Question

find the value of my so that the quadratic equating x2-4kx+k=0 has equal roots​

Answer 1

Hey there!!

D = (b^2) + 4ac

0 = ((4k)^2) + 4*1*k

0 = (16k^2) + 4k

0 = 4k(4k + 1)

0 = 4k + 1

4k= - 1

k= -1/4 (ans)

Mark my answer as the brainliest

Priyoshi❤

Answer 2

Step-by-step explanation:

We know that quadratic equation has two equal roots only when the value of discriminant is equal to zero.

Discriminant is equal to 0, is given by

b2 – 4ac = 0 …(i)

Comparing x2 – 4kx + k = 0 with general quadratic equation ax2 + bx + c = 0, we get

a = 1, b = -4k and c = k

Substituting these values in equation (i), we get

(-4k)2 – 4(1)(k) = 0

⇒ 16k2 – 4k = 0

⇒ k(16k – 4) = 0

⇒ k = 0 or 16k – 4 = 0

⇒ k = 0 or 16k = 4

⇒ k = 0 or k = 1/4

Hence, the values of k are 0 and 1/4.

__Hope_this_will_helps_uhh__