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:
- \( 2x^2 + 11x + 12 = 0 \)
- \( 4x^2 + 14x + 12 = 0 \)
- \( x^2 + 10x + 16 = 0 \)
- \( x^2 + 8x + 12 = 0 \)
The Correct Option is D
Approach Solution - 1
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: \[ x^2 + 8x + 12 = 0 \]
Approach Solution -2
We are given that 2 and 6 are the roots of the quadratic equation:
\[ ax^2 + bx + 1 = 0 \] We need to find the new quadratic equation whose roots are: \[ \frac{1}{2a + b} \quad \text{and} \quad \frac{1}{6a + b}. \]
Step 2: Use the relationship between roots and coefficients.
For the given quadratic equation \( ax^2 + bx + 1 = 0 \):
If 2 and 6 are the roots, then by Vieta’s formulas:
\[ \text{Sum of roots} = 2 + 6 = 8 = -\frac{b}{a}, \quad \text{and} \quad \text{Product of roots} = 2 \times 6 = 12 = \frac{1}{a}. \]
From the second equation, we get: \[ a = \frac{1}{12}. \] Substitute this into the first equation to find \( b \):
\[ 8 = -\frac{b}{a} = -b \times 12 \quad \Rightarrow \quad b = -\frac{2}{3}. \]
Step 3: Find the new roots.
We are told that the new roots are \( \frac{1}{2a + b} \) and \( \frac{1}{6a + b} \). Substitute \( a = \frac{1}{12} \) and \( b = -\frac{2}{3} \):
\[ 2a + b = 2\left(\frac{1}{12}\right) - \frac{2}{3} = \frac{1}{6} - \frac{4}{6} = -\frac{1}{2}, \] \[ 6a + b = 6\left(\frac{1}{12}\right) - \frac{2}{3} = \frac{1}{2} - \frac{4}{6} = -\frac{1}{6}. \] Thus, the new roots are: \[ \frac{1}{2a + b} = \frac{1}{-\frac{1}{2}} = -2, \quad \frac{1}{6a + b} = \frac{1}{-\frac{1}{6}} = -6. \]
Step 4: Form the new quadratic equation.
For roots \(-2\) and \(-6\), the quadratic equation is: \[ (x + 2)(x + 6) = 0 \quad \Rightarrow \quad x^2 + 8x + 12 = 0. \]
Final Answer:
\[ \boxed{x^2 + 8x + 12 = 0} \]
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