# Find value of $k$

For what value of $k$, are the roots of the quadratic equation $$(k+4)x^2 + (k+1)x +1 = 0$$ equal.

-
What have you tried? –  Ishan Banerjee Mar 2 '13 at 13:40
i tried b^2-4ac=0 –  Learner Mar 2 '13 at 13:41
Well, that's correct. Have more faith in yourself –  Ishan Banerjee Mar 2 '13 at 13:43
I second @IshanBanerjee, the best way to be a Learner is to try and do it, you're nearly there, go on! –  Andreas Caranti Mar 2 '13 at 13:44

$D = (k+1)^2-4(k+4)$
$0 = (k+1)^2-4(k+4) = k^2 - 2k - 15$
there are two solutions: $k = -3$ and $k = 5$