## Tuesday, June 28, 2016

### Find the real solution(s) to the system $\sqrt{4a-b^2}=\sqrt{b+2}+\sqrt{4a^2+b}$.

Find the real solution(s) to the system

$\sqrt{4a-b^2}=\sqrt{b+2}+\sqrt{4a^2+b}$.

My solution:

We know that for any two modulus functions, their sum is always greater than or equals to , we therefore have:

$\sqrt{b+2}+\sqrt{4a^2+b}\ge \sqrt{b+2+4a^2+b}$

But since

$\sqrt{4a-b^2}=\sqrt{b+2}+\sqrt{4a^2+b}$, we get:

$\sqrt{4a-b^2}\ge \sqrt{b+2+4a^2+b}=\sqrt{2b+2+4a^2}$

Squaring both sides to get rid of the square root and simplifying leads to:

$4a-b^2\ge 2b+2+4a^2$

$-4a^2+4a-1-b^2-2b-1\ge 0$

[MATH]-(4a^2-4a+1)-(b^2+2b+1)\ge 0[/MATH]

[MATH]-(2a-1)^2-(b+1)^2\ge 0[/MATH]

This is possible if and only if both $2a-1=0$ and $b+1=0$.

Therefore, there is only one pair of real solution for the system, namely $\left(a,\,b\right)=\left(\dfrac{1}{2},\,-1\right)$.