Question

If $33-x^2\le 8x$ then x does not lie in the interval

Answer 1

Given,

$33-x^2\le 8x$

$33-x^2-8x\le 0$

$-(x-3)(x+11)\le 0$

$(x-3)(x+11)\ge 0$

$x\ge -11$ or $x\ge 3$

Interval $[-11,3]\cup [3,\infty )$