Question

Solve the following inequality:
$1-\sqrt{13+3x^2}>2x.$

Answer 1

$13+3x^2\ge 0$

$x^2+\frac{13}{3}\ge 0$

$xϵR$

$1-\sqrt{13+3x^2}>2x$

$1-2x>\sqrt{13+3x^2}$

Squarring both sides

$1+4x^2-4x>13+3x^2$

$x^2-4x-12>0$

$(x-6)(x+2)>0$

$xϵ(-\infty ,-2)\cup (6,\infty )$