Three Ways to Solve a Quadratic

What is a quadratic?

A quadratic equation has the form

$ax^2 + bx + c = 0 \quad (a \neq 0)$

There are three standard methods to solve it. Knowing which to reach for is half the battle.

Method 1 — Factoring

Works cleanly when the roots are integers or simple fractions.

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

Find two numbers that multiply to 6 and add to −5: they are −2 and −3.

$x^2 - 5x + 6 = (x - 2)(x - 3) = 0$

$\therefore \quad x = 2 \quad \text{or} \quad x = 3$

Tip: Always check by substitution. $2^2 - 5(2) + 6 = 4 - 10 + 6 = 0$ ✓

Method 2 — Completing the Square

Useful when factoring doesn't work neatly, and essential for deriving the quadratic formula.

Example: Solve $x^2 - 6x + 7 = 0$

Step 1: Move the constant to the right. $x^2 - 6x = -7$

Step 2: Add $\left(\frac{b}{2}\right)^2 = \left(\frac{-6}{2}\right)^2 = 9$ to both sides. $x^2 - 6x + 9 = 2$

Step 3: Write the left side as a perfect square. $(x - 3)^2 = 2$

Step 4: Square root both sides. $x - 3 = \pm\sqrt{2} \implies x = 3 \pm \sqrt{2}$

Method 3 — The Quadratic Formula

Always works. Use it when the equation looks messy or the others fail.

$\boxed{x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}}$

Example: Solve $2x^2 + 3x - 2 = 0$

Here $a=2,\, b=3,\, c=-2$.

$x = \frac{-3 \pm \sqrt{9 + 16}}{4} = \frac{-3 \pm 5}{4}$

$x = \frac{2}{4} = \frac{1}{2} \quad \text{or} \quad x = \frac{-8}{4} = -2$

The Discriminant — $b^2 - 4ac$

The expression under the square root tells you the nature of the roots before you solve:

Which method to choose?

1. Look for whole-number factors first — if you spot them in ~5 seconds, factor.
2. Coefficient of $x^2$ is 1 and factoring fails? → Complete the square (or formula).
3. Anything else, or in an exam under time pressure? → Quadratic formula, always safe.

Practice problems

Try these without looking at the solutions:

1. $x^2 + 7x + 12 = 0$
2. $x^2 - 4x - 1 = 0$ (hint: completing the square)
3. $3x^2 - x - 2 = 0$