Question

Find the values of k for equation have real and equal roots

x² – 2(k + 1)x + k² = 0

Answer 1

Given,

x² – 2(k + 1)x + k² = 0

It’s of the form of ax² + bx + c = 0

Where, a =1, b = -2(k + 1), c = k²

For the given quadratic equation to have real roots D = b² – 4ac = 0

D = (-2(k + 1))² – 4(1)(k²) = 0

⇒ 4k² + 8k + 4 – 4k² = 0

⇒ 8k + 4 = 0

⇒ k = \(\frac{-4}{8}\)

⇒ k = \(\frac{-1}{2}\)

The value of k should \(\frac{-1}{2}\) to have real and equal roots.