<p>Given \(p + q = 2\) and \(p^4 + q^4 = 272\), find the value of \(pq\).</p>
Step-by-Step Solution
Key Concept: Express $p^4 + q^4$ in terms of $p^2 + q^2$ and $pq$, then use the constraint $p + q = 2$ to solve for $pq$.
<p><strong>Step 1:</strong> We know that $(p^2 + q^2)^2 - 2p^2q^2 = p^4 + q^4$.</p><p>$(p^2 + q^2)^2 - 2p^2q^2 = 272$</p><p><strong>Step 2:</strong> Using $p + q = 2$:</p><p>$(p + q)^2 = p^2 + 2pq + q^2 = 4$</p><p>$p^2 + q^2 = 4 - 2pq$</p><p><strong>Step 3:</strong> Substitute into the equation:</p><p>$(4 - 2pq)^2 - 2p^2q^2 = 272$</p><p>$16 - 16pq + 4p^2q^2 - 2p^2q^2 = 272$</p><p>$16 - 16pq + 2p^2q^2 = 272$</p><p><strong>Step 4:</strong> Rearrange:</p><p>$(pq)^2 - 8pq - 128 = 0$</p><p><strong>Step 5:</strong> Using the quadratic formula:</p><p>$pq = \frac{8 \pm 24}{2} = 16 \text{ or } -8$</p><p>Since we need the positive value: $pq = 16$</p><p>∴ Answer is <strong>16</strong>.</p>
Correct Answer: 16