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:

• A. $x^2-2x+2=0$
• B. $x^2-2x+8=0$
• C. $x^2-2x+136=0$
• D. $x^2-2x+16=0$

Answer 1

The correct option is D $x^2-2x+16=0$

$(p^2+q^2{)}^2-2p^2{q}^2=272$
$\Rightarrow {\left(\right(p+q{)}^2-2pq)}^2-2p^2{q}^2=272$
$\Rightarrow 16-16pq+2p^2{q}^2=272$
$\Rightarrow (pq{)}^2-8pq-128=0$
$\Rightarrow pq=16,-8$
$\Rightarrow pq=16$ (as $p,q>0$)
Now,
$x^2-(p+q)x+pq=0$
$\Rightarrow x^2-2x+16=0$