JEE PYQ: Quadratic Equation Question 21
Question 21 - 2020 (05 Sep Shift 1)
The product of the roots of the equation $9x^2 - 18|x| + 5 = 0$ is:
(1) $\frac{5}{9}$
(2) $\frac{25}{81}$
(3) $\frac{5}{27}$
(4) $\frac{25}{9}$
Show Answer
Answer: (2)
Solution
Let $|x| = y$ then
$9y^2 - 18y + 5 = 0$
$\Rightarrow 9y^2 - 15y - 3y + 5 = 0$
$\Rightarrow (3y - 1)(3y - 5) = 0$
$\Rightarrow y = \frac{1}{3}$ or $\frac{5}{3} \Rightarrow |x| = \frac{1}{3}$ or $\frac{5}{3}$
Roots are $\pm\frac{1}{3}$ and $\pm\frac{5}{3}$
$\therefore$ Product $= \frac{25}{81}$
BETA
