Question

3y²+ ky+12 =0 if root are real and equal find the value of k​

Answer 1

Step-by-step explanation:

3y²+ ky + 12 = 0 and roots are real, to find K

on comparing with standard form quadratic equation

ay² + by + c = 0, we get

a = 3, b = k and c = 12

as roots are real and equal, discriminant D = 0

=> D = b² - 4ac = 0

=> k² - 4(3)(12) = 0

=> k² - 144 = 0

=> k² = 144

=> k = ± √144

=> k = ± 12

therefore the two values of k are + 12 and - 12