Given the inequality:
$$2^{4} \sqrt{x^{2} - 3} < 2^{4} \sqrt{x + 3}$$
To solve this inequality, we must first solve the corresponding equation:
$$2^{4} \sqrt{x^{2} - 3} = 2^{4} \sqrt{x + 3}$$
Solve:
$$x_{1} = -2$$
$$x_{2} = 3$$
$$x_{1} = -2$$
$$x_{2} = 3$$
This roots
$$x_{1} = -2$$
$$x_{2} = 3$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} < x_{1}$$
For example, let's take the point
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$-2 - \frac{1}{10}$$
=
$$- \frac{21}{10}$$
substitute to the expression
$$2^{4} \sqrt{x^{2} - 3} < 2^{4} \sqrt{x + 3}$$
$$2^{4} \sqrt{\left(-1\right) 3 + \left(- \frac{21}{10}\right)^{2}} < 2^{4} \sqrt{- \frac{21}{10} + 3}$$
_____ ____
8*\/ 141 24*\/ 10
--------- < ---------
5 5
but
_____ ____
8*\/ 141 24*\/ 10
--------- > ---------
5 5
Then
$$x < -2$$
no execute
one of the solutions of our inequality is:
$$x > -2 \wedge x < 3$$
_____
/ \
-------ο-------ο-------
x_1 x_2