The Quadratic Formula and Discriminant
The Quadratic Formula
The quadratic formula is the universal tool for solving any quadratic equation, even those that cannot be factored with integers. Derived by completing the square on the general form ax² + bx + c = 0, the formula gives both solutions in one compact expression.
The discriminant, the expression b² − 4ac under the square root, tells you the nature of the solutions before you even compute them: positive means two real solutions, zero means one repeated real solution, and negative means two complex solutions.
This lesson teaches you to apply the quadratic formula accurately, interpret the discriminant, and decide when factoring or the formula is the better choice.
For $ax^2 + bx + c = 0$:
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
The Discriminant
| $\Delta$ | Nature of Solutions |
|---|---|
| $\Delta > 0$ | Two distinct real solutions |
| $\Delta = 0$ | One repeated real solution |
| $\Delta < 0$ | Two complex (non-real) solutions |
$3x^2 - 2x - 5 = 0$
$a = 3$, $b = -2$, $c = -5$.
$\Delta = 4 + 60 = 64$. $\;x = \dfrac{2 \pm 8}{6}$.
$x = \dfrac{10}{6} = \dfrac{5}{3}$ or $x = \dfrac{-6}{6} = -1$.
$x^2 + 4x + 7 = 0$
$\Delta = 16 - 28 = -12 < 0$. No real solutions.
$x^2 - 6x + 9 = 0$
$\Delta = 36 - 36 = 0$. One repeated solution: $x = \dfrac{6}{2} = 3$.
Completing the Square
To make $x^2 + bx$ a perfect square, add $\left(\dfrac{b}{2}\right)^2$ to both sides.
$x^2 + 6x + 5 = 0$
- $x^2 + 6x = -5$
- Add $\left(\dfrac{6}{2}\right)^2 = 9$: $x^2 + 6x + 9 = 4$
- $(x + 3)^2 = 4$
- $x + 3 = \pm 2 \;\Rightarrow\; x = -1$ or $x = -5$
$2x^2 - 8x + 1 = 0$
$a = 2$, $b = -8$, $c = 1$. $\Delta = 64 - 8 = 56$.
$x = \dfrac{8 \pm \sqrt{56}}{4} = \dfrac{8 \pm 2\sqrt{14}}{4} = \dfrac{4 \pm \sqrt{14}}{2}$.
Practice Problems
Show Answer Key
1. $x = 5$ or $x = -1$
2. $x = \dfrac{1}{2}$ or $x = -2$
3. $\Delta = 25 - 28 = -3 < 0$; no real solutions
4. $(x-1)^2 = 16$; $x = 5$ or $x = -3$
5. $x = -2$ or $x = -6$
6. $x = \dfrac{4}{3}$ or $x = -1$
7. $\Delta = 144 - 144 = 0$; one repeated root $x = \dfrac{3}{2}$
8. $x = -5$ (repeated)
9. $x = \dfrac{3 \pm \sqrt{5}}{2}$
10. $x = \dfrac{-2 \pm \sqrt{24}}{10} = \dfrac{-1 \pm \sqrt{6}}{5}$
11. $\Delta = 1 - 4 = -3 < 0$; two complex solutions
12. $(x-2)^2 = 3$; $x = 2 \pm \sqrt{3}$