Solving $\sqrt{x+3}+\sqrt{5-x}-2\sqrt{15+2x-x^2}=-4$

Question

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?

Answer 1

Squaring both sides was too complicated so I tried to substitute
$a = \sqrt{5−x}, ~b = \sqrt{x + 3}.$
This gave the new multivariable equation: $~a+b−2ab=−4$.

$a + b - 2ab = -4.$
$a^2 + b^2 = 8.$

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 $x = 1$.

Answer 2

Using dxiv's hint.

$a^2+b^2 = (a+b)^2 - 2ab\\ 8 = (a+b)^2 - 2ab\\ -2ab = 8 - (a+b)^2$

Substitute into

$a+b - 2ab = -4\\ (a+b) + 8 - (a+b)^2 = -4\\ (a+b)^2 -(a+b) - 12 = 0\\ (a+b - 4)(a+b + 3) = 0$

$a+b>0$ so $a+b = 4$

using this value in the equation above

$-2ab = 8 - 16\\ ab = 4$

$(y-a)(y-b) = y - (a+b)y + (ab) = y^2 -4y +4 = (y-2)(y-2)$

$a = 2, b = 2$ and $x = 1$

Answer 3

Squaring both sides works out not so bad. Move the big radical to the other side and square:

$$\sqrt{x+3}+\sqrt{5-x} = 2\sqrt{15+2x-x^2}-4$$

$$8 +2\sqrt{15+2x-x^2} = 4(15+2x-x^2)-16\sqrt{15+2x-x^2} +16$$

Let $k=\sqrt{15+2x-x^2}$ to get

$$8+2k = 4k^2-16k+16$$

$$2k^2-9k+4=0$$

$$(2k-1)(k-4)=0.$$

That gives you two quadratics to solve $k=4$ yields $x=1$. $2k=1$ yields two solutions which are extraneous.