#### Question

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

(4 - k)x^{2} + (2k + 4)x + 8k + 1 = 0

#### Solution

The given quadric equation is (4 - k)x^{2} + (2k + 4)x + 8k + 1 = 0, and roots are real and equal

Then find the value of *k.*

Here,

a = 4 - k, b = (2k + 4) and c = 8k + 1

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

Putting the value of a = 4 - k, b = (2k + 4) and c = 8k + 1

= (2k + 4)^{2} - 4 x (4 - k) x (8k + 1)

= (4k^{2} + 16k + 16) - 4(-8k^{2} + 31k + 4)

= 4k^{2} + 16k + 16 + 32k^{2} - 124k - 16

= 36k^{2} - 108k + 0

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

36k^{2} - 108k + 0 = 0

36(k^{2} - 3k) = 0

k^{2} - 3k = 0

Now factorizing of the above equation

k(k - 3) = 0

So, either

k = 0

Or

k - 3 = 0

k = 3

Therefore, the value of k = 0, 3.