#### Question

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

x^{2} - 2kx + 7k - 12 = 0

#### Solution

The given quadric equation is x^{2} - 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 = b^{2} - 4ac

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

= (-2k)^{2} - 4 x (1) x (7k - 12)

= 4k^{2} - 28k + 48

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

4k^{2} - 28k + 48 = 0

4(k^{2} - 7k + 12) = 0

k^{2} - 7k + 12 = 0

Now factorizing of the above equation

k^{2} - 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.

Is there an error in this question or solution?

Solution Find the Values Of K For Which the Roots Are Real and Equal in Each of the Following Equation: X2 - 2kx + 7k - 12 = 0 Concept: Nature of Roots.