Solve the following equation: $\sqrt{x+3}+\sqrt{5-x}-2\sqrt{15+2x-x^2}=-4$
Since real solutions are to be found, the domain of $x$ is $[-3; 5]$.
I immediately found that $15+2x-x^2$ can be factored to $(5-x)(x+3)$, this gave:
$\sqrt{x+3}+\sqrt{5-x}-2\sqrt{(5-x)(x+3)}=-4$
Squaring both sides was too complicated so I tried to substitute $a=\sqrt{5-x}$ for $a \in [0; \sqrt{8}]$ and $b=\sqrt{x+3}$ for $b \in [0; \sqrt{8}]$, this gave the new multivariable equation: $a+b-2ab=-4$
Another thing I noticed was $a^2+b^2=8$, together with the above equation, I got this system of equations: $\left\{ \begin{array}{l}
a + b – 2ab = – 4\\
{a^2} + {b^2} = 8
\end{array} \right.$
. Solving this by elimination is quite difficult for me.
This problem needs to be solved using algebra so I wonder how do I continue with this or are there any better way to solve this algebraically?
Best Answer
This is nice work, so far.
Subtracting the 1st equation above, from the 2nd equation above gives
$(a^2 + b^2 + 2ab) - (a + b) = 8 - (-4) = 12$.
Let $u = (a + b).$
Then, $u^2 - u = 12 \implies u \in \{4,-3\}.$
However, by the constraints, $~~a,b~~$ must each be positive.
Therefore, $u = -3$ must be rejected.
Therefore, $a + b = u = 4.$
This implies that $2ab = 8 = a^2 + b^2 \implies (a - b)^2 = 0.$
Therefore, $a = b = 2,~$ since (again), $~~a,b~~$ must each be positive.
This forces $x = 1$.