Question 25

Explanation

Define the first absolute value for $|x - 2|$.

$x - 2$ is positive for $x - 2 \geq 0 \rightarrow x \geq 0$.

$x - 2$ is negative for $x - 2 < 0 \rightarrow x < 2$.

So the definition is

For $|x^2 - 4|$

$x^2 - 4$ is positive for $x^2 - 4 \geq 0 \rightarrow x \leq -2 \vee x \geq 0$.

$x^2 - 4$ is negative for $x^2 - 4 < 0 \rightarrow -2 < x < 2$.

So the definition is

Based on the definitions above, the absolute value is bounded by $x = -2$ and $x = 2$. This means there are $3$ possible regions/values of $x$.

Solve the problem based on regions.

Region I when $x < -2$

Since region I is negative, the root that satisfies is $x_1 = -\sqrt{6}$.

Region II when $-2 \leq x < 2$

Since region II includes both positive and negative, all roots satisfy region II.

Region III when $x \geq 2$

Determine the discriminant value to find the type of roots.

Since $D > 0$, the roots are distinct, one positive root satisfies region III.

Therefore the total number of solutions is $4$ solutions.