Math, asked by pradnyasshukla, 27 days ago

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

Answers

Answered by biligiri
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

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