If the set of all a ∈ R, for which the equation 2 x²+(a-5) x+15=3 a has no real root, is the interval ( α, β ), and X=|x ∈ Z ; α < x < β|, then Σ_(x ∈ X) x² is equal to:

Solution

Maths sir · 4 steps

No real root means a negative discriminant. That gives a quadratic inequality in a, and you solve that one.

1. Standard form. Move 3a to the left: 2x² + (a-5)x + (15 - 3a) = 0. The equation has no real root exactly when D < 0.

   D = (a-5)² - 4 × 2 × (15-3a) = a² + 14a - 95

2. Solve the inequality. The roots of a² + 14a - 95 = 0 are a = (-14 ± √(196 + 380))/2 = (-14 ± 24)/2, which gives 5 and -19. Since the parabola opens upwards, D < 0 for -19 < a < 5, so α = -19 and β = 5.
3. Integers in between. X = {-18, -17, …, 0, …, 4}. The squares of -18 to -1 give (18 × 19 × 37)/6 = 2109. Zero adds nothing, and 1² + 2² + 3² + 4² = 30.
4. So. Σ x² = 2109 + 30 = 2139, which is option (a).

The idea to keep. For no real roots, set D < 0 and solve it as a quadratic inequality in the parameter. When summing squares over an interval that crosses zero, count the negative side and the positive side separately.

Answer from the NTA final answer key. Solution by PrepWiser.