Question

If $\alpha \ne \beta $ , ${\alpha ^2} = 5\alpha - 3$ , ${\beta ^2} = 5\beta - 3$ , then the equation whose roots are $\dfrac{\alpha }{\beta }$ and $\dfrac{\beta }{\alpha }$ , is
A. ${x^2} + 5x - 3 = 0$
B. $3{x^2} + 12x + 3 = 0$
C. $3{x^2} - 19x + 3 = 0$
D. None of these

Answer 1

Hint: To solve such a question start by writing the quadratic equation from the given terms. After that find the sum and product of the roots of the quadratic equation obtained. Then substitute these values to find the equation whose roots are $\dfrac{\alpha }{\beta }$ and $\dfrac{\beta }{\alpha }$.

Complete step-by-step solution:
Given that $\alpha \ne \beta $ . Also given that ${\alpha ^2} = 5\alpha - 3$ and ${\beta ^2} = 5\beta - 3$ .
It is asked to find an equation whose roots are $\dfrac{\alpha }{\beta }$ and $\dfrac{\beta }{\alpha }$ .
Consider ${\alpha ^2} = 5\alpha - 3$ . It can be written as
${\alpha ^2} - 5\alpha + 3 = 0$
Similarly ${\beta ^2} = 5\beta - 3$ can be written as,
${\beta ^2} - 5\beta + 3 = 0$
Writing this in quadratic equation form, we get,
$\Rightarrow$${x^2} - 5x + 3 = 0$
Hence here $\alpha $ and $\beta $ are the roots of this quadratic equation.
Therefore, it can be written that
$\Rightarrow$$\alpha + \beta = 5......(1)$ and
$\Rightarrow$$\alpha \beta = 3......(2)$
It is asked to find the equation whose roots are $\dfrac{\alpha }{\beta }$ and $\dfrac{\beta }{\alpha }$ . So,
$\Rightarrow$$\dfrac{\alpha }{\beta } + \dfrac{\beta }{\alpha } = \dfrac{{{\alpha ^2} + {\beta ^2}}}{{\alpha \beta }}$
This can be written as,
$\Rightarrow$$\dfrac{\alpha }{\beta } + \dfrac{\beta }{\alpha } = \dfrac{{{{\left( {\alpha + \beta } \right)}^2} - 2\alpha \beta }}{{\alpha \beta }}......(3)$
Substituting equation $(1)$ and equation $(2)$ in equation $(3)$ , we get
$\Rightarrow$$\dfrac{\alpha }{\beta } + \dfrac{\beta }{\alpha } = \dfrac{{{{\left( 5 \right)}^2} - 2\left( 3 \right)}}{3}$
$\Rightarrow$$\dfrac{\alpha }{\beta } + \dfrac{\beta }{\alpha } = \dfrac{{25 - 6}}{3}$
Further simplifying we get,
$\Rightarrow$$\dfrac{\alpha }{\beta } + \dfrac{\beta }{\alpha } = \dfrac{{19}}{3}$
It is known that the product of the roots will be one. That is,
$\Rightarrow$$\dfrac{\alpha }{\beta } \times \dfrac{\beta }{\alpha } = 1$
Therefore, the required equation will be,
$\Rightarrow$${x^2} - \dfrac{{19}}{3}x + 1 = 0$
Dividing throughout by $3$ we get
$\Rightarrow$$3{x^2} - 19x + 3 = 0$

Hence option C is the correct option.

Note: There are different methods to solve this question. An alternate method to solve this question is as shown below:
Suppose that
${\alpha ^2} = 5\alpha - 3......(1)$ and
${\beta ^2} = 5\beta - 3......(2)$
Subtract equation $\left( 1 \right)$ from equation $(2)$
That is,
$\Rightarrow$${\alpha ^2} - {\beta ^2} = 5\alpha - 3 - \left( {5\beta - 3} \right)$
$\Rightarrow$${\alpha ^2} - {\beta ^2} = 5\alpha - 5\beta $
Simplifying further we get,
$\Rightarrow$$\left( {\alpha - \beta } \right)\left( {\alpha + \beta } \right) = 5\left( {\alpha - \beta } \right)$
Dividing throughout by $\alpha - \beta $ we get,
$\Rightarrow$$\alpha + \beta = 5......(3)$
Next add equation $\left( 1 \right)$ and equation $(2)$
$\Rightarrow$${\alpha ^2} + {\beta ^2} = 5\alpha - 3 + \left( {5\beta - 3} \right)$
$\Rightarrow$${\alpha ^2} + {\beta ^2} = 5\left( {\alpha + \beta } \right) - 6$
Substituting equation $(3)$
$\Rightarrow$${\alpha ^2} + {\beta ^2} = 5\left( 5 \right) - 6$
$\Rightarrow$${\alpha ^2} + {\beta ^2} = 25 - 6$
Further simplifying we get,
$\Rightarrow$${\left( {\alpha + \beta } \right)^2} - 2\alpha \beta = 19$
Substituting equation $(3)$
$\Rightarrow$${\left( 5 \right)^2} - 2\alpha \beta = 19$
$\Rightarrow$$25 - 2\alpha \beta = 19$
Simplifying further, we get
$\Rightarrow$$ - 2\alpha \beta = 19 - 25$
v$ - 2\alpha \beta = - 6$
Dividing throughout by $ - 2$ , we get
$\Rightarrow$$\alpha \beta = 3......(4)$
Next,
$\Rightarrow$$\dfrac{\alpha }{\beta } + \dfrac{\beta }{\alpha } = \dfrac{{{\alpha ^2} + {\beta ^2}}}{{\alpha \beta }}$
$\Rightarrow$$\dfrac{\alpha }{\beta } + \dfrac{\beta }{\alpha } = \dfrac{{19}}{3}$
Also,
$\Rightarrow$$\dfrac{\alpha }{\beta } \times \dfrac{\beta }{\alpha } = 1$
Therefore, given below is the required equation with roots $\dfrac{\alpha }{\beta }$ and $\Rightarrow$$\dfrac{\beta }{\alpha }$
$\Rightarrow$${x^2} - \left( {\dfrac{\alpha }{\beta } + \dfrac{\beta }{\alpha }} \right)x + \left( {\dfrac{\alpha }{\beta } \times \dfrac{\beta }{\alpha }} \right) = 0$
Substituting the values we get the required equation, that is,
$\Rightarrow$${x^2} - \dfrac{{19}}{3}x + 1 = 0$
Or it can be written as
$\Rightarrow$$3{x^2} - 19x + 3 = 0$