# factorisation of a quadratic equation that is a fraction :/

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• March 21st 2013, 07:48 PM
jewelmaule
factorisation of a quadratic equation that is a fraction :/
Hey guys im stuck on an equation and just cant seem to get the right answer. I know there must be something wrong with my working. Could someone please show me how you multiply out the denominator to solve this equation.

1+ 6/x-1 = 1/x+1

Thank you in advance for any help.
• March 21st 2013, 08:18 PM
princeps
Re: factorisation of a quadratic equation that is a fraction :/
Hi , multiply equation by $(x-1)(x+1)$
• March 21st 2013, 09:44 PM
jewelmaule
Re: factorisation of a quadratic equation that is a fraction :/
Quote:

Originally Posted by princeps
Hi , multiply equation by $(x-1)(x+1)$

Thank you for your quick reply. I tried that and still cant seem to get the correct answer. I know the answer because I have checked and I cant even work backwards to get it. Ill show my working out and maybe someone will know where im going wrong..

1+ 6/x-1 = 1/x+1
1+6(x+1) = (x-1)
1+6x+6-x-1 = 0
5x+6 = 0
)
(
)
But the proper answer is= X2+5x+6=0
x=-3 & -2

I cant seem to find where x2 came from.
• March 21st 2013, 10:08 PM
princeps
Re: factorisation of a quadratic equation that is a fraction :/
You have to multiply $1$ by $(x-1)(x+1)$ also .
• March 22nd 2013, 02:29 AM
ibdutt
Re: factorisation of a quadratic equation that is a fraction :/
In such cases go by the following algorithm.
Step-1: Multiply the equation be the LCM of the denominator which in this case is (x-1)(x=1)
.(x-1)(x+1)[ 1 + 6/(x-1) = 1/(x+1) OR x^2 -1 + 6(x+1) = (x-1)
Step 2: Simplify the equation
x^2 -1 + 6x+6 = x-1
x^2 -1 + 6x+6 -x+1= 0
x^2 + 5x+6 = 0
(x+3)(x+2) = 0
etc.
• March 22nd 2013, 02:44 AM
jewelmaule
Re: factorisation of a quadratic equation that is a fraction :/
Ahhh I think I get it now! Thanks heaps for all the help.
• March 22nd 2013, 04:42 AM
Prove It
Re: factorisation of a quadratic equation that is a fraction :/
Quote:

Originally Posted by jewelmaule
Hey guys im stuck on an equation and just cant seem to get the right answer. I know there must be something wrong with my working. Could someone please show me how you multiply out the denominator to solve this equation.

1+ 6/x-1 = 1/x+1

Thank you in advance for any help.

I assume that your equation is $\displaystyle 1 + \frac{6}{x-1} = \frac{1}{x + 1}$. Some brackets where they're needed would be nice... Anyway...

\displaystyle \begin{align*} 1 + \frac{6}{x - 1} &= \frac{1}{x + 1} \\ \frac{x-1}{x-1} + \frac{6}{x-1} &= \frac{1}{x+1} \\ \frac{x+5}{x-1} &= \frac{1}{x+1} \\ \frac{(x+5)(x+1)}{(x-1)(x+1)} &= \frac{x-1}{(x-1)(x+1)} \\ (x+5)(x+1) &= x-1 \\ x^2 + 6x + 5 &= x-1 \\ x^2 +5x + 6 &= 0 \\ (x + 3)(x+2) &= 0 \\ x+ 3 = 0\textrm{ or } x+2 &= 0 \\ x = -3 \textrm{ or } x &= -2 \end{align*}
• March 22nd 2013, 02:00 PM
jewelmaule
Re: factorisation of a quadratic equation that is a fraction :/
Quote:

Originally Posted by Prove It
I assume that your equation is $\displaystyle 1 + \frac{6}{x-1} = \frac{1}{x + 1}$. Some brackets where they're needed would be nice... Anyway...

Thank you for your help. Yeah I have been trying to figure out how everyone gets their equations in that format on this forum.