The discriminant on the SAT: How It Is Tested

What the discriminant is

For a quadratic in standard form, $ax^2 + bx + c = 0$, the discriminant is $b^2 - 4ac$.

It is the part under the square root in the quadratic formula, $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. That square root is why it decides how many solutions there are: a positive number has two square roots, zero has one, and a negative number has none that are real.

The SAT does not give you this. It is not on the reference sheet, so it has to be memorized.

The three cases

Greater than 0: two real solutions. The parabola crosses the x-axis twice. For $x^2 - 6x + 5 = 0$, the discriminant is $36 - 20 = 16$, and the solutions are $x = 1$ and $x = 5$.

Equal to 0: exactly one real solution. The parabola touches the x-axis at its vertex and turns back. For $x^2 - 6x + 9 = 0$, the discriminant is $36 - 36 = 0$, and the only solution is $x = 3$.

Less than 0: no real solutions. The parabola never reaches the x-axis. For $x^2 - 6x + 10 = 0$, the discriminant is $36 - 40 = -4$.

How the SAT asks it

"Exactly one solution. What is the value of $c$?" Set $b^2 - 4ac = 0$ and solve for the constant. That is the question on this page.

"Which equation has no real solutions?" Work out $b^2 - 4ac$ for each choice and pick the one that comes out negative.

A line and a parabola that meet at exactly one point. This is the same question in disguise. Set the two equations equal, move everything to one side, and set the discriminant to 0. For $y = x^2 + 3$ and $y = 2x + k$: $x^2 - 2x + 3 - k = 0$, so $4 - 4(3 - k) = 0$ and $k = 2$.

Mistakes that cost the point

Reading $a$, $b$ and $c$ before everything is on one side. $x^2 = 6x - 9$ has $b = -6$ and $c = 9$ only after you rewrite it as $x^2 - 6x + 9 = 0$.

Dropping a sign. When $a$ is negative, $-4ac$ changes sign. In the question on this page $a = -3$, so $-4(-3)(c)$ is $+12c$.

Squaring a negative $b$ wrong. $b^2$ is never negative: $(-6)^2 = 36$.

The Desmos way

Type the quadratic as $y =$ with a slider for the constant and count where the graph meets the x-axis. For "exactly one solution," drag the slider until the vertex just touches the x-axis and read off the value. It is a good check on the algebra even when you solve it by hand.