Question

Find the value of k for which the roots of the equation 8 kx(x – 1) + 1 = 0 are real and equal.

Answer 1

as roots are equal and real

b^2-4ac=0

64k^2-32k=0

2k*k-k=0

k(2k-1)=0

2k-1=0

k=1/2

so k=1/2

Answer 2

we have to find the value of k for which the roots of the equation 8kx(x- 1) + 1 = 0 are real and equal.

we know, roots are equal only when discriminant = 0.

resolve the equation for better understanding.
8kx(x - 1) + 1 = 0
8kx² - 8kx + 1 = 0
here, coefficient of x² = 8k
coefficient of x = -8k
constant term = 1

so, Discriminant = (-8k)² - 4.8k = 0
64k² - 32k = 0
k(2k - 1) = 0
k = 0, 1/2