Question

if quadratic equation x²+k(2x+k-1)+2=0 has real and equal roots then find value of k.​

Answer 1

Answer:

k=2

Step-by-step explanation:

Given, x2+k(2x+k−1)+2=0

Simplify above equation:

x2+2kx+(k2−k+2)=0

Compare given equation with the general form of quadratic equation, which ax2+bx+c=0

here, a=1,b=2k,c=(k2−k+2)

Find discriminant:

D=b2−4ac

=(2k)2−4×1×(k2−k+2)

=4k2−4k2+4k−8

=4k−8

Since roots are real and equal (given)

Put D=0

4k−8=0

k=2