Question

Solve the following inequalities
$3\sqrt{6+x-x^2}>4x-2$

Answer 1

$3\sqrt{6+x-x^2}>4x-2$

Square both the sides

$9(6+x-x^2)>(4x-2{)}^2$

$9(6+x-x^2)>16x^2+4-16x$

$54+9x-9x^2>16x^2+4-16x$

$-25x^2+25x+50>0$

$x^2-x-2>0$

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

$(x-2)(x+1)>0$

For $x<-1$

$(x-2)(x+1)>0$

For. $x=-1$

$(x-2)(x+1)=0$

For $-1<x<2$

$(x-2)(x+1)<0$

at $x=2$

$(x-2)(x+1)=0$

for $x>2$

$(x-2)(x+1)>0$

Solution: - $(-\infty ,-1)\cup (2,\infty )$.