Find the values of k for which the quadratic equation $(3 k + 1) x^{2} + 2 (k + 1) x + 1 = 0$ has equal roots. Also, find the roots.

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Solution

The given quadric equation is $(3 k + 1) x^{2} + 2 (k + 1) x + 1 = 0$ , and roots are real and equal.

Then, find the value of k.

Here, $a = 3 k + 1, b = 2 (k + 1) and c = 1$ .

As we know that $D = b^{2} - 4 a c$

Putting the values of $a = 3 k + 1, b = 2 (k + 1) and c = 1$ .

$D = {[2 (k + 1)]}^{2} - 4 (3 k + 1) (1) = 4 (k^{2} + 2 k + 1) - 12 k - 4 = 4 k^{2} + 8 k + 4 - 12 k - 4 = 4 k^{2} - 4 k$

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

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

Therefore, the value of k is 0 or 1.

Now, for k = 0, the equation becomes

$x^{2} + 2 x + 1 = 0 \Rightarrow x^{2} + x + x + 1 = 0 \Rightarrow x (x + 1) + 1 (x + 1) = 0 \Rightarrow (x + 1)^{2} = 0 \Rightarrow x = - 1, - 1$

for k = 1, the equation becomes

$4 x^{2} + 4 x + 1 = 0 \Rightarrow 4 x^{2} + 2 x + 2 x + 1 = 0 \Rightarrow 2 x (2 x + 1) + 1 (2 x + 1) = 0 \Rightarrow (2 x + 1)^{2} = 0 \Rightarrow x = - \frac{1}{2}, - \frac{1}{2}$

Hence, the roots of the equation are $- 1 and - \frac{1}{2}$ .

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