Let p and q be two positive numbers such that | vedclass

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(D) Given $p + q = 2$ and $p^{4} + q^{4} = 272$. We know that $p^{2} + q^{2} = (p + q)^{2} - 2pq = 2^{2} - 2pq = 4 - 2pq$. Also,$p^{4} + q^{4} = (p^{2

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$x^{2} - 2x + 16 = 0$

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(D) Given $p + q = 2$ and $p^{4} + q^{4} = 272$. We know that $p^{2} + q^{2} = (p + q)^{2} - 2pq = 2^{2} - 2pq = 4 - 2pq$. Also,$p^{4} + q^{4} = (p^{2} + q^{2})^{2} - 2p^{2}q^{2} = 272$. Substituting the expression for $p^{2} + q^{2}$: $(4 - 2pq)^{2} - 2(pq)^{2} = 272$. $16 - 16pq + 4(pq)^{2} - 2(pq)^{2} = 272$. $2(pq)^{2} - 16pq - 256 = 0$. $(pq)^{2} - 8pq - 128 = 0$. Let $t = pq$. Then $t^{2} - 8t - 128 = 0$. Using the quadratic formula: $t = \frac{8 \pm \sqrt{64 - 4(1)(-128)}}{2} = \frac{8 \pm \sqrt{64 + 512}}{2} = \frac{8 \pm \sqrt{576}}{2} = \frac{8 \pm 24}{2}$. $t = 16$ or $t = -8$. Since $p$ and $q$ are positive,$pq$ must be positive,so $pq = 16$. The quadratic equation with roots $p$ and $q$ is $x^{2} - (p+q)x + pq = 0$. Substituting the values,we get $x^{2} - 2x + 16 = 0$.

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:

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