Inequalities - When to know if some expression are below/above 0? Inequality:
$$\sqrt{9x^2 - 16} > 3x + 1$$
I was thinking of...


*

*Squaring the whole inequality

*Move them between the > sign, which eventually I get (x < -17/6)


However on my answer sheet:
Left hand side - $9x^2 - 16 > 0 \implies x > 4/3 \text{ and } x < -4/3$
Right hand side - $3x + 1 > 0 \implies x > -1/3$
Question:


*

*How do we know $9x^2 - 16$ and $3x + 1$ is above $0$?

*When do we need to have 2 answers (LHS and RHS answer)

Let's apply your idea:
$$\sqrt{9x^2-16}>3x+1\implies 9x^2-16>9x^2+6x+1\implies6x<-17\implies$$
$$\implies x<-\frac{17}6\cong- 2.833\;\;\;(**)$$
But the real square root is defined iff
$$9x^2-16\ge0\iff x^2\ge\frac{16}9\iff |x|\ge\frac43\cong1.33$$
Thus, the answer in $(**)$  is within the allowed values for $\;x\;$ ...and then the left side is always non-negative, whereas the right side is negative in that ray...
