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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.

1. x² - 2x + 2 = 0
2. x² - 2x + 8 = 0
3. x² - 2x + 136 = 0
4. x² - 2x + 16 = 0

Correct answer: x² - 2x + 16 = 0

Solution

With p+q=2, p^2+q^2=4-2pq and p^4+q^4=(p^2+q^2)^2-2(pq)^2=(4-2pq)^2-2(pq)^2=272 -> 2(pq)^2-16pq-256=0 -> (pq)^2-8pq-128=0 -> pq=16 (taking the positive root). So p,q are roots of x^2-(p+q)x+pq=0 = x^2-2x+16=0.

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