If $x-\sqrt{\frac7x}=8$, find $x-\sqrt{7x}$.

Question

If $x-\sqrt{\frac7x}=8$, find $x-\sqrt{7x}$.

Answer 1

Hint:   moving the radical to one side and squaring gives $\,x^3-16 x^2 + 64 x - 7 = 0\,$. The rational root theorem finds the root $\,x=7\,$, which gives the factorization $\,(x-7)(x^2 - 9x + 1)=0\,$. But $\,x=7\,$ is not a root of the original equation, so that one must be a root of $\,x^2-9x+1=0\,$, whose roots are both real and positive.

Then $\,x - \sqrt{7x}=x\left(1-\sqrt{\dfrac{7}{x}}\right)=x\big(1-(x-8)\big)=x(9-x)=\ldots\,$