JEE PYQ: Quadratic Equation Question 36
Question 36 - 2019 (09 Apr Shift 1)
Let $p, q \in R$. If $2 - \sqrt{3}$ is a root of the quadratic equation, $x^2 + px + q = 0$, then:
(1) $p^2 - 4q + 12 = 0$
(2) $q^2 - 4p - 16 = 0$
(3) $q^2 + 4p + 14 = 0$
(4) $p^2 - 4q - 12 = 0$
Show Answer
Answer: (4)
Solution
Since $2 - \sqrt{3}$ is a root of the quadratic equation $x^2 + px + q = 0$
$\therefore 2 + \sqrt{3}$ is the other root
$\Rightarrow x^2 + px + q = [x - (2-\sqrt{3})][x - (2+\sqrt{3})]$
$= x^2 - (2+\sqrt{3})x - (2-\sqrt{3})x + (2^2 - (\sqrt{3})^2)$
$= x^2 - 4x + 1$
Now, by comparing $p = -4, q = 1$
$\Rightarrow p^2 - 4q - 12 = 16 - 4 - 12 = 0$
BETA
