# Proving $\sqrt{2a + 2\sqrt{a^2 - b}}=\sqrt{a-\sqrt{b}} + \sqrt{a+\sqrt{b}}$, where $a\geq 0,\ b\geq 0$.

Who can help me to prove this equation? $$\sqrt{2a + 2\sqrt{a^2 - b}}=\sqrt{a-\sqrt{b}} + \sqrt{a+\sqrt{b}}$$ Where, $$a\geq 0,\ b\geq 0$$

• square your equation Commented Dec 23, 2015 at 3:53
• I wouldn't call this equation linear. Commented Dec 23, 2015 at 3:56
• @Xouque55 I tried to simplify left part. But did not succeed. Commented Dec 23, 2015 at 3:56

We have $$\left(\sqrt{a-\sqrt{b}}+\sqrt{a+\sqrt{b}}\right)^2 = a-\sqrt{b}+a+\sqrt{b} +2\sqrt{a^2-b} = 2a+2\sqrt{a^2-b}$$
• why did you assume $b^2$ on the left hand side? It should be $b$, right? see the question edited by OP Commented Dec 23, 2015 at 5:13
Notice, here is the correct proof $$LHS=\sqrt{2a+2\sqrt{a^2-b}}$$ $$=\sqrt{(a-\sqrt{b})+(a+\sqrt b)+2\sqrt{a^2-(\sqrt b)^2}}$$
$$=\sqrt{\underbrace{\left(\sqrt{a-\sqrt{b}}\right)^2}_{A^2}+\underbrace{\left(\sqrt{a+\sqrt b}\right)^2}_{B^2}+2\underbrace{\sqrt{(a-\sqrt b)}}_{A}\underbrace{\sqrt{(a+\sqrt b)}}_{B}}$$ using identity, $A^2+B^2+2AB=(A+B)^2$, $$=\sqrt{\left(\sqrt{a-\sqrt b}+\sqrt{a+\sqrt b}\right)^2}$$ $$=\left|\sqrt{a-\sqrt b}+\sqrt{a+\sqrt b}\right|$$ since, $a\ge 0,\ b\ge 0$ $$=\sqrt{a-\sqrt b}+\sqrt{a+\sqrt b}$$ $$=RHS$$