Question

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

Answer 1

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

$3\sqrt{n}>\sqrt{x+3}+1$

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

$8x-4>2\sqrt{x+3}$

$4x-2>\sqrt{x-3}$

$16x^2+4-16x>x+3$

$16x^2-17x+1>0$

$16x^2-16x-x+1?0$

$16x(x-1)-1(x-1)>0$

$(x-1)(16x-1)>0$

$xϵ(-\infty ,\frac{1}{16})\cup (1,\infty )$