Question

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

Answer 1

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

$x>-3,x>1andx>\frac{1}{2}$

Hence, $x>1$

On squaring we have,

$=>x+3+x-1+2\sqrt{(x+3)(x-1)}>2x-1$

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

$=>\sqrt{x^2+2x-3}>-3$

$=>x^2+2x-3>9$

$=>x^2+2x+6>0$

$\because D<0$

Hence $x$ is satisfied for all value of $x$

Hence, $x>1$