Example

Using the Discriminant to Determine the Number of Solutions of Four Quadratic Equations

Use the discriminant b2−4acb^2 - 4ac to determine how many real solutions each of the following quadratic equations has, without fully solving them.

ⓐ 2v2−3v+6=02v^2 - 3v + 6 = 0

The equation is already in standard form. Identify a=2a = 2, b=−3b = -3, c=6c = 6. Compute the discriminant:

(−3)2−4⋅2⋅6=9−48=−39(-3)^2 - 4 \cdot 2 \cdot 6 = 9 - 48 = -39

Because the discriminant is negative, there are no real solutions.

ⓑ 3x2+7x−9=03x^2 + 7x - 9 = 0

Identify a=3a = 3, b=7b = 7, c=−9c = -9. Compute the discriminant:

(7)2−4⋅3⋅(−9)=49+108=157(7)^2 - 4 \cdot 3 \cdot (-9) = 49 + 108 = 157

Because the discriminant is positive, there are two real solutions.

ⓒ 5n2+n+4=05n^2 + n + 4 = 0

Identify a=5a = 5, b=1b = 1, c=4c = 4. Compute the discriminant:

(1)2−4⋅5⋅4=1−80=−79(1)^2 - 4 \cdot 5 \cdot 4 = 1 - 80 = -79

Because the discriminant is negative, there are no real solutions.

ⓓ 9y2−6y+1=09y^2 - 6y + 1 = 0

Identify a=9a = 9, b=−6b = -6, c=1c = 1. Compute the discriminant:

(−6)2−4⋅9⋅1=36−36=0(-6)^2 - 4 \cdot 9 \cdot 1 = 36 - 36 = 0

Because the discriminant is zero, there is one real solution.

This example illustrates all three possible outcomes of the discriminant in a single exercise. When b2−4acb^2 - 4ac is negative (parts ⓐ and ⓒ), the square root of a negative number is not real, so the Quadratic Formula produces no real solutions. When b2−4acb^2 - 4ac is positive (part ⓑ), the ±\pm in the formula yields two distinct values. When b2−4acb^2 - 4ac equals zero (part ⓓ), the ±\pm contributes nothing, and the formula produces exactly one solution.

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Updated 2026-04-21

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