Question

Solve:
$√(x-5) - √(9- x) > 1$
​

Answer 1

Hope it's helpful to you

Answer 2

$\rm :\longmapsto\: \sqrt{x - 5} - \sqrt{9 - x} > 1$

Let first define the domain of the given inequality,

$\rm :\longmapsto\:x - 5 \geqslant 0 \: \: and \: 9 - x \geqslant 0$

$\rm :\implies\:x \geqslant 5 \: \: and \: \: x \leqslant 9$

$\bf\implies \:x \in \: [5, \: 9]$

Now,

$\rm :\longmapsto\: \sqrt{x - 5} - \sqrt{9 - x} > 1$

$\rm :\longmapsto\: \sqrt{x - 5} > \sqrt{9 - x} + 1$

On squaring both sides, we get

$\rm :\longmapsto\:x - 5 > 1 + 9 - x + 2 \sqrt{9 - x}$

$\rm :\longmapsto\:x - 5 >10 - x + 2 \sqrt{9 - x}$

$\rm :\longmapsto\:x - 5 - 10 + x > 2 \sqrt{9 - x}$

$\rm :\longmapsto\:2x - 15 > 2 \sqrt{9 - x}$

On squaring both sides, we get

$\rm :\longmapsto\: {4x}^{2} + 225 - 60x > 4(9 - x)$

$\rm :\longmapsto\: {4x}^{2} + 225 - 60x > 36 - 4x$

$\rm :\longmapsto\: {4x}^{2} - 56x + 189 > 0$

Please onwards find the attachment.