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# If $\alpha$and $\beta$are the roots of $\begin{array}{*{20}{c}} {6{x^2} - 6x + 1}& = &0 \end{array}$, then the value of$\dfrac{1}{2}\left[ {a + b\alpha + c{\alpha ^2} + d{\alpha ^3}} \right] + \dfrac{1}{2}\left[ {a + b\beta + c{\beta ^2} + d{\beta ^3}} \right]$ is A) $\dfrac{1}{4}\left( {a + b + c + d} \right)$B) $\dfrac{a}{1} + \dfrac{b}{2} + \dfrac{c}{3} + \dfrac{d}{4}$C) $\dfrac{a}{2} - \dfrac{b}{2} + \dfrac{c}{3} - \dfrac{d}{4}$D) None of these

Last updated date: 09th Apr 2024
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Hint: during this question we've got given the equation whose roots area unit given. 1st of all, we are going to verify the add and therefore the product of the roots of the given equation. so we are going to modify the given expression. afterward, we are going to place these values within the expression. Hence, we are going to get an acceptable answer.

Formula Used:1) $\begin{array}{*{20}{c}} {\alpha + \beta }& = &{ - \dfrac{b}{a}} \end{array}$
2) $\begin{array}{*{20}{c}} {\alpha \beta }& = &{\dfrac{c}{a}} \end{array}$
3) $\begin{array}{*{20}{c}} {{{\left( {\alpha + \beta } \right)}^2}}& = &{{\alpha ^2} + {\beta ^2} + 2\alpha \beta } \end{array}$
4) $\begin{array}{*{20}{c}} {{{\left( {\alpha + \beta } \right)}^3}}& = &{{\alpha ^3} + {\beta ^3} + 3\alpha \beta \left( {\alpha + \beta } \right)} \end{array}$

Complete step by step solution:According to the question, we have the equation whose roots are $\alpha$and $\beta$respectively. Now we will write the equation,
$\Rightarrow \begin{array}{*{20}{c}} {6{x^2} - 6x + 1}& = &0 \end{array}$
Now we will determine the sum and the product of the roots of the equation. Therefore, we can write
$\begin{array}{*{20}{c}} { \Rightarrow \alpha + \beta }& = &{\dfrac{6}{6}} \end{array}$
$\begin{array}{*{20}{c}} { \Rightarrow \alpha + \beta }& = &1 \end{array}$ ------ (1)
And we know the product of the roots of the equation. Therefore, we can write it as
$\begin{array}{*{20}{c}} { \Rightarrow \alpha \beta }& = &{\dfrac{1}{6}} \end{array}$ -------- (2)
Now we have given an expression whose value we have to determine.
$\Rightarrow \dfrac{1}{2}\left[ {a + b\alpha + c{\alpha ^2} + d{\alpha ^3}} \right] + \dfrac{1}{2}\left[ {a + b\beta + c{\beta ^2} + d{\beta ^3}} \right]$
Now we will take the common $\dfrac{1}{2}$from the above expression. Therefore, we will get
$\Rightarrow \dfrac{1}{2}\left[ {\left( {a + b\alpha + c{\alpha ^2} + d{\alpha ^3}} \right) + \left( {a + b\beta + c{\beta ^2} + d{\beta ^3}} \right)} \right]$
Now we will combine the same variable terms. Therefore, we will get
$\Rightarrow \dfrac{1}{2}\left[ {2a + b\left( {\alpha + \beta } \right) + c\left( {{\alpha ^2} + {\beta ^2}} \right) + d\left( {{\alpha ^3} + {\beta ^3}} \right)} \right]$ ---------- (3)
Now we know the formula of the ${\left( {\alpha + \beta } \right)^2}$and ${\left( {\alpha + \beta } \right)^3}$. Therefore, we will write is as
$\begin{array}{*{20}{c}} { \Rightarrow {{\left( {\alpha + \beta } \right)}^2}}& = &{{\alpha ^2} + {\beta ^2} + 2\alpha \beta } \end{array}$
Now
$\begin{array}{*{20}{c}} { \Rightarrow {\alpha ^2} + {\beta ^2}}& = &{{{\left( {\alpha + \beta } \right)}^2} - 2\alpha \beta } \end{array}$
Similarly,
$\begin{array}{*{20}{c}} { \Rightarrow {\alpha ^3} + {\beta ^3}}& = &{{{\left( {\alpha + \beta } \right)}^3} - 3\alpha \beta \left( {\alpha + \beta } \right)} \end{array}$
Now we will put these values in equation (3). Therefore, we will get
$\Rightarrow \dfrac{1}{2}\left[ {2a + b\left( {\alpha + \beta } \right) + c\left\{ {{{\left( {\alpha + \beta } \right)}^2} - 2\alpha \beta } \right\} + d\left\{ {{{\left( {\alpha + \beta } \right)}^3} - 3\alpha \beta \left( {\alpha + \beta } \right)} \right\}} \right]$
Now we will put the value of the equation (1) and (2) in the above expression. Therefore, we will get
$\Rightarrow \dfrac{1}{2}\left[ {2a + b\left( 1 \right) + c\left\{ {{{\left( 1 \right)}^2} - 2 \times \dfrac{1}{6}} \right\} + d\left\{ {{{\left( 1 \right)}^3} - 3 \times \dfrac{1}{6}\left( 1 \right)} \right\}} \right]$
$\Rightarrow \dfrac{1}{2}\left[ {2a + b\left( 1 \right) + c\left( {\dfrac{2}{3}} \right) + d\left( {\dfrac{1}{2}} \right)} \right]$
By simplifying the on top of the expression, we are going to get
$\Rightarrow \dfrac{a}{1} + \dfrac{b}{2} + \dfrac{c}{3} + \dfrac{d}{4}$
Now we can choose the correct answer from the given option.

Option ‘B’ is correct

Note: In this question, the first point is to keep in mind that we will put the value of the sum and the product of the roots of the equation when we will get the simplified expression.