Question

The number of distinct real roots of the equation $|x| \, |x + 2| - 5|x + 1| - 1 = 0$ is _________.

Answer 1

Consider the different cases based on the value of $x$.

Case 1: $x \geq 0$

$x^2 + 2x - 5x - 1 = 0 \implies x^2 - 3x - 6 = 0$

The roots are given by:

$x = \frac{3 \pm \sqrt{9 + 24}}{2} = \frac{3 \pm \sqrt{33}}{2}$

Since $x \geq 0$, one positive root exists.

Case 2: $-1 \leq x < 0$

$-x^2 - 2x - 5x - 1 = 0 \implies x^2 + 7x + 6 = 0$

The roots are:

$x = -1, \, x = -6$

Only $x = -1$ is within the range.

Case 3: $-2 \leq x < -1$

$x^2 - 2x + 5x - 1 = 0 \implies x^2 - 3x - 4 = 0$

The roots are:

$x = \frac{-3 \pm \sqrt{9 + 16}}{2} = \frac{-3 \pm \sqrt{25}}{2}$

No root lies in the range.

Case 4: $x < -2$

$x^2 + 7x + 4 = 0$

The roots are:

$x = \frac{-7 \pm \sqrt{49 - 16}}{2} = \frac{-7 \pm \sqrt{33}}{2}$

One root lies in the range.

Total number of distinct real roots: 3