The expression under the square root in the quadratic formula, $b^2 - 4ac$, is called the discriminant. Before solving anything, it tells you how many real solutions to expect.
Why it works
The formula is $x = \frac{-b \pm \sqrt{D}}{2a}$ with $D = b^2 - 4ac$. If $D > 0$, then $\pm\sqrt{D}$ gives two different numbers: two real solutions. If $D = 0$, then $\pm 0$ is just $0$: one solution, $x = -\frac{b}{2a}$. If $D < 0$, there is no real square root at all: no real solutions. On the graph these are a parabola crossing the $x$-axis twice, touching it at the vertex, or missing it entirely.
| $b^2 - 4ac$ | Real solutions | Graph |
|---|---|---|
| positive | two | crosses the $x$-axis twice |
| zero | one | touches at the vertex |
| negative | none | never meets the $x$-axis |
Key idea
$D = b^2 - 4ac$: positive means two real solutions, zero means one, negative means none.
Worked examples
Worked example: Count the solutions
How many real solutions does $3x^2 - 2x + 4 = 0$ have?
- $D = (-2)^2 - 4(3)(4) = 4 - 48 = -44$.
- Negative, so none.
Answer: No real solutions.
Worked example: Choose a value
For which $c$ does $x^2 + 8x + c = 0$ have exactly one solution?
- One solution means $D = 0$: $64 - 4c = 0$, so $c = 16$.
- Indeed $x^2 + 8x + 16 = (x + 4)^2$.
Answer: $c = 16$
Worked example: Two irrational solutions
Without solving, describe the solutions of $x^2 - 2x - 1 = 0$.
- $D = (-2)^2 - 4(1)(-1) = 4 + 4 = 8$.
- $8 > 0$, so there are two real solutions. Since $8$ is not a perfect square, $\sqrt{8}$ is irrational and so are the solutions ($1 \pm \sqrt{2}$).
Answer: Two real, irrational solutions.
A perfect-square discriminant ($0, 1, 4, 9, \ldots$) with whole-number $a$, $b$, $c$ means the solutions are rational, which is exactly when the quadratic factors nicely over the integers.
Common mistake
Thinking a negative discriminant means you made an arithmetic slip. It is a real result: the parabola never reaches the $x$-axis, so the equation has no real solutions. Check with the graph or a few values: $3x^2 - 2x + 4$ is positive for every $x$.