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Q: find k: (k-1)x^2+kx-4=o has only one sodutium?
A: The equation (k-1)x^2+kx-4=0 has only one solution, then k^2 - 4(k-1)(-4)=0 k^2+16(k-1)=0 k^2+16k-16=0 k^2+16k+64=80 (k+8)^2=80 k+8= ±sqrt 80 k = -8 ± 4 sqrt 5
Original conversation
User: find k: (k-1)x^2+kx-4=o has only one sodutium?

Question
Asked 2/16/2009 3:32:23 PM
Updated 12/26/2014 2:53:29 AM
Edited by jeifunk [12/26/2014 2:53:26 AM], Flagged by jeifunk [12/26/2014 2:53:29 AM]
Rating
3
The equation (k-1)x^2+kx-4=0 has only one solution,
then k^2 - 4(k-1)(-4)=0
k^2+16(k-1)=0
k^2+16k-16=0
k^2+16k+64=80
(k+8)^2=80
k+8= ±sqrt 80
k = -8 ± 4 sqrt 5
Added 12/26/2014 1:15:49 AM
This answer has been confirmed as correct and helpful.
Confirmed by jeifunk [12/26/2014 2:53:28 AM]

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