SOLUTION: show that the equations y=kx-1 and y=x^2+(k-1)x+k^2 have no real solution for all values of k.

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Question 156386: show that the equations y=kx-1 and y=x^2+(k-1)x+k^2 have no real solution for all values of k.
Answer by Edwin McCravy(20064) About Me  (Show Source):
You can put this solution on YOUR website!
show that the equations y=kx-1 and y=x%5E2%2B%28k-1%29x%2Bk%5E2
have no real common solution for all values of k.


system%28y=kx-1%2C+y=x%5E2%2B%28k-1%29x%2Bk%5E2%29

Set the right side of the first equation equal to
the right side of the second equation, since both
equal to y:

kx-1=x%5E2%2B%28k-1%29x%2Bk%5E2

Distribute to remove the parentheses:

kx-1=x%5E2%2Bkx-x%2Bk%5E2

Get 0 on the left side by adding -kx%2B1
to both sides:

kx-1-kx%2B1=x%5E2%2Bkx-x%2Bk%5E2-kx%2B1

0=x%5E2-x%2Bk%5E2%2B1

Swap sides:

x%5E2-x%2Bk%5E2%2B1=0

Group the last two terms on the left in 
parentheses:

We need to find the DISCRIMINANT

DISCRIMINANT_OF_ax%5E2%2Bbx%2Bc+=+b%5E2-4ac
x%5E2-x%2B%28k%5E2%2B1%29=0

is the same as

1x%5E2-1x%2B%28k%5E2%2B1%29=0

So a=1, b=-1, c=k%5E2%2B1

DISCRIMINANT=b%5E2-4ac
DISCRIMINANT=%28-1%29%5E2-4%281%29%28k%5E2%2B1%29
DISCRIMINANT=1-4%28k%5E2%2B1%29
DISCRIMINANT=1-4k%5E2-4
DISCRIMINANT=-3-4k%5E2

There are no real solutions when the
DISCRIMINANT%3C0

and -3-4k%5E2 is ALWAYS negative,
since k%5E2 is never negative.

So there can be no real common solutions 
for any real value of k.

Edwin