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May 6th, 2013, 01:12 PM  #1 
Senior Member Joined: May 2012 Posts: 203 Thanks: 5  can someone work this out
Find the discriminant; 1/(x1)= x+k 
May 6th, 2013, 02:10 PM  #2 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,786 Thanks: 1029 Math Focus: Elementary mathematics and beyond  Re: can someone work this out 
May 6th, 2013, 03:42 PM  #3 
Senior Member Joined: May 2012 Posts: 203 Thanks: 5  Re: can someone work this out
I do not understand.. Is it not correct to rearranging to x^2+kx+xk1=0 , then plug in for b^24ac? I am really lost on this, if someone could show me the step I'd appreciate it a lot. 
May 6th, 2013, 04:23 PM  #4 
Global Moderator Joined: Dec 2006 Posts: 18,844 Thanks: 1566 
How did you get x^2 + kx + x  k  1 = 0, rather than x^2 + kx + x  k  1 = 0?

May 6th, 2013, 04:38 PM  #5 
Senior Member Joined: May 2012 Posts: 203 Thanks: 5  Re: can someone work this out
I meant  x^2+ kx +x k 1= 0, but how do you plug this into b^24ac?

May 6th, 2013, 04:55 PM  #6 
Global Moderator Joined: Dec 2006 Posts: 18,844 Thanks: 1566 
The expression b²  4ac uses the variable a for the coefficient of x², b for the coefficient of x, and c for the constant term. Hence b²  4ac = (k + 1)²  4(k + 1) = k²  2k  3. 
May 6th, 2013, 06:00 PM  #7 
Senior Member Joined: May 2012 Posts: 203 Thanks: 5  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= (k1) then I plug in (k+1)^2  4 (1)(k1)=......(k+1)^2 4(k+1) and I see now that that equals k^22k 3 
May 6th, 2013, 06:47 PM  #8 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,786 Thanks: 1029 Math Focus: Elementary mathematics and beyond  Re: can someone work this out 
May 7th, 2013, 06:40 AM  #9 
Global Moderator Joined: Dec 2006 Posts: 18,844 Thanks: 1566 
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.

May 7th, 2013, 11:32 AM  #10 
Senior Member Joined: May 2012 Posts: 203 Thanks: 5  Re: can someone work this out
I think I got it now.


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