Let p and q be two positive numbers such that p + q = 2 and p⁴+q⁴ = 272. Then p and q are roots of the equation :

Solution

Maths sir · 4 steps

Call pq = t, write p⁴+q⁴ in terms of p+q and t, and the whole question becomes a quadratic in t.

1. Build up powers. Let pq = t. Then p²+q² = (p+q)² - 2pq = 4 - 2t.

   p⁴+q⁴ = (p²+q²)² - 2(pq)² = (4-2t)² - 2t² = 2t² - 16t + 16

2. Solve for t. Set 2t² - 16t + 16 = 272, so t² - 8t - 128 = 0. This factorises as (t-16)(t+8) = 0, giving t = 16 or t = -8.
3. Pick the sign. The product of two positive numbers cannot be negative, so pq = 16. The question uses this condition only to choose between the two values of t. Strictly, x²-2x+16=0 has no real roots, because its discriminant 4 - 64 is negative.
4. So. With sum 2 and product 16, the equation is x² - (p+q)x + pq = 0, that is x² - 2x + 16 = 0, option (c).

The idea to keep. Any symmetric expression like p⁴+q⁴ can be written using only p+q and pq. Climb up one square at a time, then solve for the unknown product.

Answer from the NTA final answer key. Solution by PrepWiser.