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May 6th, 2013, 01:12 PM   #1
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can someone work this out

Find the discriminant;

1/(x-1)= -x+k
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May 6th, 2013, 02:10 PM   #2
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Re: can someone work this out

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May 6th, 2013, 03:42 PM   #3
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Re: can someone work this out

I do not understand..

Is it not correct to rearranging to x^2+kx+x-k-1=0 , then plug in for b^2-4ac? I am really lost on this, if someone could show me the step I'd appreciate it a lot.
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May 6th, 2013, 04:23 PM   #4
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How did you get x^2 + kx + x - k - 1 = 0, rather than -x^2 + kx + x - k - 1 = 0?
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May 6th, 2013, 04:38 PM   #5
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Re: can someone work this out

I meant - x^2+ kx +x -k -1= 0, but how do you plug this into b^2-4ac?
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May 6th, 2013, 04:55 PM   #6
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The expression b - 4ac uses the variable a for the coefficient of x, b for the coefficient of x, and c for the constant term.

Hence b - 4ac = (k + 1) - 4(k + 1) = k - 2k - 3.
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May 6th, 2013, 06:00 PM   #7
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Re: can someone work this out

So, I look at - x^2+ kx +x -k -1= 0.

-x^2 has coefficent of -1... so a = -1

now theres 2 x's. So one of them has a coefficient of 1, the other has the variable k hmm so b= 1 ??? b= k??? b= k+1? Im going to go with (k+1) so I can make it at least look similar to your equation...

now the constant term is clearly -1. But I havn't done anything with the free 'k' yet so thats making me anxious, maybe k is like a 'variable constant' which makes no sense to me, but then again I could make it look like yours, say c= -(k+1)

So a= -1 , b= (k+1), c= (-k-1)

then I plug in (k+1)^2 - 4 (-1)(-k-1)=......(k+1)^2 -4(k+1) and I see now that that equals k^2-2k -3
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May 6th, 2013, 06:47 PM   #8
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May 7th, 2013, 06:40 AM   #9
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Perhaps "constant" wasn't the best word to use. If there are separate terms, such as kx and x, they should be taken together, so one would take the coefficient of x in my example to be k+1. Any terms not involving x are taken together as the value for c. For your problem, a = -1, b = k + 1, and c = -k - 1 will work. Note that if you change the sign of a and c simultaneously, the value of the discriminant is unaffected.
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May 7th, 2013, 11:32 AM   #10
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Re: can someone work this out

I think I got it now.
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