# solutions for k values

• Jan 8th 2010, 05:47 PM
smmmc
solutions for k values
show that (k+1)x^2 -2x -k = 0 has a solution for all values of k.

i used b^2-4ac=0

i ended up with k=-2, not sure what the step is next

thanks
• Jan 8th 2010, 05:53 PM
mr fantastic
Quote:

Originally Posted by smmmc
show that (k+1)x^2 -2x -k = 0 has a solution for all values of k.

i used b^2-4ac=0

i ended up with k=-2, not sure what the step is next

thanks

All you have to do is show that $\displaystyle b^2 - 4ac \geq 0$ for all values of k.
• Jan 8th 2010, 07:36 PM
Soroban
Hello, smmmc!

Quote:

Show that $\displaystyle (k+1)x^2 -2x -k \:=\: 0$ has a solution for all values of $\displaystyle k$.

i used $\displaystyle b^2-4ac\:=\:0$ . . Why zero?

i ended up with k = -2 . . . . How?

$\displaystyle \text{We have: }\;\underbrace{(k+1)}_{a}x^2 + \underbrace{(-2)}_{b}x + \underbrace{(-k)}_{c} \:=\:0$

$\displaystyle b^2-4ac \;=\;(-2)^2 - 4(k+1)(-k) \;=\; 4 + 4k^2 + 4k \;=\;4(k^2 + k + 1)$

And it can be shown that $\displaystyle k^2 + k + 1$ is always positive.