#### Question

Find the values of *k* for which the roots are real and equal in each of the following equation:

(3k+1)x^{2} + 2(k + 1)x + k = 0

#### Solution

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

Then find the value of *k.*

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

As we know that D = b^{2} - 4ac

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

= (2(k + 1))^{2} - 4 x (3k + 1) x (k)

= 4(k^{2} + 2k + 1) - 4k(3k + 1)

= 4k^{2} + 8k + 4 - 12k^{2} - 4k

= -8k^{2} + 4k + 4

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

Thus,

-8k^{2} + 4k + 4 = 0

-4(2k^{2} - k - 1) = 0

2k^{2} - k - 1 = 0

Now factorizing of the above equation

2k^{2} - k - 1 = 0

2k^{2} - 2k + k - 1 = 0

2k(k - 1) + 1(k - 1) = 0

(k - 1)(2k + 1) = 0

So, either

k - 1 = 0

k = 1

Or

2k + 1 = 0

2k = -1

k = -1/2

Therefore, the value of k = 1, -1/2