If `alpha and beta` are the roots of `ax^(2)+bx+c=0 and ` if `px^(2)+qx+r=0` has roots `(1-alpha)/(alpha) and (1-beta)/(beta)` then r=

Options

A. `a+2b`
B. `a+b+c`
C. `ab+bc+ca`
D. `abc`

Answer

To solve the problem, we need to find the value of \( r \) in the quadratic equation \( px^2 + qx + r = 0 \) given that the roots of this equation are \( \frac{1-\alpha}{\alpha} \) and \( \frac{1-\beta}{\beta} \), where \( \alpha \) and \( \beta \) are the roots of the equation \( ax^2 + bx + c = 0 \).

### Step 1: Identify the roots of the first equation
The roots of the equation \( ax^2 + bx + c = 0 \) are given by:
- Sum of roots: \( \alpha + \beta = -\frac{b}{a} \)
- Product of roots: \( \alpha \beta = \frac{c}{a} \)

### Step 2: Calculate the sum of the new roots
The new roots are \( \frac{1-\alpha}{\alpha} \) and \( \frac{1-\beta}{\beta} \). We calculate their sum:
\[ \text{Sum of new roots} = \frac{1-\alpha}{\alpha} + \frac{1-\beta}{\beta} \]
Finding a common denominator:
\[ = \frac{(1-\alpha)\beta + (1-\beta)\alpha}{\alpha\beta} \]
Expanding the numerator:
\[ = \frac{\beta - \alpha\beta + \alpha - \alpha\beta}{\alpha\beta} = \frac{\alpha + \beta - 2\alpha\beta}{\alpha\beta} \]
Substituting the values of \( \alpha + \beta \) and \( \alpha \beta \):
\[ = \frac{-\frac{b}{a} - 2\frac{c}{a}}{\frac{c}{a}} = \frac{-b - 2c}{c} \]

### Step 3: Relate the sum of roots to \( q \) and \( p \)
According to Vieta's formulas, the sum of the roots of \( px^2 + qx + r = 0 \) is equal to \( -\frac{q}{p} \):
\[ -\frac{q}{p} = \frac{-b - 2c}{c} \]
Thus, we can equate:
\[ q = p \cdot \frac{b + 2c}{c} \]

### Step 4: Calculate the product of the new roots
Now we calculate the product of the new roots:
\[ \text{Product of new roots} = \left(\frac{1-\alpha}{\alpha}\right) \left(\frac{1-\beta}{\beta}\right) = \frac{(1-\alpha)(1-\beta)}{\alpha\beta} \]
Expanding the numerator:
\[ = \frac{1 - (\alpha + \beta) + \alpha\beta}{\alpha\beta} = \frac{1 + \frac{b}{a} + \frac{c}{a}}{\frac{c}{a}} = \frac{a + b + c}{c} \]

### Step 5: Relate the product of roots to \( r \) and \( p \)
According to Vieta's formulas, the product of the roots of \( px^2 + qx + r = 0 \) is equal to \( \frac{r}{p} \):
\[ \frac{r}{p} = \frac{a + b + c}{c} \]
Thus, we can equate:
\[ r = p \cdot \frac{a + b + c}{c} \]

### Step 6: Conclusion
From the above steps, we have established that:
\[ r = \frac{p(a + b + c)}{c} \]
Now, if we assume \( p = c \) (as derived from the earlier steps), we can substitute:
\[ r = \frac{c(a + b + c)}{c} = a + b + c \]

Thus, the final answer is:
\[ \boxed{a + b + c} \]