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Higher Order Equations

Let m1, m2, m...

Question

Let $m_{1}, m_{2}, m_{3}$ be the slopes of all the three straight lines represented by an equation $y^{3} + (2 a + 5) x y^{2} - 6 x^{2} y - 2 a x^{3} = 0.$ If $a, m_{1}, m_{2}, m_{3}$ are all integers, then which of the following hold(s) good ?

A

a + 3 \sum i = 1 m_{i} = - 1

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B

a + 3 \sum i = 1 m_{i} = - 5

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C

a + 3 \prod i = 1 m_{i} = 0

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D

a + 3 \prod i = 1 m_{i} = 4

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Solution

The correct option is C $a + 3 \prod i = 1 m_{i} = 0$
Given equation is $y^{3} + (2 a + 5) x y^{2} - 6 x^{2} y - 2 a x^{3} = 0$
$\Rightarrow {(\frac{y}{x})}^{3} + (2 a + 5) {(\frac{y}{x})}^{2} - 6 (\frac{y}{x}) - 2 a = 0$
Substitute $\frac{y}{x} = m$
$m^{3} + (2 a + 5) m^{2} - 6 m - 2 a = 0$
The roots of above equation are $m_{1}, m_{2}, m_{3} .$
$∴ m_{1} + m_{2} + m_{3} = - (2 a + 5)$
$m_{1} m_{2} + m_{2} m_{3} + m_{3} m_{1} = - 6$
$m_{1} m_{2} m_{3} = 2 a$

For $a + 3 \sum i = 1 m_{i} = - 1$
$\Rightarrow a - (2 a + 5) = - 1$
$\Rightarrow a = - 4$ which is an integer.
$\Rightarrow m_{1} m_{2} m_{3} = - 8$
$\Rightarrow m_{1} = 1, m_{2} = - 2, m_{3} = 4$
So, here condition is satisfied.

For $a + 3 \sum i = 1 m_{i} = - 5$
$\Rightarrow a - (2 a + 5) = - 5$
$\Rightarrow a = 0$ which is an integer.
$\Rightarrow m_{1} m_{2} m_{3} = 0$
$\Rightarrow m_{1} = 1, m_{2} = 0, m_{3} = - 6$
So, here also condition is satisfied.

For $a + 3 \prod i = 1 m_{i} = 0$
$\Rightarrow a + 2 a = 0$
$\Rightarrow a = 0$ which is an integer
So, here also condition is satisfied.

For $a + 3 \prod i = 1 m_{i} = 4$
$\Rightarrow a + 2 a = 4$
$\Rightarrow a = \frac{4}{3}$ which is not an integer
So, here condition is not satisfied.

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