Question

Find the least value of k for which the quadratic equation minus 3 x square - 6 X + K is equal to zero has real roots​

Answer 1

Answer:

k>or = -3

Step-by-step explanation:

-3x^2 -6x + k=0

From the above equation,

a=-3, b=-6, c=k

Since the equation has real root

Therefore, b^2-4ac >or = 0.........(1)

putting the values of a, b and c in (1)

(-6)^2- 4(-3)(k) >or = 0

36 - (-12k) > or = 0

36 + 12k > or = 0

12k > or = -36

k > or = -36/12

k > or = -3

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Answer 2

Answer:

k= -3

Step-by-step explanation:

-3×^2-6×+k

a=-3, b=6 , c=k

b^2-4ac=0-----(1)

6^2-4×-3×k=0

36+12k=0

12k=-36

k=-36÷12

k=-3