Question

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

Answer 1

Step-by-step explanation:

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 is ax2 + bx + c = 0

Here, a = 1, b = 2k, c = (k2 – k + 2)

Find Discriminant:

D = b2 – 4ac

= (2k)2 – 4 x 1 x (k2 – k + 2)

= 4k2 – 4k2 + 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.

Answer 2

Answer:

hope this helps you

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