We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.

+0

# Please solve

0
387
3

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

Sep 27, 2017

### Best Answer

#1
+865
0

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

You can distribute the square.

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

The square root and the square cancel.

$$2+8=10$$

.
Sep 27, 2017

### 3+0 Answers

#1
+865
0
Best Answer

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

You can distribute the square.

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

The square root and the square cancel.

$$2+8=10$$

AdamTaurus Sep 27, 2017
#2
+2338
+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
+865
+1

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

AdamTaurus  Sep 28, 2017