Find the values of k for which the quadratic equation $(3 k + 1) x^{2} + 2 (k + 1) x + 1 = 0$ has real and equal roots. [CBSE 2014]

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Solution

The given equation is $(3 k + 1) x^{2} + 2 (k + 1) x + 1 = 0$ .

This is of the form $a x^{2} + b x + c = 0$ , where a = 3k +1, b = 2(k + 1) and c = 1.

$∴ D = b^{2} - 4 a c = {[2 (k + 1)]}^{2} - 4 \times (3 k + 1) \times 1 = 4 (k^{2} + 2 k + 1) - 4 (3 k + 1) = 4 k^{2} + 8 k + 4 - 12 k - 4$
$= 4 k^{2} - 4 k$

The given equation will have real and equal roots if D = 0.

$∴ 4 k^{2} - 4 k = 0 \Rightarrow 4 k (k - 1) = 0 \Rightarrow k = 0 or k - 1 = 0 \Rightarrow k = 0 or k = 1$

Hence, 0 and 1 are the required values of k.

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