Question

The integral value(s) of $x$ satisfying $|x^2-9|+|x^2-4|=5$ is/are

• A. $-3$
• B. $2$
• C. $-2$
• D. $3$

Answer 1

Answer: (A), (B), (C), (D)

$3$

$\text{(i)}{x}^2\le 4\text{i.e.,}x\in [-2,2]$
$9-x^2+4-x^2=5\Rightarrow 2x^2=8\Rightarrow x=\pm 2$

$\text{(ii)}4<x^2<9\text{i.e.,}x\in (-3,-2)\cup (2,3)$
$9-x^2+x^2-4=5$
$\Rightarrow 5=5$ which is always true.
But there is no integer in the given domain.

$\text{(iii)}{x}^2\ge 9\text{i.e.,}x\in (-\infty ,-3]\cup [3,\infty )$
$x^2-9+x^2-4=5\Rightarrow 2x^2=18\Rightarrow x=\pm 3$

Hence, the integers satisfying the given equation are $\{-3,-2,2,3\}$