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If

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

what is the result of the following expression?

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

thanks

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    The first line gives the solution $x \approx 10.53366758297133599675066185$ which can be put into exact form using the cubic formula and the substitution $\sqrt{x} = u$ – 2017-02-27
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    The only thing I can think of that would be similar to the OP's equalities with simple form for $x$ is if the RHS of the first equality were $0$, which would yield $x = \sqrt[3]{3}$ – 2017-02-27
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    @BrevanEllefsen My bet is on the LHS being wrong, see answer below. – 2017-02-27
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    Not a duplicate, but this looks a lot like http://math.stackexchange.com/questions/2153038/how-to-solve-x-3-sqrt-frac5x-8-for-x-sqrt5x – 2017-02-28

3 Answers 3

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:

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

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    Looks like you found the question! – 2017-02-28
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    So there is a mistype in book. – 2017-02-28
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    @user55944 Likely so. It's easy to verify that if the answer is $1$ then $x=(11+3 \sqrt{13})/2$ which does not satisfy the first equation as posted (but does satisfy the one above). – 2017-02-28
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    dxiv, you asume $ t>0$ and you find $ t=-3$, but the problem can be solved by taking $x= t^2 $ thanks a lot. – 2017-02-28
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    @user55944 $-3$ is a root of the polynomial (whence the factor $t+3$), but $t=-3$ is not an eligible root of the original equation because it's negative. This allows, at the last step, to deduce that the *other* factor must be $0$ i.e. $x-3\sqrt{x}-1=0\,$. – 2017-02-28
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$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?

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    The answer is 1. But I did not find it. – 2017-02-27
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    @user55944 Why do you say the answer is 1? Calculation shows the answer to be around $0.79698$. Are you sure you've typed out the question correctly? – 2017-02-27
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    @Théophile $1$ is probably the correct answer, but to a different question than posted ;-) – 2017-02-27
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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.

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    $3=x(x-10)^2$ ?? – 2017-02-27
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    Ah yes, my bad on that one let me fix it upa bit – 2017-02-27