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

Answer: (D)

$x^2 + 8x + 12 = 0$

Answer 2

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: $x^2 + 8x + 12 = 0$