#### Question

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

(k + 1)x^{2} - 2(3k + 1)x + 8k + 1 = 0

#### Solution

The given quadric equation is (k + 1)x^{2} - 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 = b^{2} - 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(9k^{2} + 6k + 1) - 4(8k^{2} + 9k + 1)

= 36k^{2} + 24k + 4 - 32k^{2} - 36k - 4

= 4k^{2} - 12k

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

4k^{2} - 12k = 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.

Is there an error in this question or solution?

#### APPEARS IN

Solution 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 Concept: Nature of Roots.