Please help me ASAP :( I am stuck at this question!!

√ [ 2 + 3√x - x]  + √ [6 - 2√x - 3x ]   = √10 + 4√x - 5x]

Note, Max......that, by inspection, x = 1   is at least one solution.......let's work through this

Square both sides

[ 2 + 3√x - x]  + 2√ [ 2 + 3√x - x] * √ [6 - 2√x - 3x ] + [6 - 2√x - 3x ]   = 10 + 4√x - 5x

Simplify

8 + √x - 4x  + 2 √ [( 2 + 3√x - x) (6 - 2√x - 3x) ] =   10 + 4√x - 5x

2 √ [( 2 + 3√x - x) (6 - 2√x - 3x) ]  =  2 + 3√x -x

2 √  [-7 x^(3/2)+3 x^2-18 x+14 sqrt(x)+12 ]   = 2 + 3√x -x

Square both sides again

4 [-7 x^(3/2)+3 x^2-18 x+14 sqrt(x)+12 ]  = -6 x^(3/2)+x^2+5 x+12 sqrt(x)+4

-28x^(3/2) + 12x^2 - 72x + 56sqrt(x) + 48  =  -6 x^(3/2)+x^2+5 x+12 sqrt(x)+4

22x^(3/2) - 11x^2 + 77x - 44sqrt(x) - 44  =  0

Divide through by 11

2x^(3/2) - x^2 + 7x - 4sqrt(x) - 4  = 0

This is a little difficult to factor......WolframAlpha gives

-(sqrt(x)-1) (sqrt(x)+2) (x-3 sqrt(x)-2) = 0

Setting the first factor to 0  →  sqrt(x) - 1  = 0       and it's obvious that  x = 1

Setting the second factor to 0  → sqrt (x) + 2  = 0  →  sqrt (x)  = -2    and this is a non-real result

Setting the third factor to 0  →  x - 3sqrt(x) - 2  = 0

And WolframAlpha  gives  [ 13 + 3√17 ] / 2    as the other real solution

So....the real solutions are  x = 1   and x = [ 13 + 3√17 ] / 2

Just when you are working in this my friend sent me help too:

His answer( He is good at maths too :) )

$\text{Let }a = \sqrt x\\ \sqrt{2+3\sqrt x - x}+\sqrt{6 - 2\sqrt x - 3x}=\sqrt{ 10 + 4\sqrt x - 5x}\\ \sqrt{-a^2+3a+2}+\sqrt{-3a^2-2a+6}=\sqrt{-5a^2+4a+10}\\ \sqrt{(2-a^2)+3a}+\sqrt{3(2-a^2)-2a}=\sqrt{5(2-a^2)+4a}\\ \text{Let } b=2-a^2\\ \sqrt{b+3a}+\sqrt{3b-2a}=\sqrt{5b+4a}\\ \text{Let }u=\sqrt{b+3a}\text{, }v=\sqrt{3b-2a}\\ \therefore b+3a=u^2\text{, }3b-2a=v^2\\ a=\dfrac{3u^2-v^2}{11},b =\dfrac{2u^2+3v^2}{11} \\ \therefore 5b+4a=2u^2+v^2\\ \text{Substitute back into the equation}\\ u+v=\sqrt{2u^2+v^2}\\ u^2+2uv+v^2=2u^2+v^2\\ u^2-2uv=0\\ u(u-2v)=0\\ \text{If u = 0, b= -3a and b = }2-a^2\\ a=1\text{ OR }a=2(rejected)\\ \text{If u - 2v = 0}\\ u^2=4v^2\\ \therefore a=b\text{ OR }b = 2-a^2\\ \therefore a= 1\text{ Or }a=-2(Rejected)\\ \therefore x= 1$

Don't even know what is he trying to do....... Is he wrong? Because he got only 1 of your answers!!

As far as I can see there is only one real number solution: