Solveeit Logo
Login

Question

Question: If we have the roots as \(\alpha ,\beta ,\gamma \) of a cubic equation \({{x}^{3}}-7x+6=0\) then the...

If we have the roots as α,β,γ\alpha ,\beta ,\gamma of a cubic equation x3−7x+6=0{{x}^{3}}-7x+6=0 then the equation whose roots are (α+β)2,(β+γ)2,(γ+α)2{{\left( \alpha +\beta \right)}^{2}},{{\left( \beta +\gamma \right)}^{2}},{{\left( \gamma +\alpha \right)}^{2}} is
(a) x(x2+7)=36x\left( {{x}^{2}}+7 \right)=36
(b) x(x+7)=36x\left( x+7 \right)=36
(c) x(x2−7)=36x\left( {{x}^{2}}-7 \right)=36
(d) x(x−7)=36x\left( x-7 \right)=36

Explanation

Solution

Hint: To solve this question, first we will find the relation between α,β,γ\alpha ,\beta ,\gamma and the coefficients of x3,x2,x{{x}^{3}},{{x}^{2}},x and constant term. Then we will apply this relation for finding the equation which has roots (α+β)2,(β+γ)2{{\left( \alpha +\beta \right)}^{2}},{{\left( \beta +\gamma \right)}^{2}} and (γ+α)2{{\left( \gamma +\alpha \right)}^{2}} .

Complete step-by-step solution -
To start with, we are given that the roots of the equation x3−7x+6=0{{x}^{3}}-7x+6=0 are α,β\alpha ,\beta and γ\gamma . Now, we have to develop relations between roots and coefficients. We know that sum of the roots of the cubic equation is given as:
α+β+γ=−coefficient of x2coefficient of x3 ⇒α+β+γ=−01 ⇒α+β+γ=0..............(i) \begin{aligned} & \alpha +\beta +\gamma =\dfrac{-\text{coefficient of }{{x}^{2}}}{\text{coefficient of }{{x}^{3}}} \\\ & \Rightarrow \alpha +\beta +\gamma =\dfrac{-0}{1} \\\ & \Rightarrow \alpha +\beta +\gamma =0..............\left( i \right) \\\ \end{aligned}
Similarly, the sum of multiplication of two roots is given by
αβ+βγ+γα=coefficient of xcoefficient of x3 ⇒αβ+βγ+γα=−71 ⇒αβ+βγ+γα=−7..............(ii) \begin{aligned} & \alpha \beta +\beta \gamma +\gamma \alpha =\dfrac{\text{coefficient of }x}{\text{coefficient of }{{x}^{3}}} \\\ & \Rightarrow \alpha \beta +\beta \gamma +\gamma \alpha =\dfrac{-7}{1} \\\ & \Rightarrow \alpha \beta +\beta \gamma +\gamma \alpha =-7..............\left( ii \right) \\\ \end{aligned}
Similarly, the multiplication of the roots is given by:
⇒αβγ=−constant termcoefficient of x3 ⇒αβγ=−61 ⇒αβγ=−6...............(iii) \begin{aligned} & \Rightarrow \alpha \beta \gamma =\dfrac{-\text{constant term}}{\text{coefficient of }{{x}^{3}}} \\\ & \Rightarrow \alpha \beta \gamma =\dfrac{-6}{1} \\\ & \Rightarrow \alpha \beta \gamma =-6...............\left( iii \right) \\\ \end{aligned}
Now, we are going to find the equation which has (α+β)2,(β+γ)2{{\left( \alpha +\beta \right)}^{2}},{{\left( \beta +\gamma \right)}^{2}} and (γ+α)2{{\left( \gamma +\alpha \right)}^{2}} . Let us assume that the cubic equation will be
x3+bx2+cx+d=0{{x}^{3}}+b{{x}^{2}}+cx+d=0
We know that sum of the roots of the above equation is given as:
(α+β)2+(β+γ)2+(γ+α)2=−b{{\left( \alpha +\beta \right)}^{2}}+{{\left( \beta +\gamma \right)}^{2}}+{{\left( \gamma +\alpha \right)}^{2}}=-b
Now, using the identity (a+b)2=a2+b2+2ab{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab , we will expand this as:
α2+β2+2αβ+β2+γ2+2βγ+γ2+α2+2γα=−b ⇒2(α2+β2+γ2)+2(αβ+βγ+γα)=−b \begin{aligned} & {{\alpha }^{2}}+{{\beta }^{2}}+2\alpha \beta +{{\beta }^{2}}+{{\gamma }^{2}}+2\beta \gamma +{{\gamma }^{2}}+{{\alpha }^{2}}+2\gamma \alpha =-b \\\ & \Rightarrow 2\left( {{\alpha }^{2}}+{{\beta }^{2}}+{{\gamma }^{2}} \right)+2\left( \alpha \beta +\beta \gamma +\gamma \alpha \right)=-b \\\ \end{aligned}
Now, we will add and subtract 2(αβ+βγ+γα)2\left( \alpha \beta +\beta \gamma +\gamma \alpha \right) on both sides. After doing this, we will get:

& \Rightarrow 2\left( {{\alpha }^{2}}+{{\beta }^{2}}+{{\gamma }^{2}} \right)+2\left( \alpha \beta +\beta \gamma +\gamma \alpha \right)+2\left( \alpha \beta +\beta \gamma +\gamma \alpha \right)-2\left( \alpha \beta +\beta \gamma +\gamma \alpha \right)=-b \\\ & \Rightarrow 2\left[ {{\alpha }^{2}}+{{\beta }^{2}}+{{\gamma }^{2}}+2\alpha \beta +2\beta \gamma +2\gamma \alpha \right]-2\left( \alpha \beta +\beta \gamma +\gamma \alpha \right)=-b \\\ \end{aligned}$$ Using the identity ${{a}^{2}}+{{b}^{2}}+2ab+2bc+2ac={{\left( a+b+c \right)}^{2}}$ , we get: $\Rightarrow 2{{\left( \alpha +\beta +\gamma \right)}^{2}} - 2\left( \alpha \beta +\beta \gamma +\gamma \alpha \right)=-b...............\left( iv \right)$ Now from equations (i), (ii) and (iv), we get $\begin{aligned} & \Rightarrow 2{{\left( 0 \right)}^{2}}-2\left( -7 \right)=-b \\\ & \Rightarrow 14=-b \\\ & \Rightarrow b=-14 \\\ \end{aligned}$ Also, another relation is obtained by ${{\left( \alpha +\beta \right)}^{2}}{{\left( \beta +\gamma \right)}^{2}}+{{\left( \beta +\gamma \right)}^{2}}{{\left( \gamma +\alpha \right)}^{2}}+{{\left( \gamma +\alpha \right)}^{2}}{{\left( \alpha +\beta \right)}^{2}}=c..............\left( v \right)$ Now from equation (i) , we have $$\begin{aligned} & \alpha +\beta +\gamma =0 \\\ & \Rightarrow \beta +\gamma =-\alpha ..........\left( vi \right) \\\ & \Rightarrow \alpha +\gamma =-\beta ...........\left( vii \right) \\\ & \Rightarrow \alpha +\beta =-\gamma ............\left( viii \right) \\\ \end{aligned}$$ Now using the values from equations (vi), (vii) and (viii) in equation (v), we get: $\begin{aligned} & {{\left( -\gamma \right)}^{2}}{{\left( -\alpha \right)}^{2}}+{{\left( -\alpha \right)}^{2}}{{\left( -\beta \right)}^{2}}+{{\left( -\beta \right)}^{2}}{{\left( -\gamma \right)}^{2}}=c \\\ & {{\gamma }^{2}}{{\alpha }^{2}}+{{\alpha }^{2}}{{\beta }^{2}}+{{\beta }^{2}}{{\gamma }^{2}}=c..................\left( ix \right) \\\ \end{aligned}$ From equation (ii), we have $\alpha \beta +\beta \gamma +\gamma \alpha =-7$ On squaring both sides, we get: $\begin{aligned} & {{\left( \alpha \beta +\beta \gamma +\gamma \alpha \right)}^{2}}={{\left( -7 \right)}^{2}} \\\ & \Rightarrow {{\alpha }^{2}}{{\beta }^{2}}+{{\beta }^{2}}{{\gamma }^{2}}+{{\gamma }^{2}}{{\alpha }^{2}}+2\alpha {{\beta }^{2}}\gamma +2\alpha \beta {{\gamma }^{2}}+2{{\alpha }^{2}}\beta \gamma =49 \\\ & \Rightarrow {{\alpha }^{2}}{{\beta }^{2}}+{{\beta }^{2}}{{\gamma }^{2}}+{{\gamma }^{2}}{{\alpha }^{2}}+2\alpha \beta \gamma \left( \alpha +\beta +\gamma \right)=49.................\left( x \right) \\\ \end{aligned}$ From equation (i), (iii) and (x) we get $\begin{aligned} & \Rightarrow {{\alpha }^{2}}{{\beta }^{2}}+{{\beta }^{2}}{{\gamma }^{2}}+{{\gamma }^{2}}{{\alpha }^{2}}+0=49 \\\ & \Rightarrow {{\alpha }^{2}}{{\beta }^{2}}+{{\beta }^{2}}{{\gamma }^{2}}+{{\gamma }^{2}}{{\alpha }^{2}}=49 \\\ \end{aligned}$ Now we put the value in equation (ix) $\Rightarrow c=49$ Now, another relation is given by ${{\left( \alpha +\beta \right)}^{2}}{{\left( \beta +\gamma \right)}^{2}}{{\left( \gamma +\alpha \right)}^{2}}=-d..................\left( xi \right)$ From equations (xi), (vi), (vii) and (viii), we have $\begin{aligned} & \Rightarrow {{\left( -\gamma \right)}^{2}}{{\left( -\alpha \right)}^{2}}{{\left( -\gamma \right)}^{2}}=-d \\\ & \Rightarrow {{\alpha }^{2}}{{\beta }^{2}}{{\gamma }^{2}}=-d \\\ & \Rightarrow {{\left( \alpha \beta \gamma \right)}^{2}}=-d \\\ & \Rightarrow d=-{{\left( \alpha \beta \gamma \right)}^{2}} \\\ & \Rightarrow d=-{{\left( -6 \right)}^{2}} \\\ & \Rightarrow d=-36 \\\ \end{aligned}$ So, the equation required is ${{x}^{3}}-14{{x}^{2}}+49x-36=0$ On simplifying this, we get $\begin{aligned} & \Rightarrow x\left( {{x}^{2}}-14x+49 \right)=36 \\\ & \Rightarrow x{{\left( x-7 \right)}^{2}}=36\text{ }\therefore \text{Using identity }{{a}^{2}}+{{b}^{2}}+2ab={{\left( a+b \right)}^{2}} \\\ \end{aligned}$ Hence option (d) is correct. Note: The alternate method to solve this question is by hit and trial method. We will put different values and check them if the remainder is 0. We will find that $\alpha =1$ . Similarly, other roots are $\beta =3$ and $\gamma =-2$ . Using these values, we will find the other equation.