# Find the Value of K for Which the Following Equations Have Real and Equal Roots: - Mathematics

Find the value of k for which the following equations have real and equal roots:

$\left( k + 1 \right) x^2 - 2\left( k - 1 \right)x + 1 = 0$

#### Solution

The given quadric equation is $\left( k + 1 \right) x^2 - 2\left( k - 1 \right)x + 1 = 0$, and roots are real and equal

Then find the value of k.

Here,

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

As we know that D = b2 - 4ac

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

 = {-2 (k-1)}^2 - 4 xx (k-1 ) xx 1

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

=4k^2 -8k + 4 - 4k - 4'

=4k^2 - 12k + 0

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

4k^2 -  12k+ 0 =0

4k^2 - 12k = 0

Now factorizing of the above equation

4k (k-3) = 0

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

RD Sharma Class 10 Maths