Question

    I. 4p² + 17p + 15 = 0 II. 3q² + 19q +

    28 = 0 In each of these questions two equations numbered I and II are given. You have to solve both the equations and give answer, if –
    A p = q or relationship cannot be established between p and q. Correct Answer Incorrect Answer
    B p > q Correct Answer Incorrect Answer
    C p < q Correct Answer Incorrect Answer
    D p ≥ q Correct Answer Incorrect Answer
    E p ≤ q Correct Answer Incorrect Answer

    Solution

    I. 4p² + 17p + 15 = 0 4p² +12p + 5p + 15 = 0 ⟹4p (p + 3) + 5 (p + 3) = 0 ⟹ (p + 3)(4p + 5)= 0 ⟹ p= - 3 , - 5/4 II. 3q² + 19q + 28 = 0 ⟹3q² + 12q + 7q + 28 =0 ⟹3q(q + 4) + 7(q + 4)= 0 ⟹ (q + 4) (3q + 7)= 0 ⟹ q = -4, - 7/3 ∴Relationship cannot be established between p and q.

    Practice Next
    More Quadratic equation Questions

    Relevant for Exams:

    Hey, Don't leave. If you have any queries regarding any exam let us know here.

    ask-question