Solving quadratic equations

Three ways to solve a quadratic equation (factoring, completing the square and the quadratic formula) and how to choose between them.

Always start with $= 0$

Every method needs the equation in the form $ax^2 + bx + c = 0$. Move every term to one side first.

If $AB = 0$, then $A = 0$ or $B = 0$. So once the quadratic is factored, set each factor to zero.

Example 1●

Solve $2x^2 + 5x = 3$.

$2x^2 + 5x - 3 = 0$

Make one side zero.

$(2x - 1)(x + 3) = 0$

ac = −6; the numbers 6 and −1 multiply to −6 and add to 5.

$x = \tfrac12$ or $x = -3$

Example 2●●

Solve $x^2 + 6x - 2 = 0$ by completing the square.

$x^2 + 6x = 2$

Move the constant across.

$x^2 + 6x + 9 = 11$

Half of 6 is 3; add 3² = 9 to both sides.

$(x + 3)^2 = 11$

$x + 3 = \pm\sqrt{11}$, so $x = -3 \pm \sqrt{11}$.

Take the square root of both sides. Don't forget ±.

Quadratic formula

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

Example 3●●

Solve $3x^2 - 2x - 2 = 0$.

$a = 3$, $b = -2$, $c = -2$

$b^2 - 4ac = 4 + 24 = 28$

$x = \dfrac{2 \pm \sqrt{28}}{6} = \dfrac{2 \pm 2\sqrt{7}}{6} = \dfrac{1 \pm \sqrt{7}}{3}$

√28 = √4 × √7 = 2√7, then divide every term by 2.

$x \approx 1.22$ or $x \approx -0.55$

See the solutions

The solutions of the equation are the x-intercepts of y = ax² + bx + c. Set the sliders to Example 3 (a = 3, b = −2, c = −2) and compare.

Situation	Best method
Numbers that factor easily	Factoring (fastest)
Question says "exact" and $b$ is even	Completing the square or the formula
Anything else, or you are unsure	The quadratic formula (always works)

Flashcards

First step for any quadratic equation

Rearrange to $ax^2 + bx + c = 0$.

Null factor law

If $AB = 0$, then $A = 0$ or $B = 0$.

Never divide by x

You lose the solution $x = 0$. Factor instead.

The method that always works

The quadratic formula.

Practice