Quadratic Equations • Topic 2 of 4

The Quadratic Formula

When a quadratic will not factor with clean integers, the formula always works: for ax² + bx + c = 0, x = [ −b ± √(b² − 4ac) ] / (2a). The quantity under the root, D = b² − 4ac, is the discriminant, and it tells you how many real solutions exist before you compute anything: D > 0 gives two distinct real roots, D = 0 gives exactly one (a repeated) root, and D < 0 gives no real roots. That last fact is a favourite GRE Quantitative Comparison hook — if a quantity is defined as "a solution to" an equation whose discriminant is negative, the comparison may be "cannot be determined" or the setup may be impossible. Keep the ± as two separate cases and be careful with signs when b is negative: −b of −6 is +6. The on-screen calculator handles the square root, but if b² − 4ac is a perfect square you rarely need it — that is usually the exam quietly telling you the thing would have factored.

✅ Solved examples

1. Solve x² − 4x − 1 = 0.
a = 1, b = −4, c = −1. D = 16 − 4(1)(−1) = 20. x = [4 ± √20] / 2 = [4 ± 2√5] / 2 = 2 ± √5.
2. Solve 2x² + 3x − 2 = 0 with the formula.
D = 9 − 4(2)(−2) = 9 + 16 = 25, √25 = 5. x = [−3 ± 5] / 4, giving x = 2/4 = 1/2 or x = −8/4 = −2.
3. How many real solutions does x² + x + 1 = 0 have?
D = 1² − 4(1)(1) = 1 − 4 = −3 < 0. A negative discriminant means no real solutions.
4. For what value of k does x² + kx + 9 = 0 have exactly one real solution (k > 0)?
One real root means D = 0: k² − 4(1)(9) = 0, so k² = 36, k = 6 (taking the positive value).

✏️ Practice — try these, take hints as needed

1. Solve x² − 6x + 4 = 0.
a = 1, b = −6, c = 4.
D = 36 − 16 = 20.
x = [6 ± √20]/2 = 3 ± √5.
x = 3 + √5 or 3 − √5
2. Solve 3x² − 5x + 1 = 0 (leave the root in surd form).
D = 25 − 12 = 13.
x = [5 ± √13] / 6.
No further simplification.
x = (5 ± √13)/6
3. How many real roots does 4x² − 12x + 9 = 0 have?
Compute the discriminant.
D = 144 − 144.
D = 0.
Exactly one (x = 3/2)
4. Does x² + 2x + 5 = 0 have any real solution?
D = b² − 4ac.
4 − 20.
Negative discriminant.
No real solution
5. For which k does x² − kx + 16 = 0 have a repeated root (k > 0)?
Set D = 0.
k² − 64 = 0.
Take the positive root.
k = 8

📝 Topic test — 8 questions

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