#1**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**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**+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**+1 **

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

AdamTaurus
Sep 28, 2017