SAT Math: How Many Solutions Does a Quadratic Have?

b^{2} - 4 a c

Worked examples

$(- 5)^{2} - 4 (1) (6) = 1$

It's positive, so there are two real solutions.

$1^{2} - 4 (1) (1) = - 3$

It's negative, so there are no real solutions.

Set the discriminant to 0: $4^{2} - 4 (1) k = 0$ .

$16 = 4 k$ , so $k = 4$ .

For $2 x^{2} + 3 x - 1 = 0$ , the discriminant is $9 - 4 (2) (1) = 1$ .

The sign of $c$ matters. Here $c = - 1$ , so it's $9 - 4 (2) (- 1) = 17$ .

$x^{2} + 2 x + 5 = 0$ . How many real solutions?
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None. The discriminant is $- 16$ .
$x^{2} - 8 x + 16 = 0$ . How many real solutions?
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One. The discriminant is 0.
$x^{2} + k x + 9 = 0$ has one solution. Find $k$ .
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$k = 6$ or $k = - 6$

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