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 :

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 : (A) x2-2 x+2=0 (B) x2-2 x+8=0 (C) x2-2 x+136=0 (D) x2-2 x+16=0

Tardigrade Admin · Accepted Answer

Consider ( p 2+ q 2)2-2 p 2 q 2=272 (( p + q )2-2 pq )2-2 p 2 q 2=272 16-16 pq +2 p 2 q 2=272 ( pq )2-8 pq -128=0 ( pq )2-8 pq -128=0 pq =(8 ± 24/2)=16,-8 ∴ pq =16 ∴ Required equation: x 2-(2) x +16=0

Q. 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 :

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

$x^{2} - 2 x + 8 = 0$

$x^{2} - 2 x + 136 = 0$

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

Consider $(p^{2} + q^{2})^{2} - 2 p^{2} q^{2} = 272$
$((p + q)^{2} - 2 pq)^{2} - 2 p^{2} q^{2} = 272$
$16 - 16 pq + 2 p^{2} q^{2} = 272$
$(pq)^{2} - 8 pq - 128 = 0$
$(pq)^{2} - 8 pq - 128 = 0$
$pq = \frac{8 \pm 24}{2} = 16, - 8$
$∴ pq = 16$
$∴$ Required equation
$: x^{2} - (2) x + 16 = 0$

Q. Let p and q be two positive numbers such that p+q=2 and p4+q4=272. Then p and q are roots of the equation :

x2−2x+2=0

x2−2x+8=0

x2−2x+136=0

x2−2x+16=0