JEE Main 2020 · 5 September · morning · Single correct · +4 / −1
The product of the roots of the
equation 9x² - 18|x| + 5 = 0 is :
- (a) 5/9
- (b) 5/27
- (c) 25/81
- (d) 25/9
Answer and solution
Answer (c) 25/81
Solution
Maths sir · 4 steps
Put t = |x|, solve the quadratic in t, then remember each positive t gives two values of x.
- Substitute. Since x² = |x|², put t = |x| with t ≥ 0. The equation becomes 9t² - 18t + 5 = 0.
9t² - 18t + 5 = (3t - 1)(3t - 5) = 0
- Solve for t. So t = 1/3 or t = 5/3. Both are positive, so both are allowed values of |x|.
- Back to x. |x| = 1/3 gives x = ±1/3, and |x| = 5/3 gives x = ±5/3. The equation has four real roots.
- So. Product = (1/3)(-1/3)(5/3)(-5/3) = 25/81, which is option (c).
The idea to keep. With |x| in a quadratic, solve in t = |x| first. Every positive t gives two roots ± t, and a negative t gives none. Count the roots before you multiply.
Answer from the NTA final answer key. Solution by PrepWiser.