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

Guest Sep 27, 2017

Best Answer 

 #1
avatar+771 
0

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

You can distribute the square.

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

The square root and the square cancel.

\(2+8=10\)

AdamTaurus  Sep 27, 2017
 #1
avatar+771 
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
avatar+2248 
+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
avatar+771 
+1

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

AdamTaurus  Sep 28, 2017

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