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!!
