# values of k

• Sep 24th 2009, 02:11 AM
mark
values of k
hi, i'm stuck on a question and would like some help please:

find the values for k for which the quadratic equation $3x^2 + (6 - k)x + 12 = 0$ has equal roots. i used the discriminant theory and got it to $\sqrt{36 - 12k + k^2 - 144}$ = $\sqrt{-108 - 12k + k^2}$ and then $- 12k + k^2 = 108$ but i'm really not sure if any of this is right. could someone show me if this is the right way?

thanks, Mark
• Sep 24th 2009, 02:17 AM
red_dog
You have to solve the quadratic $k^2-12k-108=0$ and find k.
• Sep 24th 2009, 02:20 AM
mark
i see, where did the 8 come from?
• Sep 24th 2009, 02:25 AM
red_dog
Sorry, it's 0 not 8. I pushed the wrong button. I edited.
• Sep 24th 2009, 06:03 AM
stapel
Quote:

Originally Posted by mark
find the values for k for which the quadratic equation $3x^2 + (6 - k)x + 12 = 0$ has equal roots.

For the roots to be equal, the "plus minus" part must be zero. This means that:

. . . . . $(6\, -\, k)^2\, -\, 4(3)(12)\, =\, 0$

. . . . . $36\, -\, 12k\, +\, k^2\, -\, 144\, =\, 0$

. . . . . $k^2\, -\, 12k\, -\, 108\, =\, 0$

That's where the earlier helper got his/her equation. (Rock)

Now apply the Quadratic Formula to this equation to find the value(s) of k. (Wink)