help

**blame_canada100** · September 8th 2008, 04:30 PM

√5(2√20+√2)

**Quick** · September 8th 2008, 04:34 PM

Originally Posted by blame_canada100

√5(2√20+√2)

do you mean this: $\sqrt{5}(2\sqrt{20}+\sqrt{2})$

If so, then start by finding the perfect squares: $\sqrt{5}(2\sqrt{4\cdot 5}+\sqrt{2})$

Then spread them out: $\sqrt{5}(2\sqrt{4}\sqrt{5}+\sqrt{2})$

Then simplify: $\sqrt{5}(4\sqrt{5}+\sqrt{2})$

Now distribute: $4\sqrt{5}\sqrt{5}+\sqrt{2}\sqrt{5}$

Now combine together: $4\sqrt{25}+\sqrt{10}$

Now simplify: $4\cdot 5+\sqrt{10}$

Therefore: $\boxed{ \sqrt{5}(2\sqrt{4\cdot 5}+\sqrt{2})=20+\sqrt{10}}$

**Soroban** · September 8th 2008, 06:01 PM

Hello, blame_canada100!

Simplify: . $\sqrt{5}\left(2\sqrt{20} + \sqrt{2}\right)$

Multiply: . $\underbrace{(\sqrt{5})(2\sqrt{20})} + \underbrace{(\sqrt{5})(\sqrt{2})}$

. . . . . . . $= \;2\sqrt{100} \quad+ \quad\sqrt{10}$

. . . . . . . $=\;\; 2\cdot 10 \quad+ \quad\sqrt{10}$

Answer: . $20 + \sqrt{10}$

Thread: help

LinkBack

Thread Tools

Display

help