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Question
Find the value of k for which the following equations have real and equal roots:
\[x^2 - 2\left( k + 1 \right)x + k^2 = 0\]
Solution
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`
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