Question

The number of distinct real roots of the equation $|x + 1| |x + 3| - 4|x + 2| + 5 = 0,$ is _______.

Answer 1

Given equation:
$|x + 1||x + 3| - 4|x + 2| + 5 = 0.$
To solve this, we break it into different cases based on the values of $x$ that change the absolute values:
Case 1: $x \leq -3$
$(x + 1)(x + 3) + 4(x + 2) + 5 = 0.$
Simplifying:
$x^2 + 4x + 3 + 4x + 8 + 5 = 0 \implies x^2 + 8x + 16 = 0 \implies (x + 4)^2 = 0.$
$x = -4.$
Case 2: $-3 <x \leq -2$
$-(x + 1)(x + 3) + 4(x + 2) + 5 = 0.$
Simplifying:
$-x^2 - 4x - 3 + 4x + 8 + 5 = 0 \implies -x^2 + 10 = 0 \implies x^2 = 10.$
$x = \pm \sqrt{10}.$
Case 3: $-2 <x \leq -1$
$-(x + 1)(x + 3) - 4(x + 2) + 5 = 0.$
Simplifying:
$-x^2 - 4x - 3 - 4x - 8 + 5 = 0 \implies -x^2 - 8x - 6 = 0 \implies x^2 + 8x + 6 = 0.$
Solving using the quadratic formula:
$x = \frac{-8 \pm \sqrt{64 - 24}}{2} = \frac{-8 \pm \sqrt{40}}{2} = -4 \pm \sqrt{10}.$
Case 4: $x >-1$
$x^2 + 4x + 3 - 4x - 8 + 5 = 0.$
Simplifying:
$x^2 = 0 \implies x = 0.$
The number of distinct real roots is:
Total Solutions = 2.
Answer: 2.