Site navigation

\(\begin{array}{|c|c|c|} \hline \textbf{List I (Quadratic Equation)} & \textbf{List II (Roots)} \\ \hline \text{A. } 6x^2 + x - 12 = 0 & \text{I. } (-6, 4) \\ \hline \text{B. } 8x^2 + 16x - 10 = 182…

Question and options

List I (Quadratic Equation)List II (Roots)A. 6x2+x−12=0I. (−6,4)B. 8x2+16x−10=182II. (9,36)C. x2−45x+324=0III. (3,−12)D. 2x2−5x−3=0IV. (−32,43)\begin{array}{|c|c|c|} \hline \textbf{List I (Quadratic Equation)} & \textbf{List II (Roots)} \\ \hline \text{A. } 6x^2 + x - 12 = 0 & \text{I. } (-6, 4) \\ \hline \text{B. } 8x^2 + 16x - 10 = 182 & \text{II. } (9, 36) \\ \hline \text{C. } x^2 - 45x + 324 = 0 & \text{III. } (3, -\frac{1}{2}) \\ \hline \text{D. } 2x^2 - 5x - 3 = 0 & \text{IV. } \left(-\frac{3}{2}, \frac{4}{3}\right) \\ \hline \end{array}

Match List I with List II. Choose the correct answer from the given options:
  1. A-I, B-III, C-IV, D-II
  2. A-IV, B-I, C-II, D-III
  3. A-II, B-IV, C-III, D-I
  4. A-III, B-II, C-I, D-IV

Answer and explanation

Correct answer: B - A-IV, B-I, C-II, D-III
A: 6x^2+x-12 = (3x-4)(2x+3), roots 4/3, -3/2 -> IV. B: 8x^2+16x-192 = 0 -> x^2+2x-24 = (x+6)(x-4), roots -6, 4 -> I. C: x^2-45x+324 = (x-9)(x-36), roots 9, 36 -> II. D: 2x^2-5x-3 = (2x+1)(x-3), roots 3, -1/2 -> III. Answer: A-IV, B-I, C-II, D-III.

Practice for these exams

  • JIPMAT

Related question papers

Related exam syllabi

More solved explanations in Number series & simplification