Giving best answer, please help with this Solve 4/(x^2-8x+12) = x/(x-2) + 1/(x-6)
@Noura11
4/((x-2)(x-6))=(x(x-6)+1(x-2))/(x-2)(x-6) Denominators cancel out... 4=x(x-6)+1(x-2) 4=x^2-6x+x-2 4=x^2-5x-2 x^2-5x-6=0 hope u can do further.......
We have : \[\frac4{x^2-8x+12}=\frac x{x-2}+\frac1{x-6}\] So : \[\frac{4}{x^2-8x+12}=\frac{x(x-6)+x-2}{(x-2)(x-6)}\] So : \[\frac{4}{x^2-8x+12}=\frac{x^2-6x+x-2}{x^2-6x-2x+12}\] So : \[\frac{4}{x^2-8x+12}=\frac{x^2-5x-2}{x^2-8x+12}\] so we get simply : \[x^2-5x-2=4\] And this is simple equation, can you complete the solution from here ?
Could you tell me what you got just to be sure I got the sane answer? Not the best at this haha @Noura11
lol ! I think you should get \[x=6\text{ or } x=1\] So the solution is \[S=\{6,1\}\]
That's incorrect. If you check it, you'll realize why
x=6 is undefined...
x^2-5x-6=0 x^2-6x+x-6=0 x(x-6)+1(x-6)=0 (x+1)(x-6)=0 x=-1,6
x=6 is undefined....
why?
1/(x-6)
ya....
when you cancel the denominators, you need to make note that x cant be 2 and x cant be 6
then,-1 is the answer
correct
as it satisfies the original equation.....
right
I was just explaining what @hero was referring to
okk....thanks @zzr0ck3r
@zzr0ck3r, you don't want to see my solution to this.
When I do it, I get exactly x = -1
Without having to deal with factoring a trinomial.
would you have always solved it like that?
That's how I typically solve it, yes.
where are you from?
or where were you schooled rather?
I'm from the US. I don't know what you mean by school. But yeah, for the most part, I do develop my own methods. Some I steal from others.
im only asking because we all seem to go the exact same rout, and yours is so different.
this one I invented myself.
love it
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