Question

$| {x}^{2} - 7x - 8 | > {x}^{2} - 7x + 12 solve \: this \: inequality$
solve in equality​

Answer 1

Step-by-step explanation:

We have,

$| {x}^{2} - 7x - 8 | > {x}^{2} - 7x + 12$

We have, the following two cases,

$\tt\large\bold{•\:\:WHEN\:\:(x^2-7x-8)>0\::}$

We know, if $y>0$, then, $|y|=y$

So,

${x}^{2} - 7x - 8 > {x}^{2} - 7x + 12$

$\implies - 8 > 12$

$\bold{ \red{ \mathbb{WHICH \: \: \: IS \: \: \: NOT \: \: \: POSSIBLE}}}$

$\tt\large\bold{•\:\:WHEN\:\:(x^2-7x-8)<0\::}$

We know, if $y<0$, then, $|y|=-y$

So,

$- {x}^{2} + 7x + 8 > {x}^{2} - 7x + 12$

$\implies 2{x}^{2} - 14x + 4 < 0$

$\implies {x}^{2} - 7x + 2 < 0$

$\implies \bigg \{x - \bigg( \frac{7 - \sqrt{41} }{2} \bigg) \bigg \} \bigg \{x - \bigg( \frac{7 + \sqrt{41} }{2} \bigg) \bigg \}< 0$

$\implies x \in \bigg( \frac{7 - \sqrt{41} }{2} ,\frac{7 + \sqrt{41} }{2} \bigg)$