Quadratic equations for NDA

Quadratic equations are a small chapter with a long reach. The same few results about α and β turn up inside questions on complex numbers, sequences, coordinate geometry and maxima and minima. NDA questions on the chapter itself rarely ask you to solve an equation outright. They ask about the roots without finding them: their sum, their nature, an expression in them, or the value of a constant that makes them behave in some way.

Typical question types are:

• nature of the roots from the discriminant;
• the value of α² + β², α³ + β³, 1/α + 1/β and similar expressions;
• forming an equation whose roots are related to the roots of a given one;
• conditions: equal roots, one root double the other, roots of opposite sign;
• the least or greatest value of a quadratic expression;
• solving a quadratic inequality, or an equation that becomes quadratic after substitution.

The core results

For ax² + bx + c = 0 (a ≠ 0) with roots α and β:

• roots: x = (−b ± √D)/(2a), where D = b² − 4ac;
• sum: α + β = −b/a;
• product: αβ = c/a.

The formula comes from completing the square. Divide by a, move c/a across, and add (b/2a)² to both sides: (x + b/2a)² = (b² − 4ac)/4a². Taking square roots gives the formula.

Conversely, an equation with roots α and β is x² − (α + β)x + αβ = 0.

Nature of the roots

Assume a, b and c are real.

Two consequences are worth remembering. With real coefficients, non-real roots always come as a conjugate pair p ± iq. With rational coefficients, irrational roots come in pairs p ± √q.

Symmetric expressions

These let you evaluate expressions in α and β using only the sum and product:

• α² + β² = (α + β)² − 2αβ
• (α − β)² = (α + β)² − 4αβ, so the difference of the roots is √D divided by the absolute value of a
• α³ + β³ = (α + β)³ − 3αβ(α + β)
• 1/α + 1/β = (α + β)/αβ
• α²β + αβ² = αβ(α + β)

Equations with transformed roots

Starting from ax² + bx + c = 0 with roots α and β:

Conditions on the roots

• Equal roots: D = 0.
• One root double the other: 2b² = 9ac. (Let the roots be r and 2r: then 3r = −b/a and 2r² = c/a; eliminating r gives the condition.)
• Roots of opposite sign: c/a < 0.
• Both roots positive: D ≥ 0, −b/a > 0 and c/a > 0.
• Roots reciprocal to each other: c = a.

Sign and extreme values

The graph of y = ax² + bx + c is a parabola with its vertex at x = −b/2a, where the expression takes its extreme value, −D/4a.

• If a > 0, the parabola opens upward and −D/4a is the least value.
• If a < 0, it opens downward and −D/4a is the greatest value.
• If a > 0 and D < 0, the expression is positive for every real x.

Inequalities. If the roots are p < q and a > 0, then ax² + bx + c < 0 exactly when p < x < q, and > 0 when x < p or x > q.

Worked NDA-style MCQs

Q1. If α and β are the roots of 2x² − 3x − 4 = 0, then α³ + β³ equals:
(a) 99/8 (b) 27/8 (c) 45/8 (d) 63/8

α + β = 3/2 and αβ = −2. α³ + β³ = (3/2)³ − 3(−2)(3/2) = 27/8 + 9 = 99/8. Answer: (a).

Q2. For what values of k does x² − 2(k + 1)x + k² = 0 have real roots?
(a) k ≥ −1/2 (b) k ≤ −1/2 (c) k > 1 (d) all real k

D = 4(k + 1)² − 4k² = 4(2k + 1). Real roots need D ≥ 0, so k ≥ −1/2. Answer: (a).

Q3. If one root of x² − 6x + k = 0 is double the other, then k is:
(a) 6 (b) 8 (c) 9 (d) 12

Using 2b² = 9ac: 2 × 36 = 9k, so k = 8. Check: the roots are 2 and 4, whose sum is 6 and product 8. Answer: (b).

Q4. The equation whose roots are the reciprocals of the roots of 3x² − 5x + 2 = 0 is:
(a) 2x² − 5x + 3 = 0 (b) 2x² + 5x + 3 = 0 (c) 3x² + 5x − 2 = 0 (d) 3x² − 2x + 5 = 0

Swap a and c: 2x² − 5x + 3 = 0. Check: the original roots are 1 and 2/3, so the new ones are 1 and 3/2, with sum 5/2 and product 3/2. Answer: (a).

Q5. If p and q are rational and 2 + √3 is a root of x² + px + q = 0, then p + q equals:
(a) −3 (b) 3 (c) −5 (d) 5

The other root is 2 − √3. Sum = 4, so p = −4; product = 4 − 3 = 1, so q = 1. p + q = −3. Answer: (a).

Q6. The least value of x² − 6x + 13 for real x is:
(a) 13 (b) 9 (c) 4 (d) 3

a = 1 > 0, so the least value is −D/4a. D = 36 − 52 = −16, giving 16/4 = 4, reached at x = 3. Answer: (c).

Common mistakes

• Sign of the sum. α + β is −b/a, not b/a.
• Forgetting a ≠ 0. For kx² + 4x + 1 = 0 to be a quadratic, k cannot be 0.
• Reversing an inequality. (x − 2)(x − 5) < 0 means x lies between the roots, not outside them.
• Using D ≥ 0 when the question says "distinct". Distinct real roots need D > 0.
• Least or greatest? Check the sign of a before calling −D/4a a minimum.

Practice set

1. If α and β are the roots of x² − 4x + 1 = 0, then 1/α² + 1/β² equals: (a) 12 (b) 14 (c) 16 (d) 18
2. kx² + 4x + 1 = 0 has real and distinct roots when: (a) k < 4, k ≠ 0 (b) k > 4 (c) k ≤ 4 (d) k = 4
3. The equation whose roots are the squares of the roots of x² − 3x + 2 = 0 is: (a) x² − 5x + 4 = 0 (b) x² − 9x + 4 = 0 (c) x² + 5x + 4 = 0 (d) x² − 5x − 4 = 0
4. If the roots of x² + bx + 12 = 0 differ by 1, then b is: (a) ±5 (b) ±6 (c) ±7 (d) ±8
5. The greatest value of 5 + 4x − x² is: (a) 5 (b) 7 (c) 9 (d) 11
6. The solution set of x² − x − 6 ≥ 0 is: (a) −2 ≤ x ≤ 3 (b) x ≤ −2 or x ≥ 3 (c) x ≥ 3 (d) x ≤ −3 or x ≥ 2
7. The roots of x² − 2x + 5 = 0 are: (a) 1 ± 2i (b) −1 ± 2i (c) 2 ± i (d) 1 ± 4i
8. If α and β are the roots of x² − px + q = 0, then α²β + αβ² equals: (a) p + q (b) pq (c) p²q (d) −pq

Answers:

1. (b) 14. (α² + β²)/(αβ)² = (16 − 2)/1 = 14.
2. (a). D = 16 − 4k > 0 gives k < 4, and k ≠ 0 for a quadratic.
3. (a). The roots are 1 and 2; the squares are 1 and 4, with sum 5 and product 4.
4. (c) ±7. (α − β)² = b² − 48 = 1, so b² = 49. The roots are 3 and 4, or −3 and −4.
5. (c) 9. At x = 2: 5 + 8 − 4 = 9.
6. (b). (x − 3)(x + 2) ≥ 0, so x lies outside the roots −2 and 3.
7. (a). D = 4 − 20 = −16, so x = (2 ± 4i)/2 = 1 ± 2i.
8. (b) pq. αβ(α + β) = q × p.

What to do next

• Write the symmetric-expression list and the transformed-roots table from memory.
• Solve 30 questions from old NDA papers on this chapter, timed at 60 seconds each.
• Revise complex numbers for the D < 0 case, then move on to sequences and series.