Quadratic Equations: JEE Advanced PYQ, 2007–2026

Answer: (A)

Put t = |x − 1|, so t ≥ 0. The equation becomes t² + t − 6 = 0, which factorises as (t + 3)(t − 2) = 0, so t = 2 or t = −3. Since a modulus cannot be negative, only t = 2 is valid. Then |x − 1| = 2 gives x − 1 = 2 or x − 1 = −2, so x = 3 or x = −1. The sum of squares is 3² + (−1)² = 10. Do not accept t = −3 (that adds a false solution, giving 13). Do not drop the negative branch of the modulus (that gives 9). Do not report the sum of the solutions, 2.

Answer: (A)

For roots α + 1/β and β + 1/α, find the sum and the product in terms of S = α+β = 4 and P = αβ = 2. Sum = S + (1/α + 1/β) = S + S/P = 4 + 2 = 6. Product = (α + 1/β)(β + 1/α) = αβ + 1 + 1 + 1/(αβ) = P + 2 + 1/P = 2 + 2 + 1/2 = 9/2. The equation is x² − 6x + 9/2 = 0, which becomes 2x² − 12x + 9 = 0.

Answer: (A)

The discriminant is (2 + i)² − 4(−1 + 7i) = 3 + 4i + 4 − 28i = 7 − 24i. Since (4 − 3i)² = 16 − 9 − 24i = 7 − 24i, the square root is ±(4 − 3i). The roots are z = [(2 + i) ± (4 − 3i)]/2, which gives 3 − i and −1 + 2i. Check: the sum is 2 + i and the product is −1 + 7i. Then |3 − i|² = 10 and |−1 + 2i|² = 5, so the total is 15.

Answer: (A)

For E1: D/4 = p² − 3p − 4 = (p − 4)(p + 1) ≥ 0, so p ≤ −1 or p ≥ 4. For E2: D/4 = p² − 5p + 6 = (p − 2)(p − 3) ≥ 0, so p ≤ 2 or p ≥ 3. The E1 set lies entirely inside the E2 set, so exactly one equation has real roots when E2 is real and E1 is not, i.e. −1 < p < 4 with p ≤ 2 or p ≥ 3. That gives −1 < p ≤ 2 or 3 ≤ p < 4, with integers 0, 1, 2, 3. The count is 4.

Answer: (A)

Roots that are negatives of each other have sum zero, so k² − 5k + 6 = 0, which gives k = 2 or k = 3. The roots must also be distinct, so the product must be negative. Then the roots are ±√(−product). For k = 2, the product is k² − 9 = −5 < 0 and the roots are ±√5. For k = 3, the product is 0 and the roots are 0, 0, which are not distinct. So only k = 2 works.

Answer: (A)

Each root satisfies t² = 6t + 2. Multiply by t⁸ to get t¹⁰ = 6t⁹ + 2t⁸. Apply this to t = α and t = β and subtract: α¹⁰ − β¹⁰ = 6(α⁹ − β⁹) + 2(α⁸ − β⁸), so a₁₀ = 6a₉ + 2a₈. Then a₁₀ − 2a₈ = 6a₉, and (a₁₀ − 2a₈)/(2a₉) = 6a₉/(2a₉) = 3.

Answer: (C)

Use (α−β)² = α² + β² − 2αβ: 16 = 40 − 2αβ, so αβ = 12. Then (α+β)² = α² + β² + 2αβ = 40 + 24 = 64. Positive roots give α + β = 8. The equation is x² − 8x + 12 = 0, with roots 2 and 6.

Answer: (A)

Rationalise: (3 + √2)/(3 − √2) = (3 + √2)²/(9 − 2) = (11 + 6√2)/7. Since p and q are rational, the irrational roots occur as conjugates, so the other root is (11 − 6√2)/7. The sum of roots is 22/7 = −p, so p = −22/7. The product is (121 − 72)/49 = 1 = q. Hence p + q = −22/7 + 1 = −15/7.

Answer: (A)

Apply Vieta's relations with roots p and q. Sum of roots = −(coefficient of x) = −p, so p + q = −p. Product of roots = constant term = q, so pq = q. Since q ≠ 0, divide to get p = 1. Then q = −2p = −2. Hence p − q = 1 − (−2) = 3. Check: x² + x − 2 = (x + 2)(x − 1) has roots 1 and −2, which are p and q.

Answer: (B)

For x² − 5x + 3 = 0, α + β = 5 and αβ = 3. The new sum is α + β + 1/α + 1/β = 5 + 5/3 = 20/3. The new product is (α + 1/β)(β + 1/α) = αβ + 1 + 1 + 1/(αβ) = 3 + 2 + 1/3 = 16/3. The equation is x² − (20/3)x + 16/3 = 0, which becomes 3x² − 20x + 16 = 0.