New Register for Workshop on How to start for NABARD 2025 (Stepwise Plan) - Apr 26, 7 PM

    Question

    Solve the three equations and construct a new equation in variable 'c' (reduced to the lowest possible factor) using roots of equations I, II, and III, following the instructions provided below.

    Let 's' represent the sum of the highest root of

    equations I and III, and 'r' denote the product of the lowest root of equation I and the highest root of equation II.Determine the new equation if the roots of this equation are 's'and 'r'.
    A x² - 14x + 40 = 0 Correct Answer Incorrect Answer
    B x² - 18x + 72 = 0 Correct Answer Incorrect Answer
    C x² - 24x + 80 = 0 Correct Answer Incorrect Answer
    D x² - 13x + 58 = 0 Correct Answer Incorrect Answer
    E none of these Correct Answer Incorrect Answer

    Solution

    ATQ, 4 - (20/a) + (28/a) = a2 = (4/a2) + (9/a) - 3 7 - (29/a) + (24/a2) = 0 7a2 - 29a + 24 = 0 (7a - 8)(a - 3) = 0 a = 8/7, 3 Equation II b2 - 4b + 2 = 1/8 × (2b - 5) 8b2 - 32b + 16 = 2b - 5 8b2 - 34b + 21 = 0 (4b - 3)(2b - 7) = 0 b = 3/4, 7/2 Equation III, c2 + 10.5 = 17c/2 2c2 - 17c + 21 = 0 (2c - 3)(c - 7) = 0 c = 3/2, 7 s = 3 + 7 = 10 r = (8/7) × (7/2) = 4 New Equation (x - 10)(x - 4) = 0 x2 - 14x + 40 = 0

    Practice Next

    Relevant for Exams: