Question

Find the range of $n$, satisfying the inequation $\left(2n+7\right)<{\left(n+3\right)}^2.$

Answer 1

$2n+7<{(n+3)}^2$

$\Rightarrow 2n+7<n^2+6n+9$

$\Rightarrow 2n+7-n^2-6n-9<0$

$\Rightarrow -n^2-4n-2<0$

$\Rightarrow n^2+4n+2>0$

$\Rightarrow n^2+2\times 2n+4-4+2>0$

$\Rightarrow {(n+2)}^2-2>0$

$\Rightarrow (n+2-\sqrt{2})(n+2+\sqrt{2})>0$

$\Rightarrow (n-(-2+\sqrt{2}))(n-(-2-\sqrt{2}))>0$

$\therefore n\in (-\infty ,-2-\sqrt{2})\cup (-2+\sqrt{2},\infty )$