Quadratic Equations • Topic 1 of 4

Roots & Nature of Roots

The roots of ax² + bx + c = 0 are x = [−b ± √(b² − 4ac)] / 2a, and the quantity under the root, D = b² − 4ac, is the discriminant — it tells you the nature of the roots before you compute anything. If D > 0 there are two distinct real roots; if D = 0 the roots are real and equal (a perfect square, x = −b/2a); if D < 0 there are no real roots, just a complex conjugate pair. CAT loves to ask "for what value of k does this equation have equal/real roots?" — that is purely a discriminant condition, so set D = 0 or D ≥ 0 and solve for k, never the full formula. One more high-value fact: if a, b, c are rational and D is a perfect square, the roots are rational (the equation factorises); otherwise they are irrational and come in conjugate surd pairs p ± √q.

✅ Solved examples

1. Find the nature of the roots of 2x² − 4x + 5 = 0.
D = (−4)² − 4(2)(5) = 16 − 40 = −24 < 0 ⇒ no real roots (a complex conjugate pair).
2. For what value of k does x² − kx + 9 = 0 have equal roots?
Equal roots ⇒ D = 0 ⇒ k² − 4(1)(9) = 0 ⇒ k² = 36 ⇒ k = ±6.
3. Solve 3x² − 11x + 6 = 0 using the formula.
D = 121 − 72 = 49, √D = 7. x = (11 ± 7)/6 ⇒ x = 3 or x = 2/3.
4. For what values of k does x² + (k − 3)x + k = 0 have real roots?
Need D ≥ 0: (k − 3)² − 4k ≥ 0 ⇒ k² − 10k + 9 ≥ 0 ⇒ (k − 1)(k − 9) ≥ 0 ⇒ k ≤ 1 or k ≥ 9.

✏️ Practice — try these, take hints as needed

1. Nature of roots of x² − 6x + 9 = 0?
Compute D = b² − 4ac.
36 − 36.
D = 0 means what?
Real and equal (x = 3)
2. For x² + 4x + k = 0 to have real roots, find the range of k.
Need D ≥ 0.
16 − 4k ≥ 0.
Solve for k.
k ≤ 4
3. Solve 5x² + 6x + 1 = 0.
D = 36 − 20 = 16.
√D = 4.
x = (−6 ± 4)/10.
x = −1/5 or x = −1
4. For what k does kx² − 4x + 1 = 0 have equal roots (k ≠ 0)?
D = 0.
16 − 4k = 0.
Solve.
k = 4
5. Are the roots of x² − 5x + 7 = 0 real?
D = 25 − 28.
D = −3.
Sign of D?
No, D < 0 (complex roots)

📝 Topic test — 8 questions

Auto-graded with full solutions; saved to your dashboard. Use the calculator and formula sheet (top-right) any time.

Loading questions…