# Solving Algebra involving surds

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.

• What's wrong with $$x=\left(\sqrt{124}+\sqrt{20}-\sqrt{80}\right)^2$$ are you sure this is the equation you mean?
– user304329
Commented Apr 8, 2017 at 16:24
• Well it's simply (approximatively) $$11.35 + 4.47 = 8.94 + \sqrt{x}$$ Which is quite trival to solve, right ? Commented Apr 8, 2017 at 16:26
• @vrugtehagel: probably the question wants this expression to be simplified using the distributive law. Commented Apr 8, 2017 at 16:27
• @zubzub: That is an unhelpful comment. Presumably the question is not about finding an approximate solution. Commented Apr 8, 2017 at 16:28
• @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. Commented Apr 8, 2017 at 16:58

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

The original equation is

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

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

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

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

$$x=45$$

• 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. Commented Apr 8, 2017 at 16:36
• It is indeed not the correct answer. The question (now) states to find $x$ when $\sqrt{125}+\sqrt{20}=\sqrt{80}+\sqrt{x}$. Commented Apr 8, 2017 at 16:59
• The question was edited. I used to say $124$ instead of $125$. I'll edit my answer.
– user304329
Commented Apr 8, 2017 at 17:00