Solving Quadratic Equations

When $a = 1$: $x^2 + bx + c = 0$

Find two numbers that multiply to $c$ and add to $b$.

$x^2 + 7x + 12 = 0$ → $(x + 3)(x + 4) = 0$ → $x = -3$ or $x = -4$

When $a \neq 1$: $ax^2 + bx + c = 0$

Find two numbers that multiply to $ac$ and add to $b$, then split the middle term.

$2x^2 + 5x - 3 = 0$: $ac = -6$. Numbers: $6$ and $-1$. $2x^2 + 6x - x - 3 = 0$ → $2x(x + 3) - 1(x + 3) = 0$ → $(2x - 1)(x + 3) = 0$ $x = \frac{1}{2}$ or $x = -3$

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

Works for any quadratic, even when factorising is difficult.

Example

$3x^2 - 2x - 4 = 0$: $a=3, b=-2, c=-4$

$x = \frac{2 \pm \sqrt{4 + 48}}{6} = \frac{2 \pm \sqrt{52}}{6} = \frac{2 \pm 2\sqrt{13}}{6} = \frac{1 \pm \sqrt{13}}{3}$

$x^2 + bx + c = 0$ → $(x + \frac{b}{2})^2 - (\frac{b}{2})^2 + c = 0$

$x^2 + 6x + 2 = 0$ → $(x + 3)^2 - 9 + 2 = 0$ → $(x + 3)^2 = 7$ → $x = -3 \pm \sqrt{7}$

$\Delta = b^2 - 4ac$

• $\Delta > 0$: two distinct real roots
• $\Delta = 0$: one repeated root
• $\Delta < 0$: no real roots