Question

Solve the following inequality:
$\sqrt{1-x^2}+1<\sqrt{3-x^2}.$

Answer 1

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

1. For definition of roots,

$1-x^2\ge 0$

$\Rightarrow x^2-1\le 0$

$\Rightarrow (x-1)(x+1)\le 0$-Equation 1

2. For definition of another square root,

$3-x^2\ge 0$

$\Rightarrow x^2-3\le 0$

$\Rightarrow (x-\sqrt{3})(x+\sqrt{3})\le 0$-Equation 2

Squaring the main inequality now,

$\Rightarrow (\sqrt{1-x^2}+1{)}^2<3-x^2$

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

$\Rightarrow \sqrt{1-x^2}<\frac{1}{2}$

On again squaring,

$1-x^2<\frac{1}{4}$

$x^2>\frac{3}{4}$

$\Rightarrow (x-\frac{\sqrt{3}}{2})(x+\frac{\sqrt{3}}{2})>0$-Equation 3

From Equation 1, 2, 3

$x\in \left(\frac{\sqrt{-}3}{2}\right),\left(\frac{\sqrt{3}}{2}\right)$