JEE PYQ: Quadratic Equation Question 6
Question 6 - 2021 (24 Feb Shift 1)
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:
(1) $x^2 - 2x + 2 = 0$
(2) $x^2 - 2x + 8 = 0$
(3) $x^2 - 2x + 136 = 0$
(4) $x^2 - 2x + 16 = 0$
Show Answer
Answer: (4)
Solution
$(p^2 + q^2)^2 - 2p^2q^2 = 272$
$((p + q)^2 - 2pq)^2 - 2p^2q^2 = 272$
$16 + 16pq + 2p^2q^2 = 272$ (wait, let’s redo)
$(pq)^2 - 8pq - 128 = 0$
$pq = \frac{8 \pm 24}{2} = 16, -8$
$pq = 16$
Now $x^2 - (p+q)x + pq = 0$
$x^2 - 2x + 16 = 0$
BETA
