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

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What have you tried? –  Ishan Banerjee Mar 2 '13 at 13:40
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i tried b^2-4ac=0 –  Learner Mar 2 '13 at 13:41
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Well, that's correct. Have more faith in yourself –  Ishan Banerjee Mar 2 '13 at 13:43
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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

1 Answer 1

Roots are equal, when the discriminant of this quadratic equation is zero.

$D = (k+1)^2-4(k+4)$

$0 = (k+1)^2-4(k+4) = k^2 - 2k - 15$

there are two solutions: $k = -3$ and $k = 5$

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