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I'm a high school student and I'm currently studying a lot of Numerical Sets and Radicals then I came across with a problem that wants me to explain why $\sqrt{4+2\sqrt{3}} = 1+\sqrt{3}$, but I don't know how to do, this was my equation: $\sqrt{4} + \sqrt{2}\times\sqrt[4]{3}$ = $2 + \sqrt[4]{2^2\times3}$ = $2 + \sqrt[4]{2^2\times2+1}$ = $2 + \sqrt[4]{2^3 + 1}$. Where is the problem?

iszwnc
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  • @projectilemotion I'm improving the question a little more to make it clearer, just a second. Thanks for the reply by the way. – iszwnc Apr 20 '17 at 22:44
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    In general $\sqrt {a+b};\ne \sqrt a;+\sqrt b.$ ... E.g. $2=1+1=\sqrt 1 ;+\sqrt 1\ne \sqrt {1+1}.$ We have $\sqrt {4+2\sqrt 3}; \ne \sqrt 4+\sqrt {2\sqrt 3}.$ – DanielWainfleet Apr 21 '17 at 05:37
  • Why shouldn't? They are both positive and they have the same square, hence they are equal. – Jack D'Aurizio Apr 21 '17 at 10:34

2 Answers2

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To eliminate a square root, you need a square.

In $$\sqrt{4+2\sqrt3},$$ $3$ is obviously not a square and we can't simplify it. But is $4+2\sqrt3$ a square ?

Here we need to make the educated guess that this expression could be the square of $$a+b\sqrt3,$$ i.e.

$$a^2+2\sqrt3ab+3b^2.$$

This is a good candidate as we get an integer term and another which is a multiple of $\sqrt3$.

By inspection, we immediately see that $a=b=1$ works, hence

$$\sqrt{4+2\sqrt3}=1+\sqrt3.$$

  • I know it may sounds obvious for you but, for me it doesn't could you explain it in more details? Or put some references that will help me understand it? – iszwnc Apr 20 '17 at 22:57
  • @iszwnc: I already added more details than I could. –  Apr 21 '17 at 07:03
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Your first step is incorrect: $$\sqrt{4+2\sqrt{3}}\neq \sqrt{4}+\sqrt{2}\times \sqrt[4]{3}$$ In general: $$\sqrt{a+b}\neq \sqrt{a}+\sqrt{b}$$ Where $a,b \in \mathbb{R}^+$.

For example, $a=4, b=9$ does not satisfy.

Here is a hint for a correct approach:

Notice that one may write: $$4+2\sqrt{3}=1+2\sqrt{3}+3=1^2+2\sqrt{3}+(\sqrt{3})^2$$ Now use the fact that: $$(a+b)^2=a^2+2ab+b^2$$

projectilemotion
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  • But why did you cut the square in $\sqrt{4+2}$ and started directly from $4+2$? And if $\sqrt{a + b}\neq\sqrt{a} + \sqrt{b}$, you're saying that it's absurd to sum roots? – iszwnc Apr 20 '17 at 23:16
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    projectilemotion merely squared both sides of the original equation. In fact once you square both sides you can see the two sides are equal $4+2\sqrt{3}=1+2\sqrt{3}+3 \Rightarrow 4+2\sqrt{3} = 4+2\sqrt{3}$ by adding the one and three on the RHS. – Χpẘ Apr 20 '17 at 23:22