Find Real Solutions: Solving $\sqrt{4a-b^2}=\sqrt{b+2}+\sqrt{4a^2+b}$

[anemone]

Find the real solution(s) to the system

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

 

[lfdahl]

My solution:

Spoiler

Squaring and rearranging terms yields:\[4a-b^2 = b+2+4a^2+b + 2\sqrt{(b+2)(4a^2+b)}\\\\ -(b^2+2b+1)-4(a^2-a+\frac{1}{4}) = 2\sqrt{(b+2)(4a^2+b)}\]

- under the conditions:

$4a-b^2 \geq 0 \wedge b \geq -2$ or $a \geq \left ( \frac{b}{2} \right )^2 \wedge b \geq -2$.The LHS $\le 0$:

\[-(b+1)^2-4(a-\frac{1}{2})^2=-\left ( (b+1)^2+4(a-\frac{1}{2})^2 \right ) \le 0\]The RHS $\ge 0$, so the only possibility is: $b = -1$ and $a = \frac{1}{2}$. Checking the RHS: $4a^2+b = 0$. OK.

 

[anemone]

My solution:

Spoiler

Squaring and rearranging terms yields:\[4a-b^2 = b+2+4a^2+b + 2\sqrt{(b+2)(4a^2+b)}\\\\ -(b^2+2b+1)-4(a^2-a+\frac{1}{4}) = 2\sqrt{(b+2)(4a^2+b)}\]

- under the conditions:

$4a-b^2 \geq 0 \wedge b \geq -2$ or $a \geq \left ( \frac{b}{2} \right )^2 \wedge b \geq -2$.The LHS $\le 0$:

\[-(b+1)^2-4(a-\frac{1}{2})^2=-\left ( (b+1)^2+4(a-\frac{1}{2})^2 \right ) \le 0\]The RHS $\ge 0$, so the only possibility is: $b = -1$ and $a = \frac{1}{2}$. Checking the RHS: $4a^2+b = 0$. OK.

Very well done, lfdahl and thanks for participating!(Cool)