Question

For what value of the K the following quadratic equation will have real roots

4x^2-3kx+1=0

Answer 1

Hey

Here is your answer,

4x^2 - 3kx + 1 = 0

b^2-4ac = 0

(3k)^2 - 4 x 4 x 1 = 0
9k^2 = 16
k^2 = 16/9
k = √16/9
k = 4/3

Hope it helps you!

Answer 2

Here is your answer


4x²-3kx+1 = 0

a = 4 , b = -3k and c = 1


D = 0

(b²-4ac ) = 0

(-3k)² - 4×4×1 = 0

9k² - 16 = 0

k² = 16/9

k = ✓16/9 = 4/3.


Hope it helps you ^_^