Question

If 2 and 6 are the roots of the equation \( ax^2 + bx + 1 = 0 \), then the quadratic equation, whose roots are \( \frac{1}{2a + b} \) and \( \frac{1}{6a + b} \), is:

• A. \( 2x^2 + 11x + 12 = 0 \)
• B. \( 4x^2 + 14x + 12 = 0 \)
• C. \( x^2 + 10x + 16 = 0 \)
• D. \( x^2 + 8x + 12 = 0 \)

Answer 1

The Correct Option is D

Given Roots and Sum/Product Relations: 
Since \(2\) and \(6\) are roots of the equation \(ax^2 + bx + 1 = 0\), we know:
\[ \text{Sum of roots} = 2 + 6 = 8 = -\frac{b}{a} \]
\[ \text{Product of roots} = 2 \times 6 = 12 = \frac{1}{a} \] 
From the product, we get \(a = \frac{1}{12}\). 

Finding \(b\): 
Substitute \(a = \frac{1}{12}\) into the sum of roots equation:
\[ -\frac{b}{\frac{1}{12}} = 8 \implies -12b = 8 \implies b = -\frac{2}{3} \] 
Constructing the New Quadratic Equation: 
The roots of the new quadratic equation are \(\frac{1}{2a+b}\) and \(\frac{1}{6a+b}\). 

Substitute \(a = \frac{1}{12}\) and \(b = -\frac{2}{3}\):
\[ 2a + b = 2 \times \frac{1}{12} - \frac{2}{3} = \frac{1}{6} - \frac{2}{3} = \frac{1}{6} - \frac{4}{6} = -\frac{1}{2} \] \[ 6a + b = 6 \times \frac{1}{12} - \frac{2}{3} = \frac{1}{2} - \frac{2}{3} = \frac{3}{6} - \frac{4}{6} = -\frac{1}{6} \] 
Thus, the roots of the new equation are \(-2\) and \(-6\). 

Forming the Equation with Roots \(-2\) and \(-6\): 
A quadratic equation with roots \(-2\) and \(-6\) is: \[ x^2 + 8x + 12 = 0 \]