Question

find the value of k which the roots of the equation 8ky(y-1) +1 =0 are real and equal

Answer 1

don't know.............

Answer 2

Answer:

The value of k is 0 and 1/2.

Step-by-step explanation:

Given : The roots of the equation $8ky(y-1) +1 =0$ are real and equal.

To find : The value of k

Solution :

Solving the given equation,

$8ky(y-1) +1 =0$

$8ky^2-8ky +1 =0$

Comparing with quadratic equation, $ax^2+bx+c=0$

a=8k, b=-8k , c=1

When root are real discriminant is zero.

$D=b^2-4ac$

$0=(-8k)^2-4(8k)(1)$

$0=64k^2-32k$

$0=32k(2k-1)$

$32k=0,2k-1=0$

$k=0,k=\frac{1}{2}$

Therefore, The value of k is 0 and 1/2.