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

1 Answer/Comment

s

Question

Asked 2/16/2009 3:32:23 PM

Updated 12/26/2014 2:53:29 AM

1 Answer/Comment

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

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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