Question

Let $\alpha$ and $\beta$ be the roots of the equation $px^2 + qx - r = 0$, where $p \neq 0$. If $p, q,$ and $r$ be the consecutive terms of a non-constant G.P. and \[\frac{1}{\alpha} + \frac{1}{\beta} = \frac{3}{4},\] then the value of $(\alpha - \beta)^2$ is:

• A. \(\frac{80}{9}\)
• B. 9
• C. \(\frac{20}{3}\)
• D. 8

Answer 1

The Correct Option is A

The given quadratic equation is: \[ px^2 + qx - r = 0 \quad \text{with roots } \alpha \text{ and } \beta. \]
The coefficients are given as: \[ p = A, \quad q = AR, \quad r = AR^2. \] 
Substituting these values into the equation: \[ Ax^2 + ARx - AR^2 = 0. \] 
Dividing throughout by $A$, we get: \[ x^2 + Rx - R^2 = 0. \] 
From the roots of the quadratic equation, we know: \[ \frac{1}{\alpha} + \frac{1}{\beta} = \frac{3}{4}. \] 

Using the relationship between the roots and the coefficients of a quadratic equation: \[ \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha \beta}. \] 
Substituting the values for $\alpha + \beta = -R$ and $\alpha \beta = -R^2$, we get: \[ \frac{\alpha + \beta}{\alpha \beta} = \frac{-R}{-R^2} = \frac{R}{R^2} = \frac{1}{R}. \] 
Equating this to the given value: \[ \frac{1}{R} = \frac{3}{4}. \] 
From this, we find: \[ R = \frac{4}{3}. \] 

Now, we calculate $(\alpha - \beta)^2$ using the formula: \[ (\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta. \] 
Substituting the known values: \[ \alpha + \beta = -R \quad \text{and} \quad \alpha \beta = -R^2, \] 
we get: \[ (\alpha - \beta)^2 = (-R)^2 - 4(-R^2). \] 

Simplify this expression: \[ (\alpha - \beta)^2 = R^2 - 4(-R^2) = R^2 + 4R^2 = 5R^2. \] 
Substituting $R = \frac{4}{3}$ into the equation: \[ (\alpha - \beta)^2 = 5 \left(\frac{4}{3}\right)^2 = 5 \cdot \frac{16}{9} = \frac{80}{9}. \] 
Thus, the final answer is: \[ (\alpha - \beta)^2 = \frac{80}{9}. \]