Let p and q be two positive numbers such that p+q=2 and p^4+q^4=272.

Question

Let $p$ and $q$ be two positive numbers such that $p+q=2$ and $p^4+q^4=272$. Then $p$ and $q$ are roots of the equation:

Accepted Answer

$(p^2+q^2)^2-2p^2q^2=272$, and $p^2+q^2=(p+q)^2-2pq=4-2pq$.

$$(4-2pq)^2-2(pq)^2=272 \implies 16-16pq+2(pq)^2-2(pq)^2 \text{[expand carefully]}$$

Expanding: $16-16pq+4(pq)^2-2(pq)^2=272 \implies (pq)^2-8pq-128=0 \implies pq=\dfrac{8\pm24}2=16 \text{ or } -8.$

Since $p,q>0$, $pq=16$. Required equation: $x^2-2x+16=0$.

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