Question

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

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

Answer 1

The given equation is  $$x^2+k(2x+k-1)+2=0$$

$$\Rightarrow x^2+2kx+k(k-1)+2=0$$

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

$$D=b^2-4ac$$

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

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

$$\Rightarrow D=4k^2-4k^2+4k-8$$

$$\Rightarrow D=4k-8=4(k-2)$$

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