#### Question

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

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

#### Solution

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

Then find the value of *k.*

Here,

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

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

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

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

= {4(k^{2} + 2k + 1)} - 16(k + 4)

= 4k^{2} + 8k + 4 - 16k - 64

= 4k^{2} - 8k - 60

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

4k^{2} - 8k - 60 = 0

4(k^{2} - 2k - 15) = 0

k^{2} - 2k - 15 = 0

Now factorizing of the above equation

k^{2} - 2k - 15 = 0

k^{2} + 3k - 5k - 15 = 0

k(k + 3) - 5(k + 3) = 0

(k + 3)(k - 5) = 0

So, either

k + 3 = 0

k = -3

Or

k - 5 = 0

k = 5

Therefore, the value of k = -3, 5.