#### Question

Find the values of *k* for which the quadratic equation

\[\left( 3k + 1 \right) x^2 + 2\left( k + 1 \right)x + 1 = 0\] has equal roots. Also, find the roots.

#### Solution

The given quadric equation is \[\left( 3k + 1 \right) x^2 + 2\left( k + 1 \right)x + 1 = 0\] and roots are real and equal.

Then, find the value of *k.*

Here*, *

*\[a = 3k + 1, b = 2(k + 1) \text { and } c = 1\].*

As we know that

\[D = b^2 - 4ac\]

Putting the values of \[a = 3k + 1, b = 2(k + 1) \text { and } c = 1\]

\[D = \left[ 2\left( k + 1 \right) \right]^2 - 4\left( 3k + 1 \right)\left( 1 \right)\]

\[ = 4( k^2 + 2k + 1) - 12k - 4\]

\[ = 4 k^2 + 8k + 4 - 12k - 4\]

\[ = 4 k^2 - 4k\]

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

Thus,

\[4 k^2 - 4k = 0\]

\[\Rightarrow 4k(k - 1) = 0\]

\[ \Rightarrow k = 0 \text { or } k - 1 = 0\]

\[ \Rightarrow k = 0 \text { or } k = 1\]

Therefore, the value of k is 0 or 1.

Now, for k = 0, the equation becomes

\[x^2 + 2x + 1 = 0\]

\[ \Rightarrow x^2 + x + x + 1 = 0\]

\[ \Rightarrow x(x + 1) + 1(x + 1) = 0\]

\[ \Rightarrow (x + 1 )^2 = 0\]

\[ \Rightarrow x = - 1, - 1\]

for k = 1, the equation becomes

\[4 x^2 + 4x + 1 = 0\]

\[ \Rightarrow 4 x^2 + 2x + 2x + 1 = 0\]

\[ \Rightarrow 2x(2x + 1) + 1(2x + 1) = 0\]

\[ \Rightarrow (2x + 1 )^2 = 0\]

\[ \Rightarrow x = - \frac{1}{2}, - \frac{1}{2}\]

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