Question

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

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

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

(k + 1)x² – 2(k – 1)x + 1 = 0

Answer 1

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.