The range of real number k for which the equation x2 – 3x + k = 0 has two distinct real roots in [–1, 2], is

The given quadratic equation is x² - 3x + k = 0.

For this equation to have two distinct real roots, the discriminant (b² - 4ac) must be greater than 0.

Here, a = 1, b = -3, and c = k.

So, the discriminant is (-3)² - 4(1)(k) = 9 - 4k.

For two distinct real roots, 9 - 4k > 0.

Solving this inequality gives k < 9/4 or k < 2.25.

Now, the roots of the equation are given by the quadratic formula x = [-b ± sqrt(b² - 4ac)] / 2a.

Substituting the values of a, b, and c gives x = [3 ± sqrt(9 - 4k)] / 2.

For the roots to lie in the interval [-1, 2], the values of x obtained from the quadratic formula must satisfy -1 ≤ x ≤ 2.

Substituting x = -1 and x = 2 into the quadratic formula gives two inequalities:

1. 3 - sqrt(9 - 4k) ≥ -2, which simplifies to sqrt(9 - 4k) ≤ 5, and further simplifies to 9 - 4k ≤ 25, and finally to k ≥ -4.

2. 3 + sqrt(9 - 4k) ≤ 4, which simplifies to sqrt(9 - 4k) ≤ 1, and further simplifies to 9 - 4k ≤ 1, and finally to k ≥ 2.

The intersection of the solutions to these inequalities is the range of k for which the equation has two distinct real roots in [-1, 2].

So, the range of k is [-4, 2.25].