JIPMAT 2022 (QA) - Match List I with List II. Choose the correct answer from the options given below: | PYQs + Solutions

JIPMAT 2022

Algebra

Quadratic Equations

Conceptual

\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 = 202 & \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 options given below:

A-I, B-III, C-IV, D-II

A-IV, B-I, C-II, D-III

A-II, B-IV, C-III, D-I

A-III, B-II, C-I, D-IV

Correct Option: 2

Let's take:

6x^2 + x - 12 = 0.

Using the quadratic formula:

\newline

x = \frac{-1 \pm \sqrt{1+288}}{12} = \frac{-1 \pm \sqrt{289}}{12} = \frac{-1 \pm 17}{12}

Thus the solutions are:

\newline

x = \frac{16}{12} = \frac{4}{3}

and

x = \frac{-18}{12} = -\frac{3}{2}.

These match the pair in the fourth row.

\newline

\left(-\frac{3}{2}, \frac{4}{3}\right)

Hence, A-IV, option B is the correct answer!

Note: The third equation was typed incorrectly by NTA themselves.

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JIPMAT 2023

Given below are two statements:

Statement (I): (x² + 3x + 1) = (x - 2)² is not a quadratic equation.

Statement (II): The nature of roots of quadratic equations x² + 2x√3 + 3 = 0 are real and equal.

In light of the above statements, choose the most appropriate answer from the options given below.