$\left(\sqrt{8}+\sqrt{2}\right)^2$ = 18 why??

Question

I would like to now why

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

is equal to $18$? Please provide me the proccess. Thank you.

Answer 1

$$ (\sqrt{8}+\sqrt{2})^2 = (\sqrt{8})^2+2\sqrt{2}\sqrt{8}+(\sqrt{2})^2=8+2\sqrt{2\cdot 8}+2 = 10 + 2 \cdot 4 = 18$$

Answer 2

Hint: $\sqrt{8 \cdot 2} = \sqrt{16} = 4$.

Answer 3

Hint: $(a+b)^2 = a^2 + b^2 + 2ab$

Answer 4

$$\left(\sqrt{8}+\sqrt{2}\right)^2=\left(2\sqrt{2}+\sqrt{2}\right)^2=\left(3\sqrt{2}\right)^2=9\times2=18$$

Answer 5

Hint:

$$\begin{align} \sqrt{8}+\sqrt{2}&=2\sqrt{2}+\sqrt{2}\\ &=(2+1)\sqrt{2}\\ &=3\sqrt{2} \end{align}$$

Thus,

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

Answer 6

Take out the common factor of $\sqrt2$. $$ \begin{align*} (\sqrt8+\sqrt2)^2 &=(2\sqrt2+\sqrt2)^2\\ &=\sqrt2^2(2+1)^2\\ &=2\times3^2\\ &=\cdots\\ \end{align*}$$

Answer 7

Hint :

$$ (a+b)^2=(\color{red}a+\color{blue}b)(a+b)=\color{red}a(a+b)+\color{blue}b(a+b)=a^2+2ab+b^2. $$

Answer 8

It does look counter-intuitive, doesn't it? But if you step through the multiplication, I think it will become crystal clear to you. First of all, $$(\sqrt{8} + \sqrt{2})^2 = (\sqrt{8} + \sqrt{2})(\sqrt{8} + \sqrt{2})$$, right? Applying FOIL (First, Outer, Inner, Last), we get $$\sqrt{8} \sqrt{8} = 8$$ $$\sqrt{8} \sqrt{2} = 4$$ $$\sqrt{2} \sqrt{8} = 4$$ $$\sqrt{2} \sqrt{2} = 2$$ And then $8 + 4 + 4 + 2 = 18$.