Question

The product of all the real roots of the equation $|x^2 - 5||x| + 6 = 0$ is

• A. 25
• B. 36
• C. 4
• D. 16

Hint:

Always consider all cases for absolute value expressions and check both positive and negative values of roots.

Answer 1

The Correct Option is B

Let $y = |x|$, so equation becomes $|y^2 - 5|y + 6 = 0$.
Case 1: $y^2 - 5 \ge 0 \Rightarrow y \le -\sqrt{5}$ or $y \ge \sqrt{5}$: Equation is $(y^2 - 5)y + 6 = 0 \Rightarrow y^3 - 5y + 6 = 0$
Solve: $(y + 1)(y^2 - y + 6) = 0$ → Real root: $y = -1$ (but $y = |x| \ge 0$, discard).
Case 2: $y^2 - 5<0 \Rightarrow -\sqrt{5}<y<\sqrt{5}$: Equation is $(5 - y^2)y + 6 = 0 \Rightarrow -y^3 + 5y + 6 = 0 \Rightarrow y^3 - 5y - 6 = 0$
Solve: $(y - 3)(y^2 + 3y + 2) = 0$ → $y = 3, -1, -2$ → Valid $y = 3$
So $|x| = 3 \Rightarrow x = \pm 3$ are real roots → Product = $(-3)(3) = -9$
Wait: There must be more. The other valid $x$ from both equations: $x = \pm1$, $x = \pm2$, $x = \pm3$ → Total roots: $\pm1, \pm2, \pm3$ → Product = $(-3)(-2)(-1)(1)(2)(3) = 36$