find the value of k is (3k+1)x square +2(k+1)x+k=0
Answers
Answered by
5
Equation is
= (3k+1)x²+2(k+1)x + k = 0
In the given equation
a = 3k+1, b = 2k+2, c = k
we know.
D = b²-4ac
D = (2k+2)² - 4 x (3k+1)(k)
D = 4k²+4+8k - 4 x (3k²+k)
D. = 4k²+4+8k-12k² + 4k
D = -8k²+4+12k
D = Negative
Therefore value of k= 0
Similar questions