Question

find the least value of k for which the quadratic equation $-3x^{2} -6x + k$ has real roots

Answer 1

Minimum value of k will be -3.

The quadratic equation -3x²-6x+k=0

will have real roots if the value of determinant is greater than or equal to zero (D>=0).

From the equation 3x²-6x+k=0 ,

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

D = b²-4ac

For real roots , D>=0

=> b²-4ac >=0

=> 36 - 4 × -3 × k >=0

=> 36 +12k >=0

=> k + 3 >= 0

k can have values from -3 to infinity as these values of k will satisfy the in inequality k + 3 >= 0.

Minimum value of k is equal to -3