Question

Solve : $\sqrt{2+x-x^2}>x-4$

Answer 1

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

Squaring both sides,

$2+x-x^2>(x-4{)}^2$

$2+x-x^2>x^2+16-8x$

$-2x^2+9x-14>0$

$2x^2-9x+14>0$

$x>\frac{9\pm \sqrt{81-4\times 2\times -14}}{4}$

$x>\frac{9\pm \sqrt{193}}{4}$