Question

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

Answer 1

Hope the answer helped you.

Answer 2

Step-by-step explanation:

Given -

p(x) = x² - (2k + 1)x + k² = 0

here,

a = 1

b = -(2k + 1) = (-2k - 1)

c = k²

Now,

As we know that :-

• D = b² - 4ac = 0

» (-2k - 1)² - 4×1×k² = 0

» 4k² + 1 + 4k - 4k² = 0

» 1 + 4k = 0

» 4k = -1

• » k = -1/4

Hence,

The value of k is -1/4

Formula Used -

• D = b² - 4ac

Some related formulas :-

• x = -b ± √b² - 4ac/2a