Question

Solve the following inequalities.
$\sqrt{1-x}+\sqrt{2x+3}<5$

Answer 1

For, $\sqrt{1-x}+\sqrt{2x+3}<5$

On squaring , we get

$1-x+2x+3+2\sqrt{1-x}\sqrt{2x+3}<25$

$x+4+2\sqrt{1-x}\sqrt{2x+3}<25$

$21-x>2\sqrt{1-x}\sqrt{2x+3}$

On squaring, we get

$441+x^2-42x>2(2x+3-2x^2-3x)$

$441+x^2-42x>-2x+6-4x^2$

$5x^2-40x+435>0$

$x^2-8x+87>0$

Since, $D$ for $x^2-8x-87>0$ is less than zero, we can say that, $\forall x\in R$, this equality holds true and also, $x\ge \frac{-3}{2}$ and $x\le 1$. So, $x\in [\frac{-3}{2},1]$