Question

If $\alpha ,\beta ,\gamma $ are the roots of the equation $2{x^3} - 3{x^2} + 6x + 1 = 0$, then ${\alpha ^2} + {\beta ^2} + {\gamma ^2}$ is equal to:
A) $ - \dfrac{{15}}{4}$
B) $\dfrac{{15}}{4}$
C) $\dfrac{9}{4}$
D) $4$

Answer 1

Hint: According to given in the question we have to determine the value of ${\alpha ^2} + {\beta ^2} + {\gamma ^2}$ when $\alpha ,\beta ,\gamma $ are the roots of the equation $2{x^3} - 3{x^2} + 6x + 1 = 0$ so, first of all we have to find the sum of the roots $\alpha ,\beta ,\gamma $with the help of the formula as given below:

Formula used: $ \Rightarrow \alpha + \beta + \gamma = - \dfrac{b}{a}................(A)$
Now, we have to obtain the multiplication of $\alpha \beta + \beta \gamma + \gamma \alpha $ with the help of the formula as given below:
$ \Rightarrow \alpha \beta + \beta \gamma + \gamma \alpha = \dfrac{c}{a}.................(B)$
Where, $\alpha ,\beta ,\gamma $are the roots for the cubic expression$a{x^3} + b{x^2} + cx + d = 0$
Now, to solve the expression we have to use the formula as given below:
$ \Rightarrow {\alpha ^2} + {\beta ^2} + {\gamma ^2} = (\alpha + \beta + \gamma ) - 2(\alpha \beta + \beta \gamma + \gamma \alpha ).........................(C)$

Complete step-by-step answer:
Step 1: First of all we have to compare the given expression $2{x^3} - 3{x^2} + 6x + 1 = 0$with the cubic expression $a{x^3} + b{x^2} + cx + d = 0$to obtain the value of (a, b, c and d) as given below:
$a = 2,b = - 3,c = 6,d = 1$
Step 1: Now, we have to find the sum of the roots $\alpha ,\beta ,\gamma $with the help of the formula (A) as given in the solution hint. On substituting all the values in the formula (A),
$
   \Rightarrow \alpha + \beta + \gamma = - \dfrac{{( - 3)}}{2} \\
   \Rightarrow \alpha + \beta + \gamma = \dfrac{3}{2}....................(1)
 $
Step 3: Now, with the help of the formula (B) we have to obtain the product of the roots for the given expression $2{x^3} - 3{x^2} + 6x + 1 = 0$ hence,
$
   \Rightarrow \alpha \beta + \beta \gamma + \gamma \alpha = \dfrac{6}{2} \\
   \Rightarrow \alpha \beta + \beta \gamma + \gamma \alpha = 3...............(2)
 $
Step 4: Now, to find the value of ${\alpha ^2} + {\beta ^2} + {\gamma ^2}$we have to substitute the sum and product of roots as obtained from the step 2 and step 3 in the formula (B) as mentioned in the solution hint.
$
\Rightarrow {\alpha ^2} + {\beta ^2} + {\gamma ^2} = {\left( {\dfrac{3}{2}} \right)^2} - 2 \times 3 \\
\Rightarrow {\alpha ^2} + {\beta ^2} + {\gamma ^2} = \dfrac{9}{4} - 6
$
Now, to solve the obtained expression we have to apply the L.C.M hence,
$
\Rightarrow {\alpha ^2} + {\beta ^2} + {\gamma ^2} = \dfrac{{9 - 24}}{4} \\
\Rightarrow {\alpha ^2} + {\beta ^2} + {\gamma ^2} = \dfrac{{ - 15}}{4}
$
Final solution: Hence, with the help of formula (A) and (B) we have obtained the value of ${\alpha ^2} + {\beta ^2} + {\gamma ^2} = \dfrac{{ - 15}}{4}$.

Therefore option (A) is the correct answer.

Note: If the given expression/equation is cubic then only three roots can be obtained as (a, b, and c) from the given cubic expression/equation.
If the given expression/equation is quadratic then only two roots can be obtained as (a, and b) from the given cubic expression/equation.