Solving a Root Equation for the Variable r

[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