Question

Meera and Sarika attempted to solve a quadratic equation in x. Sarika made a mistake in reading the coefficient of x and obtained the roots as 12 and 16. Meera made a mistake in reading the constant term of the equation. She obtained the roots as 22 and 6. The corrects root(s) is (are) :
(A) -16 
(B) -12 
(C) 12 
(D) 16 
Choose the correct answer from the options given below :

• (A) and (B) only
• (A) and (C) only
• (C) and (D) only
• (B) and (D) only

Answer 1

The Correct Option is C

To solve this problem, we need to understand what's happening with the mistake in the reading of the quadratic equation by Sarika and Meera, and then determine the correct roots.

1. Let's start by considering the standard form of a quadratic equation, which is \(ax^2 + bx + c = 0\). The roots of this equation can be found using Vieta's formulas: 
   • The sum of the roots, \(r_1 + r_2\), is \(-\frac{b}{a}\).
   • The product of the roots, \(r_1 \times r_2\), is \(\frac{c}{a}\).
2. Sarika obtained roots 12 and 16. Therefore:
   • Sum of roots = \(12 + 16 = 28\), which gives \(-\frac{b}{a} = 28\).
   • Product of roots = \(12 \times 16 = 192\), which gives \(\frac{c}{a} = 192\).
3. Meera obtained roots 22 and 6. Therefore:
   • Sum of roots = \(22 + 6 = 28\), which again gives \(-\frac{b}{a} = 28\).
   • Product of roots = \(22 \times 6 = 132\), which implies an incorrect reading of \(c\).
4. Although both obtained different products (Sarika: 192, Meera: 132), they have the same sum \(28\). Therefore, the correct sum of the roots is \(28\), pointing out a consistent effect on the \(b\) term.
5. The correct equation should have integers that multiply to give \(192\) and add to \(28\). The integer factors of 192 closest to a sum of 28 are 12 and 16, the roots found by Sarika.
6. Thus, the correct roots of the original quadratic equation are 12 and 16.

Therefore, the correct option is (C) and (D) only, corresponding to the roots 12 and 16.