The number of real values of k for which the equation 2x²-6x+ k = 0 has two distinct roots in [0, 1] is041infinitely many

The given quadratic equation is 2x² - 6x + k = 0.

For a quadratic equation ax² + bx + c = 0 to have two distinct real roots, the discriminant (b² - 4ac) must be greater than 0.

In this case, a = 2, b = -6, and c = k. So, the discriminant is (-6)² - 42k = 36 - 8k.

We need this to be greater than 0 for two distinct real roots. So, 36 - 8k > 0. Solving this inequality gives k < 4.5.

However, we also need the roots to be in the interval [0, 1]. The roots of a quadratic equation ax² + bx + c = 0 are given by (-b ± √(b² - 4ac)) / (2a).

Substituting the values a = 2, b = -6, and c = k into this formula gives the roots as (6 ± √(36 - 8k)) / 4.

For these roots to be in the interval [0, 1], we need 0 ≤ (6 ± √(36 - 8k)) / 4 ≤ 1.

Solving these inequalities gives the range of k for which the roots are in [0, 1].

So, the number of real values of k for which the equation has two distinct roots in [0, 1] depends on the solution to these inequalities.