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(square root of 2+square root of 8)^2

Guest Sep 27, 2017

#1
+648
0

$$(\sqrt{2}+\sqrt{8})^2$$

You can distribute the square.

$$\sqrt{2}^2+\sqrt{8}^2$$

The square root and the square cancel.

$$2+8=10$$

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#1
+648
0

$$(\sqrt{2}+\sqrt{8})^2$$

You can distribute the square.

$$\sqrt{2}^2+\sqrt{8}^2$$

The square root and the square cancel.

$$2+8=10$$

#2
+1491
+3

Uh oh! Squaring a binomial does not work that way!

The error was made at the step

$$\left(\sqrt{2}+\sqrt{8}\right)^2\Rightarrow\sqrt{2}^2+\sqrt{8}^2$$

This is incorrect.

$$(a+b)^2\neq a^2+b^2$$

How do I know that? Well, if you are ever unsure, try doing it yourself.

$$(a+b)^2=(a+b)(a+b)=a(a+b)+b(a+b)=a^2+ab+ab+b^2=a^2+2ab+b^2\neq a^2+b^2$$

Therefore, when you square a binomial, $$(a+b)^2=a^2+2ab+b^2$$.

Therefore, you can do the following:

 $$\left(\textcolor{red}{\sqrt{2}}+\textcolor{blue}{\sqrt{8}}\right)^2$$ Use the rule that $$(\textcolor{red}{a}+\textcolor{blue}{b})^2=\textcolor{red}{a}^2+2\textcolor{red}{a}\textcolor{blue}{b}+\textcolor{blue}{b}^2$$ $$\textcolor{red}{\sqrt{2}}^2+2\textcolor{red}{\sqrt{2}}\textcolor{blue}{\sqrt{8}}+\textcolor{blue}{\sqrt{8}}^2$$ Simplify the radicals. $$2+2\sqrt{8*2}+8$$ Here, I used another algabraic rule that $$\sqrt{a}*\sqrt{b}=\sqrt{ab}\hspace{1mm}\text{if}\hspace{1mm}a,b<0$$ $$2+2\sqrt{16}+8$$ Continue simplifying to the answer. $$2+2*4+8$$ $$2+8+8$$ $$18$$
TheXSquaredFactor  Sep 27, 2017
#3
+648
+1

Oh! Thank you! Well, I see where I went wrong, so thank you! Good job answering their question properly.