Question

hey how to solve it for x

Answer 1

we can write it as follows:
(3x^2 + 1)<2^2
3x^2 + 1 < 4
3x^2 < 4-1
3x^2 < 3
x^2 <1
x<1

Answer 2

$\sqrt{(3x^2+1)} \ \textless \ 2$

Squaring both sides,

$3x^2+1 \ \textless \ 4$
$3x^2 \ \textless \ 4-1$
$3x^2 \ \textless \ 3$
$x^2 \ \textless \ \frac{3}{3}$
$x^2 \ \textless \ 1$
$x \ \textless \ \sqrt{1}$

Square root of 1 can be +1 or -1. Therefore,
x lies between -∞ and -1 or +1

As, values from -∞ to -1 are common, x ∈ (-∞,1).