Question

If $\alpha$ and $\beta$ are the roots of

$ax^2 + bx + c = 0$ then the equation with roots $\frac{1}{a\alpha + b}$ and $\frac{1}{a\beta + b}$ is

• A. $acx^2 - bx + 1 = 0$
• B. $acx^2 + bx - 1 = 0$
• C. $acx^2 - bx - 1 = 0$
• D. $bx^2 - ax + c = 0$

Answer 1

Answer: (A)

1. $acx^2 - bx + 1 = 0$

Answer 2

Given the quadratic equation $ax^2 + bx + c = 0$ with roots $\alpha$ and $\beta$. From the equation $a\alpha^2 + b\alpha + c = 0$, we can write $\alpha(a\alpha + b) = -c$, which means $a\alpha + b = -c/\alpha$. Similarly, $a\beta + b = -c/\beta$. The new roots are $y_1 = \frac{1}{a\alpha + b} = \frac{1}{-c/\alpha} = -\frac{\alpha}{c}$ and $y_2 = \frac{1}{a\beta + b} = -\frac{\beta}{c}$. The sum of the new roots is $S' = y_1 + y_2 = -\frac{\alpha+\beta}{c}$. Since $\alpha+\beta = -b/a$, we have $S' = -\frac{-b/a}{c} = \frac{b}{ac}$. The product of the new roots is $P' = y_1 y_2 = \left(-\frac{\alpha}{c}\right)\left(-\frac{\beta}{c}\right) = \frac{\alpha\beta}{c^2}$. Since $\alpha\beta = c/a$, we have $P' = \frac{c/a}{c^2} = \frac{1}{ac}$. The new quadratic equation is $y^2 - S'y + P' = 0$. Substituting $S'$ and $P'$:

$y^2 - \frac{b}{ac}y + \frac{1}{ac} = 0$

Multiply by $ac$ to clear denominators:

$acy^2 - by + 1 = 0$

Replacing $y$ with $x$ gives $acx^2 - bx + 1 = 0$.