Question

Find the value of K For which the equation x^2+k(2x+k-1)+2=0 has real and equal roots

Answer 1

Answer: k=2

Step-by-step explanation:

x^2 + k(2x+k-1)+2

= x^2 + 2kx + (k^2-k+2)

When the roots are equal b^2=4ac

So (2k)^2=4*1*(k^2-k+2)

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

4k-8=0

4k=8

k=2

So the polynomial is x^2+4x+4=(x+2)^2