Question

Solve the following equations, $\sqrt{x+8+2\sqrt{x+7}}+\sqrt{x+1-\sqrt{x+7}}=4.$

Answer 1

Given,

$\sqrt{(x+8)+2\sqrt{x+7}}+\sqrt{x+1-\sqrt{x+7}}=4$

$\sqrt{x+7+1+2\sqrt{x+7}}+\sqrt{x+1-\sqrt{x+7}}=4$

$\sqrt{(\sqrt{x+7}{)}^2+1^2+2\times 1\times \sqrt{x+7}}+\sqrt{x+1-\sqrt{x+7}}=4$

Using identity $(a+b{)}^2=a^2+b^2+2ab$, we get,

$\sqrt{(\sqrt{x+7}+1{)}^2}+\sqrt{x+1-\sqrt{x+7}}=4$

$\sqrt{x+7}+1+\sqrt{x+1-\sqrt{x+7}}=4$

$(\because$ $\sqrt{x+7}+1$ is always positive, no need of modulus)

$\sqrt{x+7}-3=-\sqrt{x+1-\sqrt{x+7}}$

Squaring on both sides

$x+7+9-6\sqrt{x+7}=x+1-\sqrt{x+7}$

$7+9-1=-\sqrt{x+7}+6\sqrt{x+7}$

$15=5\sqrt{x+7}$

$3=\sqrt{x+7}$

On squaring both the sides,we get,

$x+7=9$

$x=2$

Therefore $x=2$ is solution of given equation.