Question

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

(k + 1)x² - 2(3k + 1)x + 8k + 1 = 0

Answer 1

The given quadric equation is (k + 1)x² - 2(3k + 1)x + 8k + 1 = 0, and roots are real and equal

Then find the value of k.

Here,

a = k + 1, b = -2(3k + 1)x and c = 8k + 1

As we know that D = b² - 4ac

Putting the value of a = k + 1, b = -2(3k + 1)x and c = 8k + 1

= (-2(3k + 1))² - 4 x (k + 1) x (8k + 1)

= 4(9k² + 6k + 1) - 4(8k² + 9k + 1)

= 36k² + 24k + 4 - 32k² - 36k - 4

= 4k² - 12k

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

4k² - 12k = 0

k² - 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.