Question

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

x² - 2kx + 7k - 12 = 0

Answer 1

The given quadric equation is x² - 2kx + 7k - 12 = 0, and roots are real and equal

Then find the value of k.

Here,

a = 1, b = -2k and c = 7k - 12

As we know that D = b² - 4ac

Putting the value of a = 1, b = -2k and c = 7k - 12

= (-2k)² - 4 x (1) x (7k - 12)

= 4k² - 28k + 48

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

4k² - 28k + 48 = 0

4(k² - 7k + 12) = 0

k² - 7k + 12 = 0

Now factorizing of the above equation

k² - 7k + 12 = 0

k2 - 4k - 3k + 12 = 0

k(k - 4) - 3(k - 4) = 0

(k - 3)(k - 4) = 0

So, either

k - 3 = 0

k = 3

Or

k - 4 = 0

k = 4

Therefore, the value of k = 4, 3.