of the equation is now a perfect square trinomial. This means we can factor the trinomial into a binomial squared: $$x^2 + 20x + 100 = 25$$ $$(x + 10)^2 = 25$$ Now we can use our square root property to solve the equation: $$\sqrt{(x + 10)^2} = \pm \sqrt{25}$$ $$x + 10 = \pm 5$$ This leads to two equations to solve: $$x + 10 = 5$$ $$x = -5$$ $$x + 10 = -5$$ $$x = -15$$ Our two solutions: $$x = -5 \hspace{.5em} or \hspace{.5em} x = -15$$ Let's look at a few more examples.
Example 1: Solve each equation $$4x^2 - 16x - 20 = 0$$ Step 1) We want our ax2 and bx terms on one side and the constant on the other.
We will add 20 to each side of the equation: $$4x^2 - 16x = 20$$ Step 2) Make sure the coefficient for the squared term is 1.
In this case, we have 4x2, which means we have a coefficient of 4. We will divide each part of the equation by 4: $$\frac{4}{4}x^2 - \frac{16}{4}x = \frac{20}{4}$$ $$x^2 - 4x = 5$$ Step 3) Complete the square » add one-half of the coefficient of the first-degree term squared to both sides.
Our first-degree term is 4x, therefore, the coefficient is 4. We want to multiply this number by 1/2 or divide by 2: $$4 \cdot \frac{1}{2} = 2$$ Now square the result (2): $$2^2 = 4$$ Now we add this value of 4 to both sides of the equation: $$x^2 - 4x + 4 = 5 + 4$$ Simplify: $$x^2 - 4x + 4 = 9$$ Step 4) Solve the equation.
The left side of the equation is now a perfect square trinomial. This means we can factor the trinomial into a binomial squared: $$x^2 - 4x + 4 = 9$$ $$(x - 2)^2 = 9$$ Now we can use our square root property to solve the equation: $$\sqrt{(x - 2)^2} = \pm \sqrt{9}$$ $$x - 2 = \pm 3$$ This leads to two equations to solve: $$x - 2 = 3$$ $$x = 5$$ $$x - 2 = -3$$ $$x = -1$$ Our two solutions: $$x = 5 \hspace{.5em} or \hspace{.5em} x = -1$$ Example 2: Solve each equation $$8x^2 + 16x - 90 = 0$$ Step 1) We want our ax2 and bx terms on one side and the constant on the other.
We will add 90 to each side of the equation: $$8x^2 + 16x = 90$$ Step 2) Make sure the coefficient for the squared term is 1.
In this case, we have 8x2, which means we have a coefficient of 8. We will divide each part of the equation by 8: $$\frac{8}{8}x^2 + \frac{16}{8}x = \frac{90}{8}$$ $$x^2 + 2x = \frac{45}{4}$$ Step 3) Complete the square » add one-half of the coefficient of the first-degree term squared to both sides.
Our first-degree term is 2x, therefore, the coefficient is 2. We want to multiply this number by 1/2 or divide by 2: $$2 \cdot \frac{1}{2} = 1$$ Now square the result (2): $$1^2 = 1$$ Now we add this value of 1 to both sides of the equation: $$x^2 + 2x + 1 = \frac{45}{4} + 1$$ Simplify: $$x^2 + 2x + 1 = \frac{45}{4} + \frac{4}{4}$$ $$x^2 + 2x + 1 = \frac{49}{4}$$ Step 4) Solve the equation.
The left side of the equation is now a perfect square trinomial. This means we can factor the trinomial into a binomial squared: $$x^2 + 2x + 1 = \frac{49}{4}$$ $$(x + 1)^2 = \frac{49}{4}$$ Now we can use our square root property to solve the equation: $$\sqrt{(x + 1)^2} = \pm \sqrt{\frac{49}{4}}$$ $$x + 1 = \pm \frac{7}{2}$$ This leads to two equations to solve: $$x + 1 = \frac{7}{2}$$ $$x = \frac{7}{2} - 1$$ $$x = \frac{7}{2} - \frac{2}{2}$$ $$x = \frac{5}{2}$$ $$x + 1 = -\frac{7}{2}$$ $$x = -\frac{7}{2} - 1$$ $$x = \frac{-7}{2} - \frac{2}{2}$$ $$x = \frac{-9}{2}$$ $$x = -\frac{9}{2}$$ Our two solutions: $$x = \frac{5}{2} \hspace{.5em} or \hspace{.5em} x = -\frac{9}{2}$$ Example 3: Solve each equation $$4x^2 + 13x - 20 = -8$$ Step 1) We want our ax2 and bx terms on one side and the constant on the other.
We will add 20 to each side of the equation: $$4x^2 + 13x = 12$$ Step 2) Make sure the coefficient for the squared term is 1.
In this case, we have 4x2, which means we have a coefficient of 4. We will divide each part of the equation by 4: $$\frac{4}{4}x^2 + \frac{13}{4}x = \frac{12}{4}$$ $$x^2 + \frac{13}{4}x = 3$$ Step 3) Complete the square » add one-half of the coefficient of the first-degree term squared to both sides.
Our first-degree term is (13/4)x, therefore, the coefficient is 13/4. We want to multiply this number by 1/2 or divide by 2: $$\frac{13}{4} \cdot \frac{1}{2} = \frac{13}{8}$$ Now square the result (2): $$\left(\frac{13}{8}\right)^2 = \frac{169}{64}$$ Now we add this value of 169/64 to both sides of the equation: $$x^2 + \frac{13}{4}x + \frac{169}{64} = 3 + \frac{169}{64}$$ Simplify: $$x^2 + \frac{13}{4}x + \frac{169}{64} = \frac{192}{64} + \frac{169}{64}$$ $$x^2 + \frac{13}{4}x + \frac{169}{64} = \frac{361}{64}$$ Step 4) Solve the equation.
The left side of the equation is now a perfect square trinomial. This means we can factor the trinomial into a binomial squared: $$x^2 + \frac{13}{4}x + \frac{169}{64} = \frac{361}{64}$$ $$\left(x + \frac{13}{8}\right)^2 = \frac{361}{64}$$ Now we can use our square root property to solve the equation: $$\sqrt{\left(x + \frac{13}{8}\right)^2} = \pm \sqrt{\frac{361}{64}}$$ $$x + \frac{13}{8} = \pm \frac{19}{8}$$ This leads to two equations to solve: $$x + \frac{13}{8} = \frac{19}{8}$$ $$x = \frac{19}{8} - \frac{13}{8}$$ $$x = \frac{6}{8}$$ $$x = \frac{3}{4}$$ $$x + \frac{13}{8} = -\frac{19}{8}$$ $$x = -\frac{19}{8} - \frac{13}{8}$$ $$x = -\frac{32}{8}$$ $$x = -4$$ Our two solutions: $$x = \frac{3}{4} \hspace{.5em} or \hspace{.5em} x = -4$$