Question

The solution Set of the in equation VX2 +6x+5>8-x
is​

Answer 1

$\large\underline{\sf{Solution-}}$

➢ Given inequality is

$\rm :\longmapsto\: \sqrt{ {x}^{2} + 6x + 5} > 8 - x$

Now, for inequality to be exist,

$\rm :\longmapsto\: {x}^{2} + 6x + 5 \geqslant 0$

$\rm :\longmapsto\: {x}^{2} + 5x + x + 5 \geqslant 0$

$\rm :\longmapsto\:x(x + 5) + 1(x + 5) \geqslant 0$

$\rm :\longmapsto\:(x + 5)(x + 1) \geqslant 0$

$\bf\implies \:x \leqslant - 5 \: \: or \: \: x \geqslant - 1 - - - (1)$

Now,

Consider the given inequality,

$\rm :\longmapsto\: \sqrt{ {x}^{2} + 6x + 5} > 8 - x$

On squaring both sides, we get

$\rm :\longmapsto\: {x}^{2} + 6x + 5 > {(8 - x)}^{2}$

$\rm :\longmapsto\: {x}^{2} + 6x + 5 > 64 + {x}^{2} - 16x$

$\rm :\longmapsto\: 6x + 5 > 64 - 16x$

$\rm :\longmapsto\:6x + 16x > 64 - 5$

$\rm :\longmapsto\:22x > 59$

$\bf\implies \:x > \dfrac{59}{22} - - - (2)$

So, from equation (1) and (2), we concluded that

$\bf\implies \:x > \dfrac{59}{22}$

$\bf\implies \:x \: \in \: \bigg(\dfrac{59}{22}, \: \infty \bigg)$

Additional Information :-

Let us consider two real numbers a and b such that a > b,

$\rm :\longmapsto\:(x - a)(x - b) < 0 \implies \: b < x < a$

$\rm :\longmapsto\:(x - a)(x - b) \leqslant 0 \implies \: b \leqslant x \leqslant a$

$\rm :\longmapsto\:(x - a)(x - b) > 0 \implies \: x < b \: \: or \: \: x > a$

$\rm :\longmapsto\:(x - a)(x - b) \geqslant 0 \implies \: x \leqslant b \: \: or \: \: x \geqslant a$