Question

$\frac{\sqrt{3+x}+\sqrt{3-x}}{\sqrt{3+x}-\sqrt{3-x}}=2$ then x is equal to

Answer 1

$\frac{\sqrt{3+x}+\sqrt{3-x}}{\sqrt{3+x}-\sqrt{3-x}}=2$

$⟹\sqrt{3+x}+\sqrt{3-x}=2(\sqrt{3+x}-\sqrt{3-x})$

$⟹\sqrt{3+x}+\sqrt{3-x}=2\sqrt{3+x}-2\sqrt{3-x}$

$⟹\sqrt{3+x}=3\sqrt{3-x}$

$⟹3+x=9(3-x)$

$⟹3+x=27-9x$

$⟹10x=24$

$⟹x=2.4$