Adding and Subtracting Radicals

**chandler** · February 12th 2011, 06:53 AM

How would you simplify and do these problems?

(square root of) 2/5 + (square root of) 1/2

8 - (square root of) 5/7

I understand the concept... I just always get these wrong for some reason, so please be detailed so I can understand! Thanks!

**Plato** · February 12th 2011, 07:01 AM

Originally Posted by chandler

How would you simplify and do these problems?

(square root of) 2/5 + (square root of) 1/2

8 - (square root of) 5/7

I understand the concept... I just always get these wrong for some reason, so please be detailed so I can understand! Thanks!

Are there two separate questions there?

Is the first one $\sqrt {\dfrac{2}{5}} + \sqrt {\dfrac{1}{2}}~?$

**chandler** · February 12th 2011, 07:23 AM

Yes, these are two different equations.

Also... How do you make the square root symbol?

**Prove It** · February 12th 2011, 07:26 AM

Originally Posted by chandler

How would you simplify and do these problems?

(square root of) 2/5 + (square root of) 1/2

8 - (square root of) 5/7

I understand the concept... I just always get these wrong for some reason, so please be detailed so I can understand! Thanks!

$\displaystyle \sqrt{\frac{2}{5}} + \sqrt{\frac{1}{2}} = \sqrt{\frac{4}{10}} + \sqrt{\frac{5}{10}}$

$\displaystyle = \frac{\sqrt{4}}{\sqrt{10}} + \frac{\sqrt{5}}{\sqrt{10}}$

$\displaystyle = \frac{\sqrt{4} + \sqrt{5}}{\sqrt{10}}$

$\displaystyle = \frac{2 + \sqrt{5}}{\sqrt{10}}$

$\displaystyle = \frac{\sqrt{10}(2 + \sqrt{5})}{10}$

$\displaystyle = \frac{2\sqrt{10} + \sqrt{50}}{10}$

$\displaystyle = \frac{2\sqrt{10} + \sqrt{25}\sqrt{2}}{10}$

$\displaystyle = \frac{2\sqrt{10} + 5\sqrt{2}}{10}$ .

**Plato** · February 12th 2011, 07:38 AM

Originally Posted by chandler

How do you make the square root symbol?

[tex]\sqrt {\dfrac{2}{5}} + \sqrt {\dfrac{1}{2}}[/tex] gives $\sqrt {\dfrac{2}{5}} + \sqrt {\dfrac{1}{2}}$

**chandler** · February 12th 2011, 08:08 AM

One more thing...

Please do this one too: $\sqrt {\dfrac{1}{2}} + \sqrt {\dfrac{1}{2}}$ =

**Plato** · February 12th 2011, 08:12 AM

Originally Posted by chandler

One more thing...

Please do this one too: $\sqrt {\dfrac{1}{2}} + \sqrt {\dfrac{1}{2}}$ =

$\sqrt {\dfrac{1}{2}}={\dfrac{\sqrt2}{2}}$

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Adding and Subtracting Radicals

Yes, two seperate

Thanks guys!

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