Question

Match List I with List II :

List I (Quadratic equations)	List II (Roots)
(A) \(12x^2 - 7x + 1 = 0\)	(I) \((-13, -4)\)
(B) \(20x^2 - 9x + 1 = 0\)	(II) \(\left(\frac{1}{3}, \frac{1}{4}\right)\)
(C) \(x^2 + 17x + 52 = 0\)	(III) \((-4, -\frac{3}{2})\)
(D) \(2x^2 + 11x + 12 = 0\)	(IV) \(\left(\frac{1}{5}, \frac{1}{4}\right)\)

Choose the correct answer from the options given below :

• A. (A)-(II), (B)-(I), (C)-(III), (D)-(IV)
• B. (A)-(II), (B)-(III), (C)-(IV), (D)-(I)
• C. (A)-(II), (B)-(IV), (C)-(III), (D)-(I)
• D. (A)-(II), (B)-(IV), (C)-(I), (D)-(III)

Hint:

In exams, check the signs. If all coefficients in the quadratic equation are positive (like C and D), the roots must be negative. If the middle term is negative and the constant is positive (like A and B), the roots must be positive. This helps eliminate choices instantly.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
To find the roots of a quadratic equation of the form \(ax^2 + bx + c = 0\), we can use the factorization method (splitting the middle term) or the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
Step 2: Key Formula or Approach:
We split the middle term \(b\) into two parts such that their sum is \(b\) and their product is \(a \times c\).
Step 3: Detailed Explanation:
For (A): \(12x^2 - 7x + 1 = 0\)
Product = \(12 \times 1 = 12\), Sum = \(-7\). The factors are \(-4\) and \(-3\).
\[ 12x^2 - 4x - 3x + 1 = 0 \] \[ 4x(3x - 1) - 1(3x - 1) = 0 \implies (4x - 1)(3x - 1) = 0 \] Roots: \(x = \frac{1}{4}, \frac{1}{3}\). (Matches II)
For (B): \(20x^2 - 9x + 1 = 0\)
Product = \(20 \times 1 = 20\), Sum = \(-9\). The factors are \(-5\) and \(-4\).
\[ 20x^2 - 5x - 4x + 1 = 0 \] \[ 5x(4x - 1) - 1(4x - 1) = 0 \implies (5x - 1)(4x - 1) = 0 \] Roots: \(x = \frac{1}{5}, \frac{1}{4}\). (Matches IV)
For (C): \(x^2 + 17x + 52 = 0\)
Product = \(52\), Sum = \(17\). The factors are \(13\) and \(4\).
\[ x^2 + 13x + 4x + 52 = 0 \] \[ x(x + 13) + 4(x + 13) = 0 \implies (x + 4)(x + 13) = 0 \] Roots: \(x = -4, -13\). (Matches I)
For (D): \(2x^2 + 11x + 12 = 0\)
Product = \(2 \times 12 = 24\), Sum = \(11\). The factors are \(8\) and \(3\).
\[ 2x^2 + 8x + 3x + 12 = 0 \] \[ 2x(x + 4) + 3(x + 4) = 0 \implies (2x + 3)(x + 4) = 0 \] Roots: \(x = -\frac{3}{2}, -4\). (Matches III)
Step 4: Final Answer:
By matching the calculations, we find the sequence: (A)-(II), (B)-(IV), (C)-(I), (D)-(III).