Question

8. Find the value of k for which the equation x2 + k (2x + k – 1) + 2 = 0 has real and equal roots.​

Answer 1

Answer:

k=2

Explanation:

Given, x

2

+k(2x+k−1)+2=0

Simplify above equation:

x

2

+2kx+(k

2

−k+2)=0

Compare given equation with the general form of quadratic equation, which ax

2

+bx+c=0

here, a=1,b=2k,c=(k

2

−k+2)

Find discriminant:

D=b

2

−4ac

=(2k)

2

−4×1×(k

2

−k+2)

=4k

2

−4k

2

+4k−8

=4k−8

Since roots are real and equal (given)

Put D=0

4k−8=0

k=2

Hence, the value of k is 2.