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Question: If \(\alpha ,\beta ,\gamma \) are the roots of the equation \(2{x^3} - 3{x^2} + 6x + 1 = 0\), then \...

If α,β,γ\alpha ,\beta ,\gamma are the roots of the equation 2x3−3x2+6x+1=02{x^3} - 3{x^2} + 6x + 1 = 0, then α2+β2+γ2{\alpha ^2} + {\beta ^2} + {\gamma ^2} is equal to:
A) −154 - \dfrac{{15}}{4}
B) 154\dfrac{{15}}{4}
C) 94\dfrac{9}{4}
D) 44

Explanation

Solution

According to given in the question we have to determine the value of α2+β2+γ2{\alpha ^2} + {\beta ^2} + {\gamma ^2} when α,β,γ\alpha ,\beta ,\gamma are the roots of the equation 2x3−3x2+6x+1=02{x^3} - 3{x^2} + 6x + 1 = 0 so, first of all we have to find the sum of the roots α,β,γ\alpha ,\beta ,\gamma with the help of the formula as given below:

Formula used: ⇒α+β+γ=−ba................(A) \Rightarrow \alpha + \beta + \gamma = - \dfrac{b}{a}................(A)
Now, we have to obtain the multiplication of αβ+βγ+γα\alpha \beta + \beta \gamma + \gamma \alpha with the help of the formula as given below:
⇒αβ+βγ+γα=ca.................(B)\Rightarrow \alpha \beta + \beta \gamma + \gamma \alpha = \dfrac{c}{a}.................(B)
Where, α,β,γ\alpha ,\beta ,\gamma are the roots for the cubic expressionax3+bx2+cx+d=0a{x^3} + b{x^2} + cx + d = 0
Now, to solve the expression we have to use the formula as given below:
⇒α2+β2+γ2=(α+β+γ)−2(αβ+βγ+γα).........................(C)\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 2x3−3x2+6x+1=02{x^3} - 3{x^2} + 6x + 1 = 0with the cubic expression ax3+bx2+cx+d=0a{x^3} + b{x^2} + cx + d = 0to obtain the value of (a, b, c and d) as given below:
a=2,b=−3,c=6,d=1a = 2,b = - 3,c = 6,d = 1
Step 1: Now, we have to find the sum of the roots α,β,γ\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),
⇒α+β+γ=−(−3)2 ⇒α+β+γ=32....................(1) \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 2x3−3x2+6x+1=02{x^3} - 3{x^2} + 6x + 1 = 0 hence,
⇒αβ+βγ+γα=62 ⇒αβ+βγ+γα=3...............(2) \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 α2+β2+γ2{\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.
⇒α2+β2+γ2=(32)2−2×3 ⇒α2+β2+γ2=94−6 \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,
⇒α2+β2+γ2=9−244 ⇒α2+β2+γ2=−154 \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 α2+β2+γ2=−154{\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.