Question

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

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

Answer 1

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

Then find the value of k.

Here,

a = 1, b=2(k + 1)and,c = k²

As we know that `D = b^2 - 4ac` 

Putting the value of  a = 1,b = -2(k+1)and, c = k²

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

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

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

`= 8k + 4`

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

8k + 4 = 0

8k = - 4

`k=(-4)/8`

` = (-1)/2`

Therefore, the value of `k = (-1)/2`