Question

If x $\in$ ( $\frac{5a-2}{5}$ , a ) for $\sqrt{({-x}^2+4x-3)}>6-2x$ Find a.

Answer 1

$\sqrt{({-x}^2+4x-3)}>6-2x$
$\Rightarrow -x^2+4x-3>(6-2x{)}^2$
$\Rightarrow 5x^2-28x+39<0$
$\Rightarrow (5x-13)(x-3)<0$
$\Rightarrow x\in (\frac{13}{5},3)$
But given $x\in (\frac{5a-2}{5},a)$
So, on comparing we get
$a=3$