# Substitution to transform to quadratic form

• Nov 6th 2012, 04:58 PM
hexrei
Substitution to transform to quadratic form
Can anyone help me with this?

$\displaystyle 3=\frac{1}{(x+1)^2} + \frac{2}{(x+1)}$

I sub $\displaystyle u={(x+1)^-1}$

then flip them and rearrange so I get $\displaystyle u^2+\frac{u}{2}-3=0$

then mutiply all by two to get rid of fraction.

then I factor and end up with $\displaystyle (2u-3)(u+2)=0$

I think the problems begin when I sub back in the $\displaystyle {(x+1)^-1}$... doing something wrong and not getting book's solutions.

btw those -1's are supposed to be superscripted, not sure why its not working :/
• Nov 6th 2012, 05:09 PM
skeeter
Re: Substitution to transform to quadratic form
if $\displaystyle u = \frac{1}{x+1}$ , then the correct substitution would be ...

$\displaystyle 3 = u^2 + 2u$

try again ...
• Nov 6th 2012, 05:11 PM
HallsofIvy
Re: Substitution to transform to quadratic form
$\displaystyle \frac{1}{a}+ \frac{1}{b}$ "flipped" does NOT give "a+ b".
• Nov 6th 2012, 05:17 PM
hexrei
Re: Substitution to transform to quadratic form
Quote:

Originally Posted by skeeter
if $\displaystyle u = \frac{1}{x+1}$ , then the correct substitution would be ...

$\displaystyle 3 = u^2 + 2u$

try again ...

I appreciate the help :) and of course I'm going to try again. Why would I have taken the time to post it otherwise :)
• Nov 6th 2012, 06:11 PM
hexrei
Re: Substitution to transform to quadratic form
OK, so I have wrapped my head around that early error I made.

I'm getting tripped up after the factorization still though, and the book unfortunately has example subbing with negative exponents with single variable but none with a constant and a variable plus negative exponent.

So after I sub $\displaystyle x+1^{-1}$ back in. I will only list one of the zeros since I think once I see what I am missing the second shouldn't be hard.

my factorization is now $\displaystyle (u+3)(u-1)=0$

which after subbing back in the first one yields me

$\displaystyle \frac{1}{x+1} + 3$

from there I tried finding a common denominator

$\displaystyle \frac{1}{x+1} + \frac{3x+3}{x+1}$

$\displaystyle \frac{3x+4}{x+1}$

But not sure if that was correct, or what to do next. Can someone give me a clue please?
• Nov 6th 2012, 06:24 PM
MarkFL
Re: Substitution to transform to quadratic form
When is a rational expression equal to zero?
• Nov 6th 2012, 06:27 PM
hexrei
Re: Substitution to transform to quadratic form
Quote:

Originally Posted by MarkFL2
When is a rational expression equal to zero?

oooh..... when the numerator is zero? So x would be -4/3. Got it. Thanks :)