Rationalizing

**Morzilla** · December 16th 2007, 01:32 PM

I am not sure if I took down the correct notes but...

In rationalizing denominators, we multiply the denominator on both the top and bottom as such

$\frac{9}{\sqrt{2+7}}$ X $\frac{(\sqrt{2-7})}{(\sqrt{2-7})}<br />$

= $\frac{9(\sqrt{2-7})}{(\sqrt{4-7})(\sqrt{2-7})(\sqrt{2-49})}$

$\frac{9(\sqrt{2-7})}{-47}$ in order to get rid of the negative we multiply by (-1) and get $\frac{-9(\sqrt{2-7})}{47}$

again did I write my notes in the wrong way?!

The problem I'm trying to work out is $\frac{\sqrt{6}}{\sqrt{5+9}}$

Thanks

**Soroban** · December 16th 2007, 01:53 PM

Hello, Morzilla!

Your work makes no sense . . .

Then at the end you add:

. . The problem I'm trying to work out is: . ${\color{blue}\frac{\sqrt{6}}{\sqrt{5+9}}}$

If that's true, we have: . $\frac{\sqrt{6}}{\sqrt{14}} \:=\:\sqrt{\frac{6}{14}} \:=\:\sqrt{\frac{3}{7}} \:=\:\sqrt{\frac{3}{7}\cdot\frac{7}{7}} \:=\:\sqrt{\frac{21}{49}} \:=\:\frac{\sqrt{21}}{\sqrt{49}} \:=\:\frac{\sqrt{21}}{7}$

Why are you messing around with conjugates?

**Morzilla** · December 16th 2007, 02:10 PM

Originally Posted by Soroban

Hello, Morzilla!

Your work makes no sense . . .

Then at the end you add:

. . The problem I'm trying to work out is: . ${\color{blue}\frac{\sqrt{6}}{\sqrt{5+9}}}$

If that's true, we have: . $\frac{\sqrt{6}}{\sqrt{14}} \:=\:\sqrt{\frac{6}{14}} \:=\:\sqrt{\frac{3}{7}} \:=\:\sqrt{\frac{3}{7}\cdot\frac{7}{7}} \:=\:\sqrt{\frac{21}{49}} \:=\:\frac{\sqrt{21}}{\sqrt{49}} \:=\:\frac{\sqrt{21}}{7}$

Why are you messing around with conjugates?

........cuz, I'm a dummy

!

....(O.x) ok wait, the answer shown is $\frac{\sqrt{30}-9\sqrt{6}}{-76}$ .....I must admit that I am now 100% lost! Sorry for that.......

**Soroban** · December 16th 2007, 06:34 PM

Hello, Morzilla!

Okay, I think I get the idea.
It would help if you copied the problem correctly.

And I don't like the way they left their answer . . . very childish!

Rationalize: . $\frac{\sqrt{6}}{\sqrt{5} + 9}$

Rationalize . . . multiply top and bottom by the conjugate . . .

$\frac{\sqrt{6}}{\sqrt{5} + 9}\cdot\frac{\sqrt{5} - 9}{\sqrt{5} - 9} \;=\;\frac{\sqrt{6}(\sqrt{5} - 9)}{(\sqrt{5})^2 - 9^2} \;=\;\frac{\sqrt{6}\!\cdot\!\sqrt{5} - \sqrt{6}\!\cdot\!9}{5 - 81} \;=\;\frac{\sqrt{30} - 9\sqrt{6}}{-76}$

Multiply by $\frac{-1}{-1}\!:\;\;\frac{-1(\sqrt{30} - 9\sqrt{6})}{-1(-76)} \;=\;\boxed{\frac{9\sqrt{6} - \sqrt{30}}{76}}$

I don't know anyone past the age of puberty who leaves a negative denomiantor.
.

**Morzilla** · December 16th 2007, 09:17 PM

Originally Posted by Soroban

Hello, Morzilla!

Okay, I think I get the idea.
It would help if you copied the problem correctly.

And I don't like the way they left their answer . . . very childish!

Rationalize . . . multiply top and bottom by the conjugate . . .

$\frac{\sqrt{6}}{\sqrt{5} + 9}\cdot\frac{\sqrt{5} - 9}{\sqrt{5} - 9} \;=\;\frac{\sqrt{6}(\sqrt{5} - 9)}{(\sqrt{5})^2 - 9^2} \;=\;\frac{\sqrt{6}\!\cdot\!\sqrt{5} - \sqrt{6}\!\cdot\!9}{5 - 81} \;=\;\frac{\sqrt{30} - 9\sqrt{6}}{-76}$

Multiply by $\frac{-1}{-1}\!:\;\;\frac{-1(\sqrt{30} - 9\sqrt{6})}{-1(-76)} \;=\;\boxed{\frac{9\sqrt{6} - \sqrt{30}}{76}}$

I don't know anyone past the age of puberty who leaves a negative denomiantor.
.

So I did write the wrong notes, what a dingus I am!!!

Thank you so much for all the help!......yeah I never met them either, i just see his ugly ass every morning in the mirror

!!

Well Tomorrow shall be the day.....Again thanks for everything and Happy Festivous!!!

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