Question

If the roots of the quadratic equation $ax^2+bx+c=0$ are $\frac{k}{k+1}$ and $\frac{k+1}{k+2}$, then $\frac{a+b+c}{a}$ equals

• A. $\frac{1}{k^2+3k+2}$
• B. $\frac{k^2+3k+2}{1}$
• C. $\frac{k+1}{k+2}$
• D. $\frac{k}{k+1}$

Answer 1

Answer: (A)

$\frac{1}{k^2+3k+2}$

Answer 2

Let the roots of the quadratic equation $ax^2+bx+c=0$ be $\alpha = \frac{k}{k+1}$ and $\beta = \frac{k+1}{k+2}$.

We are asked to find the value of $\frac{a+b+c}{a}$. We can rewrite this expression as: $\frac{a+b+c}{a} = \frac{a}{a} + \frac{b}{a} + \frac{c}{a} = 1 + \frac{b}{a} + \frac{c}{a}$ From Vieta's formulas, we know that for a quadratic equation $ax^2+bx+c=0$: Sum of roots: $\alpha + \beta = -\frac{b}{a}$ Product of roots: $\alpha \beta = \frac{c}{a}$

Substituting these into the expression for $\frac{a+b+c}{a}$: $\frac{a+b+c}{a} = 1 - (\alpha + \beta) + \alpha \beta$ This expression is algebraically equivalent to $(1-\alpha)(1-\beta)$: $(1-\alpha)(1-\beta) = 1 - \beta - \alpha + \alpha\beta = 1 - (\alpha + \beta) + \alpha\beta$ Now, let's calculate $1-\alpha$ and $1-\beta$ using the given roots: $1 - \alpha = 1 - \frac{k}{k+1} = \frac{(k+1) - k}{k+1} = \frac{1}{k+1}$ $1 - \beta = 1 - \frac{k+1}{k+2} = \frac{(k+2) - (k+1)}{k+2} = \frac{k+2-k-1}{k+2} = \frac{1}{k+2}$ Now, we multiply these two results: $\frac{a+b+c}{a} = (1-\alpha)(1-\beta) = \left(\frac{1}{k+1}\right) \left(\frac{1}{k+2}\right)$ $\frac{a+b+c}{a} = \frac{1}{(k+1)(k+2)}$ Expanding the denominator, we get: $(k+1)(k+2) = k^2 + 2k + k + 2 = k^2 + 3k + 2$ Therefore, $\frac{a+b+c}{a} = \frac{1}{k^2 + 3k + 2}$