I am looking for solution of the following question?

Question

If

$x-\sqrt{\frac{3}{x}} =10$,

what is the result of the following expression?

$x-3\sqrt{x} =? $

thanks

Answer 1

The answer is 1.

Based on OP's comment copied above, the posted problem is most likely mistyped. The following answers the question assuming that the given equation was, instead:

$$\;x-\frac{3}{\sqrt{x}} =10\,$$

Let $t=\sqrt{x} \gt 0\,$, then the equation becomes $t^3-10t-3=0\,$. Using the rational root theorem it is easy to find the root $t=-3\,$, then the LHS factors as:

$$(t+3)(t^2-3t-1)=0 \quad \iff \quad (\sqrt{x}+3)(x-3\sqrt{x}-1)=0 $$

The first factor is strictly positive $\sqrt{x}+3 \gt 0\,$, which leaves:

$$ x-3\sqrt{x}-1 = 0 \quad \iff \quad x-3\sqrt{x}=1$$

Answer 2

First, lets get everything in a nice order, so $\sqrt{3\over x} = x-10$

Now, let's square both sides to solve for x fully.

${3\over x} = (x-10)^2$ so this means that $3=x(x-10)^2$.

Solve for x here and then just plug it into the next equation.

This is of course based on the assumption these two are a system of equations, you have it tagged as saying system of equations so I went with this.

Answer 3

$x-3\sqrt{x}=x(1-\dfrac{3}{\sqrt{x}})=x(1-\sqrt{3}\sqrt{\dfrac{3}{x}})=(10-\sqrt{\dfrac{3}{x}})(1-\sqrt{3}\sqrt{\dfrac{3}{x}})=\\ 10-10\sqrt{3}\sqrt{\dfrac{3}{x}}-\sqrt{\dfrac{3}{x}}+\sqrt{3}\dfrac{3}{x} $

Does it works?