The least positive value of 'a' for which the equation
2x² + (a – 10)x + 33/2 = 2a has real roots is

Solution

Maths sir · 4 steps

Move 2a to the left first, then demand discriminant ≥ 0. The factorisation does the rest.

1. Standard form. Bring 2a across: 2x² + (a-10)x + (33/2 - 2a) = 0. So A = 2, B = a - 10 and C = 33/2 - 2a.
2. Real roots condition. Real roots need D = B² - 4AC ≥ 0. Here 4AC = 8(33/2 - 2a) = 132 - 16a.

   D = (a-10)² - (132 - 16a) = a² - 4a - 32 = (a-8)(a+4)

3. Solve the inequality. (a-8)(a+4) ≥ 0 holds when a ≤ -4 or a ≥ 8. Positive values of a between 0 and 8 make D negative, so they are ruled out.
4. So. The least positive value of a is 8. At a = 8, D = 0 and the equation has equal real roots.

The idea to keep. Always write the quadratic as Ax² + Bx + C = 0 before using the discriminant. Every term with the parameter must sit inside A, B or C.

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