Question:
Let $p$ and $q$ be two positive number such that $p+q=2$ and $p^{4}+q^{4}=272 .$ Then $p$ and $q$ are roots of the equation :
Correct Option: , 4
Solution:
$\left(p^{2}+q^{2}\right)^{2}-2 p^{2} q^{2}=272$
$\left((p+q)^{2}-2 p q\right)^{2}-2 p^{2} q^{2}=272$
$16+16 p q+2 p^{2} q^{2}=272$
$(p q)^{2}-8 p q-128=0$
$\mathrm{pq}=\frac{8 \pm 24}{2}=16,-8$
$\mathrm{pq}=16$
Now
$x^{2}-(p+q) x+p q=0$
$x^{2}-2 x+16=0$
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