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