# Solve |x²-3|<3 . This looks simple but i could not get the right answer. The answer is (-√5,-1)U(1,√5). How to solve this inequality?

Jul 15, 2015

The solution is that the inequality should be $\left\mid {x}^{2} - 3 \right\mid < \textcolor{red}{2}$

#### Explanation:

As usual with absolute values, split into cases:

Case 1: ${x}^{2} - 3 < 0$

If ${x}^{2} - 3 < 0$ then $\left\mid {x}^{2} - 3 \right\mid = - \left({x}^{2} - 3\right) = - {x}^{2} + 3$
and our (corrected) inequality becomes:

$- {x}^{2} + 3 < 2$

Add ${x}^{2} - 2$ to both sides to get $1 < {x}^{2}$

So $x \in \left(- \infty , - 1\right) \cup \left(1 , \infty\right)$

From the condition of the case we have

${x}^{2} < 3$, so $x \in \left(- \sqrt{3} , \sqrt{3}\right)$

Hence:

$x \in \left(- \sqrt{3} , \sqrt{3}\right) \cap \left(\left(- \infty , - 1\right) \cup \left(1 , \infty\right)\right)$

$= \left(- \sqrt{3} , - 1\right) \cup \left(1 , \sqrt{3}\right)$

Case 2: ${x}^{2} - 3 \ge 0$

If ${x}^{2} - 3 \ge 0$ then $\left\mid {x}^{2} - 3 \right\mid = {x}^{2} + 3$ and our (corrected) inequality becomes:

${x}^{2} - 3 < 2$

Add $3$ to both sides to get:

${x}^{2} < 5$, so $x \in \left(- \sqrt{5} , \sqrt{5}\right)$

From the condition of the case we have

${x}^{2} \ge 3$, so $x \in \left(- \infty , - \sqrt{3}\right] \cup \left[\sqrt{3} , \infty\right)$

Hence:

$x \in \left(\left(- \infty , - \sqrt{3}\right] \cup \left[\sqrt{3} , \infty\right)\right) \cap \left(- \sqrt{5} , \sqrt{5}\right)$

$= \left(- \sqrt{5} , - \sqrt{3}\right] \cup \left[\sqrt{3} , \sqrt{5}\right)$

Combined:

Putting case 1 and case 2 together we get:

$x \in \left(- \sqrt{5} , - \sqrt{3}\right] \cup \left(- \sqrt{3} , - 1\right) \cup \left(1 , \sqrt{3}\right) \cup \left[\sqrt{3} , \sqrt{5}\right)$

$= \left(- \sqrt{5} , - 1\right) \cup \left(1 , \sqrt{5}\right)$