Question

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

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

Answer 1

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

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

So, a = 1, b = 2k, c = k(k − 1) + 2
We know 

\[D = b^2 - 4ac\]

\[\Rightarrow D = \left( 2k \right)^2 - 4 \times 1 \times \left[ k\left( k - 1 \right) + 2 \right]\]

\[ \Rightarrow D = 4 k^2 - 4\left[ k^2 - k + 2 \right]\]

\[ \Rightarrow D = 4 k^2 - 4 k^2 + 4k - 8\]

\[ \Rightarrow D = 4k - 8 = 4\left( k - 2 \right)\]

For equal roots, D = 0
Thus, 4(k − 2) = 0
So, k = 2.