Question

If $\alpha$, $\gamma$ are the roots of $ax^2 - 4x + 1 = 0$ and $\beta$, $\delta$ are the roots of $bx^2 - 6x + 1 = 0$ and $\alpha$, $\beta$, $\gamma$, $\delta$ are in H.P. then which of the following is correct?

• A. $\beta = \frac{1}{2}$
• B. $\gamma = \frac{1}{3}$
• C. b = 8
• D. a = 3

Answer 1

Answer: (A), (B), (C), (D)

All options are correct

Answer 2

The problem involves properties of roots of quadratic equations and harmonic progression (H.P.).

Let the roots of the quadratic equation $ax^2 - 4x + 1 = 0$ be $\alpha$ and $\gamma$. For a quadratic equation $Px^2 + Qx + R = 0$, the roots $x_1, x_2$ satisfy $x_1+x_2 = -Q/P$ and $x_1x_2 = R/P$. In this case, $\alpha + \gamma = \frac{4}{a}$ and $\alpha\gamma = \frac{1}{a}$.

Let the roots of the quadratic equation $bx^2 - 6x + 1 = 0$ be $\beta$ and $\delta$. Similarly, $\beta + \delta = \frac{6}{b}$ and $\beta\delta = \frac{1}{b}$.

To find the equation whose roots are the reciprocals of the roots of $Px^2 + Qx + R = 0$, we replace $x$ with $1/y$. This gives $P(1/y)^2 + Q(1/y) + R = 0$, which simplifies to $R y^2 + Q y + P = 0$.

For $ax^2 - 4x + 1 = 0$, the equation whose roots are $1/\alpha$ and $1/\gamma$ is $1 \cdot y^2 - 4y + a = 0$, i.e., $y^2 - 4y + a = 0$. From this, we have:

$\frac{1}{\alpha} + \frac{1}{\gamma} = 4$

$\frac{1}{\alpha} \cdot \frac{1}{\gamma} = a$

For $bx^2 - 6x + 1 = 0$, the equation whose roots are $1/\beta$ and $1/\delta$ is $1 \cdot y^2 - 6y + b = 0$, i.e., $y^2 - 6y + b = 0$. From this, we have:

$\frac{1}{\beta} + \frac{1}{\delta} = 6$

$\frac{1}{\beta} \cdot \frac{1}{\delta} = b$

It is given that $\alpha, \beta, \gamma, \delta$ are in H.P. (Harmonic Progression). This means their reciprocals, $\frac{1}{\alpha}, \frac{1}{\beta}, \frac{1}{\gamma}, \frac{1}{\delta}$, are in A.P. (Arithmetic Progression). Let these terms be $A_1, A_2, A_3, A_4$ respectively. So, $A_1 = \frac{1}{\alpha}$, $A_2 = \frac{1}{\beta}$, $A_3 = \frac{1}{\gamma}$, $A_4 = \frac{1}{\delta}$.

From the sums derived above:

$A_1 + A_3 = 4$

$A_2 + A_4 = 6$

In an A.P., if the common difference is $d$, then:

$A_2 = A_1 + d$

$A_3 = A_1 + 2d$

$A_4 = A_1 + 3d$

Substitute these into the sum equations:

$(A_1) + (A_1 + 2d) = 4 \Rightarrow 2A_1 + 2d = 4 \Rightarrow A_1 + d = 2$ (Equation 1)

$(A_1 + d) + (A_1 + 3d) = 6 \Rightarrow 2A_1 + 4d = 6 \Rightarrow A_1 + 2d = 3$ (Equation 2)

Now we solve the system of linear equations for $A_1$ and $d$: Subtract Equation 1 from Equation 2:

$(A_1 + 2d) - (A_1 + d) = 3 - 2$

$d = 1$

Substitute $d=1$ into Equation 1:

$A_1 + 1 = 2 \Rightarrow A_1 = 1$

Now we have the first term $A_1=1$ and the common difference $d=1$. We can find all the terms of the A.P.:

$A_1 = \frac{1}{\alpha} = 1 \Rightarrow \alpha = 1$

$A_2 = \frac{1}{\beta} = A_1 + d = 1 + 1 = 2 \Rightarrow \beta = \frac{1}{2}$

$A_3 = \frac{1}{\gamma} = A_1 + 2d = 1 + 2(1) = 3 \Rightarrow \gamma = \frac{1}{3}$

$A_4 = \frac{1}{\delta} = A_1 + 3d = 1 + 3(1) = 4 \Rightarrow \delta = \frac{1}{4}$

Finally, we can find the values of $a$ and $b$:

$a = \frac{1}{\alpha} \cdot \frac{1}{\gamma} = A_1 \cdot A_3 = 1 \times 3 = 3$

$b = \frac{1}{\beta} \cdot \frac{1}{\delta} = A_2 \cdot A_4 = 2 \times 4 = 8$

Now let's check the given options:

1. $\beta = \frac{1}{2}$: This matches our calculated value. (Correct)
2. $\gamma = \frac{1}{3}$: This matches our calculated value. (Correct)
3. $b = 8$: This matches our calculated value. (Correct)
4. $a = 3$: This matches our calculated value. (Correct)

All four options are correct.