Let `alpha` and `beta` are roots of the equation `px^2+qx-r=0` where `pne0`. If p, q, r are the consecutive term of non-constant G.P and `1/alpha+1/beta=3/4`, then value of `(alpha-beta)^2` is:

Options

A. `40/9`
B. `80/9`
C. `60/18`
D. `-80/9`

Answer

To solve the problem step-by-step, we start with the given information:

1. **Identify the roots and relationships**:
Let \( \alpha \) and \( \beta \) be the roots of the quadratic equation \( px^2 + qx - r = 0 \). According to Vieta's formulas, we have:
\[ \alpha + \beta = -\frac{q}{p} \quad \text{(1)} \]
\[ \alpha \beta = -\frac{r}{p} \quad \text{(2)} \]

2. **Use the given condition**:
We are given that:
\[ \frac{1}{\alpha} + \frac{1}{\beta} = \frac{3}{4} \]
This can be rewritten using the relationships from Vieta's formulas:
\[ \frac{\alpha + \beta}{\alpha \beta} = \frac{3}{4} \]
Substituting equations (1) and (2) into this gives:
\[ \frac{-\frac{q}{p}}{-\frac{r}{p}} = \frac{3}{4} \]
Simplifying this, we find:
\[ \frac{q}{r} = \frac{3}{4} \quad \text{(3)} \]

3. **Establish the relationship in G.P.**:
Since \( p, q, r \) are consecutive terms of a non-constant geometric progression (G.P.), we can express the ratios:
\[ \frac{q}{p} = k \quad \text{and} \quad \frac{r}{q} = k \quad \text{for some } k \]
From (3), we have:
\[ \frac{q}{r} = \frac{3}{4} \implies \frac{r}{q} = \frac{4}{3} \]
Thus, we can express \( r \) in terms of \( q \):
\[ r = \frac{4}{3}q \quad \text{(4)} \]

4. **Substituting back into the G.P. ratios**:
From \( \frac{q}{p} = k \) and \( \frac{r}{q} = k \), we can express \( p \) in terms of \( q \):
\[ p = \frac{q}{k} \quad \text{(5)} \]
Substituting (4) into (3) gives:
\[ \frac{q}{\frac{4}{3}q} = \frac{3}{4} \implies \frac{3}{4} = \frac{3}{4} \text{ (which is consistent)} \]

5. **Finding \( (α - β)^2 \)**:
We know:
\[ (\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha \beta \]
Substituting from (1) and (2):
\[ (\alpha - \beta)^2 = \left(-\frac{q}{p}\right)^2 - 4\left(-\frac{r}{p}\right) \]
Simplifying this gives:
\[ = \frac{q^2}{p^2} + \frac{4r}{p} \]

6. **Substituting \( r \) from (4)**:
Using \( r = \frac{4}{3}q \):
\[ = \frac{q^2}{p^2} + \frac{4 \cdot \frac{4}{3}q}{p} = \frac{q^2}{p^2} + \frac{16q}{3p} \]

7. **Expressing everything in terms of \( q \)**:
From \( \frac{q}{p} = k \), we can replace \( p \) with \( \frac{q}{k} \):
\[ = \frac{q^2}{\left(\frac{q}{k}\right)^2} + \frac{16q}{3 \cdot \frac{q}{k}} = k^2 + \frac{16k}{3} \]

8. **Finding the value of \( k \)**:
Since \( \frac{q}{p} = \frac{4}{3} \), we can set \( k = \frac{4}{3} \):
\[ = \left(\frac{4}{3}\right)^2 + \frac{16 \cdot \frac{4}{3}}{3} = \frac{16}{9} + \frac{64}{9} = \frac{80}{9} \]

Thus, the final value of \( (\alpha - \beta)^2 \) is:
\[ \boxed{\frac{80}{9}} \]