Solving a Root Equation for the Variable r

[arildno]

okay, so whenever you've got a surd in your denominator, you are to rationalize it. Fine by me; mind you, that is THEIR choice, not everybody's else's choice.

Remember that $$1=\frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}-\sqrt{2}}$$
See if you can use that to derive their answer.

 

[thomas49th]

$$\frac{3}{(\sqrt{6} + \sqrt{2})}$$

$$\frac{3(\sqrt{6} - \sqrt{2})}{(\sqrt{6} + \sqrt{2})(\sqrt{6} - \sqrt{2})}$$

use smilie face method on denominator

root 6 x root 6 = 6
root 2 x - root 2 = - 2

root 6 x - root 2 = - root 12
root 6 x root 2 = root 12

- root 12 and root 12 cancel each other out. 6 - 2 = 4



leaving you with
$$\frac{3(\sqrt{6} - \sqrt{2})}4$$

am I right or am I right?

 

[arildno]

Indeed you are right. Here is a true smilie for you:

 

[thomas49th]

Okay. I've got some more problems i wish to solve

Express
$$\frac{1}{x - 2} + \frac{2}{x+4}$$
as a single algergraic faction

 

[arildno]

So, what is the least common multiple of the denominators?

 

[thomas49th]

would it be 2?

 

[arildno]

Does x-2 divide 2?? Does x+4 divide 2??

 

[thomas49th]

is it (x-2)(x+4) = x² -8 + 2x

 

[thomas49th]

ahh me thinks you cross multiply?

$$\frac{x+4 + 2(x-2)}{(x-2)(x+4)}$$

 

[arildno]

I do not cross-multiply here either, I EXPAND both fractions in the manner you've just done.

 

[thomas49th]

To add the fractions I need to find a common denominator right?
What ever I do to the denominator I must do to the numerator right?
So how do I find the lost common mulitple of x-2 and x+4?


Thanks

 

[arildno]

You've done it just fine, as I've said.
Just don't call it cross-multiplication.

 

[thomas49th]

so I am at $$\frac{3x}{x^{2}+2x-8}$$
now where to? That's not the final answer is it?

 

[arildno]

Looks final to me, given your task to find a SINGLE algebraic fraction identical to the given sum.

 

[thomas49th]

Ive looked this up on an old test paper, and apparently the answer is 1/3

The final step before the answer I've written

3x / ((x-2)(x+4))

but I don't know how I got 1/3...do you?

 

[arildno]

That's impossible.
you have most likely miscopied the original problem.

 

[thomas49th]

Ok, must of done. Now interestly the next question on the paper is

Hence or otherwise sove

$$\frac{1}{x-2} + \frac{2}{x+4} = \frac{1}{3}$$

so I will EXPAND the fractions (or cross multiply)?
this will give me

9x = x² - 2x + 4x - 8
so this is a quadratic. - 9x

x² -7x + 8 = 0
(x + 1)(x-8) = 0
so x = -1 or x= 8

that correct?

 

[arildno]

Which is a totally different issue altogether!
What you have there is an EQUATION, what you said before was that that equality was an IDENTITY (which is NOT correct).

 

[thomas49th]

aha i see
to solve it then

$$\frac{1}{x-2} + \frac{2}{x+4} = \frac{1}{3}$$

so I will EXPAND the fractions (or cross multiply)?
this will give me

9x = x² - 2x + 4x - 8
so this is a quadratic. - 9x

x² -7x + 8 = 0
(x + 1)(x-8) = 0
so x = -1 or x= 8

that correct?

 

[arildno]

Seems so, yes.

 

[thomas49th]

O, right first time...

Now here's a hard one I don't get

Find the value of

m when $$\sqrt{128} = 2^{m}$$

I no straight away from binary that 2^7 is 128 does that help?

 

[arildno]

Indeed it helps!
Remember how roots can be written as exponents..

 

[thomas49th]

$$\frac{7}{2}$$

but how would i solve it normally?

 

[arildno]

Indeed, that is what m equals, as soon as you get the LateX right..

 

[thomas49th]

but say if i didn't know about binary how would i got about solving it
somthing to do with surds isn't it?

 

[HallsofIvy]

In fact, perhaps your original post contained a typo. You went from
A = πr² + r²root(k²-1) to r³ = (2A/π+root(k²-1)
Did you really mean the r³, rather than r²?

 

[thomas49th]

i believe your right