How many different solutions for this

Question

$\sqrt{3x^2 + 2x + 5} + \sqrt{x^2-4x+5} = \sqrt{2x^2 - 2x + 2} + \sqrt{2x^2+8}$

My attempt :

All the equation inside sqrt has $D<0$. Let it be $\sqrt A + \sqrt B = \sqrt C + \sqrt D$

2 possibilities $$\sqrt A = \sqrt C \cap \sqrt B = \sqrt D$$ Find $x$ that match this condition. $\sqrt A = \sqrt C $ i get $x=-3 \cup x=-5$ $\sqrt B = \sqrt D $ i get $x=-3 \cup x=-1$ $$\sqrt A = \sqrt D \cap \sqrt B = \sqrt C$$ Find $x$ that match this condition. $\sqrt A = \sqrt D $ i get $x=-3 \cup x=1$ $\sqrt B = \sqrt C $ i get $x=-3 \cup x=1$

2 solutions.

Another posibility $\sqrt A + \sqrt B = \sqrt C + \sqrt D$ ${[\sqrt A + \sqrt B]}^2 = {[\sqrt C + \sqrt D]}^2$ This part is complicated.

Is there easier way to solve it?

Answer 1

Let $a=\sqrt{3x^2+2x+5}, b=\sqrt{2x^2+8}, c=\sqrt{2x^2-2x+2}, d=\sqrt{x^2-4x+5}$

$\implies a-b=c-d$

Observe that $$3x^2+2x+5-(2x^2+8)=x^2+2x-3=2x^2-2x+2-(x^2-4x+5)$$

$\implies a^2-b^2=c^2-d^2$

$\implies a+b=\dfrac{c-d}{a-b}\cdot c+d=c+d$

Now add and subtract to find $$a=c$$

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

Square both sides