Question

Prove that the roots of x² + (1-k)x-3-0 are real for all real values of k.​

Answer 1

The discriminant is positive, so there are 2 distinct real roots

Explanation:

The discriminant of quadratic ax²+bx+c=0 is given by ...

d = b² -4ac

The value of the discriminant for the given equation is ...

d = (1 -k)² -4(1)(k-3) = 1 -2k +k² -4k +12

d = k² -6k +13 = (k -3)² +4

The squared term in the sum cannot be negative, so the value of the discriminant is at least +4. For any positive value of the discriminant, the quadratic will have two real roots.