I am having some trouble with this one. I think exponentiation would work but i can't seem to get it right. the answer is x=36. could someone show an intermediate step so i can see which form, radical or fractional exponent to use. thanks.

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- January 17th 2011, 11:43 AMskokerRational Exponent Equation.
I am having some trouble with this one. I think exponentiation would work but i can't seem to get it right. the answer is x=36. could someone show an intermediate step so i can see which form, radical or fractional exponent to use. thanks.

- January 17th 2011, 11:57 AMe^(i*pi)
Multiply through by

or, in standard form:

This is a quadratic equation (and factors)

Edit: Note that the domain demands that so discard any negative solution - January 17th 2011, 01:45 PMskoker
I tried to do a normal quadratic factor and get (x=4,x=9) which is not right. quadratic formula does not seem to give 36 also. I also tried to exponentiation but it just shifts the radical on x. how to factor this?

- January 17th 2011, 01:52 PMe^(i*pi)
- using the quadratic formula would give you so you'd need to square the value obtained to get x. Hence if you got 6 from the quadratic formula you'd have done it right

Perhaps it's easier if you let which would give - January 17th 2011, 01:56 PMpickslides
From

make

Finish him off! - January 17th 2011, 02:23 PMskoker
ah I got it thanks. I got which looked right for a second. Is there a rule where you can't get 2 negative factors if the quadratic has negative terms, or something like that?

- January 17th 2011, 02:38 PMe^(i*pi)
That's not right. The last term will be +6 instead of -6.

Me and Pickslides have given the correct factorisation in posts 4 and 5 respectively although I use trial and error to get it right when factoring - January 17th 2011, 02:41 PMskoker
yah. you are correct :) thanks for you help.

- January 17th 2011, 02:47 PMevanator
You will have two values for .

You need to get rid of the negative one because for your purposes we say that the square root of a negative number is undefined. So there is one correct value for here.

I think that's what you were asking about, but I'm not sure.