Question

Solve the following inequality:
$\sqrt{4-\sqrt{1-x}}-\sqrt{2-x}>0.$

Answer 1

$\sqrt{4-\sqrt{1-x}}-\sqrt{2-x}>0$

$\Rightarrow \sqrt{4-\sqrt{1-x}}>\sqrt{2-x}$

On squaring, $4-\sqrt{1-x}>2-x$

$4-2+x>\sqrt{1-x}$

On squaring again,

$(2+x{)}^2>1-x$

$x^2+4+4x-1+x>0$

$x^2+5x+3>0$

From Equation 1, for definition of roots,

1. $1-x\ge 0$

2. $2-x\ge 0$

3. $4-\sqrt{1-x}\ge 0$

$\Rightarrow 4\ge \sqrt{1-x}$

$\Rightarrow 16\ge 1-x$

$\Rightarrow x\ge -15$

From Equation 2, $(x+\frac{5+\sqrt{1}3}{2})(x-\frac{5-\sqrt{1}3}{2})>0$

$x\in (\frac{-5-\sqrt{1}3}{2},\frac{5+\sqrt{1}3}{2})$