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Could someone explain this?

$$\sqrt{125} + \sqrt{20} = \sqrt{80} + \sqrt{x}$$

Solve for $x$.

Much appreciated. Please explain in full detail as it's quite hard to understand otherwise.

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    $\begingroup$ What's wrong with $$x=\left(\sqrt{124}+\sqrt{20}-\sqrt{80}\right)^2$$ are you sure this is the equation you mean? $\endgroup$
    – user304329
    Commented Apr 8, 2017 at 16:24
  • $\begingroup$ Well it's simply (approximatively) $$ 11.35 + 4.47 = 8.94 + \sqrt{x} $$ Which is quite trival to solve, right ? $\endgroup$
    – Zubzub
    Commented Apr 8, 2017 at 16:26
  • $\begingroup$ @vrugtehagel: probably the question wants this expression to be simplified using the distributive law. $\endgroup$
    – Cheerful Parsnip
    Commented Apr 8, 2017 at 16:27
  • $\begingroup$ @zubzub: That is an unhelpful comment. Presumably the question is not about finding an approximate solution. $\endgroup$
    – Cheerful Parsnip
    Commented Apr 8, 2017 at 16:28
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    $\begingroup$ @Kaynex Common sense. In spite of being tagged "linear algebra", most probably this is a question for junior high school or so. Unless it is a very unusual thing, they are not interested in doing rational approximations to rather irrational numbers, but as Grumpy mentioned the wanted solution most probably is about rationalizing irrational expressions. $\endgroup$
    – DonAntonio
    Commented Apr 8, 2017 at 16:58

4 Answers 4

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Your equation is equivalent to $$5\sqrt{5}+2\sqrt{5}=4\sqrt{5}+\sqrt{x}$$ So $$\sqrt{x}=3\sqrt{5}=\sqrt{45}\implies x=45$$

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The original equation is

$$5\sqrt5+2\sqrt5=4\sqrt5+\sqrt x\implies\sqrt x=3\sqrt5=\sqrt{45}\ldots\ldots$$

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We can solve for $x$ by doing the following.

$$\sqrt{125}+\sqrt{20}=\sqrt{80}+\sqrt{x},$$ $$5\sqrt{5}+2\sqrt{5}=4\sqrt{5}+\sqrt{x},$$ $$3\sqrt{5}=\sqrt{x},$$ $$(\sqrt{x})^2=(3\sqrt{5})^2,$$ $$x=45.$$

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Well, first we should note that $$\sqrt{80}=\sqrt{16\cdot5}=\sqrt{16}\sqrt{5}=4\sqrt{5}$$ similarly, $$\sqrt{20}=\sqrt{4\cdot5}=\sqrt{4}\sqrt{5}=2\sqrt{5}$$ and $$\sqrt{125}=\sqrt{25\cdot5}=\sqrt{25}\sqrt{5}=5\sqrt{5}$$

so that the equation becomes

$$5\sqrt{5} + 2\sqrt{5} = 4\sqrt{4} + \sqrt{x}$$

Now subtract $4\sqrt{5}$ from both sides to obtain $$5\sqrt{5} - 2\sqrt{5} =\sqrt{x}$$

Which simplifies to

$$3\sqrt{5} = \sqrt{x}$$

And note that

\begin{align} 3\sqrt{5}&= \sqrt{x}\\ (3\sqrt{5})^2&=x\\ 3^2\sqrt{5}^2&=x\\ 9\cdot 5&=x\\ 45&=x \end{align}

So that we find the solution

$$x=45$$

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  • $\begingroup$ Hi mate, I don't know the correct answer but I definitely know that it shouldn't be that. It should be all under one square root symbol. $\endgroup$
    – Charlie
    Commented Apr 8, 2017 at 16:36
  • $\begingroup$ It is indeed not the correct answer. The question (now) states to find $x$ when $\sqrt{125}+\sqrt{20}=\sqrt{80}+\sqrt{x}$. $\endgroup$
    – Freeman Cheng
    Commented Apr 8, 2017 at 16:59
  • $\begingroup$ The question was edited. I used to say $124$ instead of $125$. I'll edit my answer. $\endgroup$
    – user304329
    Commented Apr 8, 2017 at 17:00

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