Let p and q be two real numbers such that p + q = 3 and p⁴ + q⁴ = 369. Then ( 1/p + 1/q )⁻² is equal to _________.

Solution

Maths sir · 5 steps

Write everything in terms of p + q and pq, then let the word real decide which pq survives.

1. Rewrite the target. 1/p + 1/q = (p+q)/pq = 3/t, where t = pq. So once we know t, the answer is (3/t)⁻² = t²/9.
2. Reach fourth powers. p² + q² = (p+q)² - 2pq = 9 - 2t. Square once more and subtract 2p²q² to reach the fourth powers.

   p⁴ + q⁴ = (9 - 2t)² - 2t² = 2t² - 36t + 81 = 369

3. Solve for pq. This gives t² - 18t - 144 = 0, so (t - 24)(t + 6) = 0 and t = 24 or t = -6.
4. Keep real roots. p and q are the roots of x² - 3x + t = 0, so we need 9 - 4t ≥ 0. With t = 24 we get 9 - 96 < 0, which gives no real roots. Only t = -6 works, and then 9 + 24 = 33 > 0.
5. So. 1/p + 1/q = 3/-6 = -1/2, and (-1/2)⁻² = 4.

The common slip: 64. Keeping pq = 24 without checking for real roots gives (3/24)⁻² = 64. For that value, x² - 3x + 24 = 0 has no real roots.

The idea to keep. For symmetric expressions in two numbers, write everything in terms of the sum and the product. If the product comes out with two values, check the discriminant to see which one gives real numbers.

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