+9673 A solve-for-x equation: https://web2.0calc.com/questions/equation_99695
+129852 √ [ 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
+9673 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!!
