9.2 Solve Quadratic Equations by Completing the Square

Completing the Square

Completing the square transforms a quadratic equation into a form you can solve by taking a square root. It works on any quadratic equation, even ones that don't factor neatly. The technique also converts quadratics into vertex form, which you'll need for graphing parabolas.

Binomial to Perfect Square Trinomial

The core idea behind completing the square is turning a binomial like $x^2 + bx$ into a perfect square trinomial, which is an expression that factors as a single binomial squared.

Here's how: take half the coefficient of $x$, then square it. Add that value to the expression.

• Start with $x^2 + bx$
• Compute $\left(\frac{b}{2}\right)^2$
• Add it: $x^2 + bx + \left(\frac{b}{2}\right)^2 = \left(x + \frac{b}{2}\right)^2$

Example: For $x^2 + 6x$, half of 6 is 3, and $3^2 = 9$. So $x^2 + 6x + 9 = (x + 3)^2$.

This works because $(x + 3)^2 = x^2 + 6x + 9$ when you expand it. You're just reverse-engineering that pattern.

Completing the Square with Leading Coefficient 1

When the equation has the form $x^2 + bx + c = 0$ (leading coefficient already 1), follow these steps:

1. Move the constant to the right side. Subtract $c$ from both sides: $x^2 + bx = -c$

2. Complete the square. Calculate $\left(\frac{b}{2}\right)^2$ and add it to both sides: $x^2 + bx + \left(\frac{b}{2}\right)^2 = -c + \left(\frac{b}{2}\right)^2$

3. Factor the left side as a perfect square: $\left(x + \frac{b}{2}\right)^2 = -c + \left(\frac{b}{2}\right)^2$

4. Take the square root of both sides (don't forget the $\pm$): $x + \frac{b}{2} = \pm\sqrt{-c + \left(\frac{b}{2}\right)^2}$

5. Isolate $x$ by subtracting $\frac{b}{2}$: $x = -\frac{b}{2} \pm \sqrt{-c + \left(\frac{b}{2}\right)^2}$

Example: Solve $x^2 + 6x + 5 = 0$

1. $x^2 + 6x = -5$
2. Half of 6 is 3, and $3^2 = 9$. Add 9 to both sides: $x^2 + 6x + 9 = -5 + 9$
3. $(x + 3)^2 = 4$
4. $x + 3 = \pm 2$
5. $x = -3 + 2 = -1$ or $x = -3 - 2 = -5$

Completing the Square for Any Leading Coefficient

When $a \neq 1$ in $ax^2 + bx + c = 0$, you need one extra step at the start: divide everything by $a$.

1. Divide both sides by $a$ to make the leading coefficient 1: $x^2 + \frac{b}{a}x + \frac{c}{a} = 0$

2. Move the constant to the right side: $x^2 + \frac{b}{a}x = -\frac{c}{a}$

3. Complete the square. Calculate $\left(\frac{b}{2a}\right)^2$ and add it to both sides: $x^2 + \frac{b}{a}x + \left(\frac{b}{2a}\right)^2 = -\frac{c}{a} + \left(\frac{b}{2a}\right)^2$

4. Factor the left side: $\left(x + \frac{b}{2a}\right)^2 = -\frac{c}{a} + \left(\frac{b}{2a}\right)^2$

5. Take the square root of both sides: $x + \frac{b}{2a} = \pm\sqrt{-\frac{c}{a} + \left(\frac{b}{2a}\right)^2}$

6. Isolate $x$: $x = -\frac{b}{2a} \pm \sqrt{-\frac{c}{a} + \left(\frac{b}{2a}\right)^2}$

Example: Solve $2x^2 + 12x + 7 = 0$

1. Divide by 2: $x^2 + 6x + \frac{7}{2} = 0$
2. $x^2 + 6x = -\frac{7}{2}$
3. Half of 6 is 3, and $3^2 = 9$. Add 9 to both sides: $x^2 + 6x + 9 = -\frac{7}{2} + 9 = \frac{11}{2}$
4. $(x + 3)^2 = \frac{11}{2}$
5. $x + 3 = \pm\sqrt{\frac{11}{2}} = \pm\frac{\sqrt{22}}{2}$
6. $x = -3 \pm \frac{\sqrt{22}}{2}$

A common mistake here: students divide only some terms by $a$ instead of every term. Make sure you divide the entire equation.

Other Methods for Solving Quadratic Equations

Completing the square always works, but it's not always the fastest route. Here's how it compares to the other methods:

• Factoring: Quickest when the quadratic factors cleanly over integers (like $x^2 + 6x + 5 = (x+5)(x+1)$). Not every quadratic factors neatly, though.
• Quadratic formula: Works on any quadratic, just like completing the square. The formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ is actually derived from completing the square on the general equation $ax^2 + bx + c = 0$.
• Discriminant: The expression $b^2 - 4ac$ (the part under the square root in the quadratic formula) tells you what kind of solutions to expect before you solve:
  • $b^2 - 4ac > 0$: two distinct real solutions
  • $b^2 - 4ac = 0$: one repeated real solution
  • $b^2 - 4ac < 0$: two complex (non-real) solutions