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

Given that \(\alpha\) and \(\beta\) are the roots of the quadratic equation \(px^2 + qx - r = 0\), where \(p \neq 0\), and it is known that \(p, q,\) and \(r\) form consecutive terms of a non-constant Geometric Progression (G.P.). Additionally, it is provided that \(\frac{1}{\alpha} + \frac{1}{\beta} = \frac{3}{4}\). We are to find the value of \((\alpha - \beta)^2\).

Let's start with the properties of the roots of a quadratic equation. For the equation \(px^2 + qx - r = 0\), the roots \(\alpha\) and \(\beta\) satisfy:

• \(\alpha + \beta = -\frac{q}{p}\) (Sum of roots)
• \(\alpha \beta = -\frac{r}{p}\) (Product of roots)

From the given equation, we have \(\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha \beta} = \frac{3}{4}\).

Substituting the expressions for \(\alpha + \beta\) and \(\alpha \beta\), we get:

\[\frac{-\frac{q}{p}}{-\frac{r}{p}} = \frac{3}{4}\]

Simplifying gives:

\[\frac{q}{r} = \frac{3}{4}\]

Since \(p, q, r\) are in G.P., let's denote the common ratio by \(k\). Therefore, we have:

• \(\frac{q}{p} = k\)
• \(\frac{r}{q} = k\)

From \(\frac{q}{r} = \frac{3}{4}\), we find:

\[k^2 = \frac{3}{4}\]

Now, since \(q = kp\) and \(r = k^2p\), substituting in \(\frac{q}{r} = \frac{3}{4}\) verifies that:

\[\frac{kp}{k^2p} = \frac{1}{k} = \frac{3}{4}\]

Solving further \(k = \frac{4}{3}\), thus, \(k^2 = \frac{3}{4}\).

For the discriminant part to find \((\alpha - \beta)^2\) given by:

\[(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta\]

Substituting known values:

\[= \left(-\frac{q}{p}\right)^2 - 4\left(-\frac{r}{p}\right)\]

Using \(q = \frac{4}{3}p\) and \(r = \left(\frac{4}{3}\right)^2 p\):

\[= \left(\frac{4}{3}\right)^2 \cdot p^2/p^2 - 4\cdot \frac{\left(\frac{4}{3}\right)^2 p}{p}\]

 

\[= \frac{16}{9} - 4 \cdot \frac{16}{9}\]

 

\[= \frac{16}{9} - \frac{64}{9}\]

 

\[= \frac{80}{9}\]

Thus, the correct value is \(\frac{80}{9}\).

Answer 2

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: \[ (\alpha - \beta)^2 = \frac{80}{9}. \]