Question

$ax^2+bx+c=0$ with roots $\alpha$ and $\beta$. Then $(\frac{1}{a\alpha+b}+\frac{1}{a\beta+b})=$

• A. $\frac{c}{ab}$
• B. $\frac{b}{ac}$
• C. $\frac{a}{bc}$
• D. None

Answer 1

Answer: (B)

$\frac{b}{ac}$

Answer 2

Given the quadratic equation $ax^2+bx+c=0$ with roots $\alpha$ and $\beta$.

From $a\alpha^2+b\alpha+c=0$, we have $a\alpha^2+b\alpha = -c$. Dividing by $\alpha$ (assuming $\alpha \neq 0$, which is true if $c \neq 0$), we get $a\alpha+b = -c/\alpha$.

Similarly, from $a\beta^2+b\beta+c=0$, we have $a\beta+b = -c/\beta$.

The expression to evaluate is $(\frac{1}{a\alpha+b}+\frac{1}{a\beta+b})$.

Substitute the derived terms: $(\frac{1}{-c/\alpha} + \frac{1}{-c/\beta}) = (-\frac{\alpha}{c} - \frac{\beta}{c}) = -\frac{\alpha+\beta}{c}$.

Using Vieta's formulas, the sum of the roots is $\alpha+\beta = -b/a$.

Substitute this value into the expression: $-\frac{-b/a}{c} = \frac{b/a}{c} = \frac{b}{ac}$.