Question

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

Answer 1

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

Complete step-by-step solution -
To start with, we are given that the roots of the equation ${{x}^{3}}-7x+6=0$ are $\alpha ,\beta $ and $\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:
$\begin{align}
  & \alpha +\beta +\gamma =\dfrac{-\text{coefficient of }{{x}^{2}}}{\text{coefficient of }{{x}^{3}}} \\
 & \Rightarrow \alpha +\beta +\gamma =\dfrac{-0}{1} \\
 & \Rightarrow \alpha +\beta +\gamma =0..............\left( i \right) \\
\end{align}$
Similarly, the sum of multiplication of two roots is given by
$\begin{align}
  & \alpha \beta +\beta \gamma +\gamma \alpha =\dfrac{\text{coefficient of }x}{\text{coefficient of }{{x}^{3}}} \\
 & \Rightarrow \alpha \beta +\beta \gamma +\gamma \alpha =\dfrac{-7}{1} \\
 & \Rightarrow \alpha \beta +\beta \gamma +\gamma \alpha =-7..............\left( ii \right) \\
\end{align}$
Similarly, the multiplication of the roots is given by:
$\begin{align}
  & \Rightarrow \alpha \beta \gamma =\dfrac{-\text{constant term}}{\text{coefficient of }{{x}^{3}}} \\
 & \Rightarrow \alpha \beta \gamma =\dfrac{-6}{1} \\
 & \Rightarrow \alpha \beta \gamma =-6...............\left( iii \right) \\
\end{align}$
Now, we are going to find the equation which has ${{\left( \alpha +\beta \right)}^{2}},{{\left( \beta +\gamma \right)}^{2}}$ and ${{\left( \gamma +\alpha \right)}^{2}}$ . Let us assume that the cubic equation will be
${{x}^{3}}+b{{x}^{2}}+cx+d=0$
We know that sum of the roots of the above equation is given as:
${{\left( \alpha +\beta \right)}^{2}}+{{\left( \beta +\gamma \right)}^{2}}+{{\left( \gamma +\alpha \right)}^{2}}=-b$
Now, using the identity ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ , we will expand this as:
$\begin{align}
  & {{\alpha }^{2}}+{{\beta }^{2}}+2\alpha \beta +{{\beta }^{2}}+{{\gamma }^{2}}+2\beta \gamma +{{\gamma }^{2}}+{{\alpha }^{2}}+2\gamma \alpha =-b \\
 & \Rightarrow 2\left( {{\alpha }^{2}}+{{\beta }^{2}}+{{\gamma }^{2}} \right)+2\left( \alpha \beta +\beta \gamma +\gamma \alpha \right)=-b \\
\end{align}$
Now, we will add and subtract $2\left( \alpha \beta +\beta \gamma +\gamma \alpha \right)$ on both sides. After doing this, we will get:
\[\begin{align}
  & \Rightarrow 2\left( {{\alpha }^{2}}+{{\beta }^{2}}+{{\gamma }^{2}} \right)+2\left( \alpha \beta +\beta \gamma +\gamma \alpha \right)+2\left( \alpha \beta +\beta \gamma +\gamma \alpha \right)-2\left( \alpha \beta +\beta \gamma +\gamma \alpha \right)=-b \\
 & \Rightarrow 2\left[ {{\alpha }^{2}}+{{\beta }^{2}}+{{\gamma }^{2}}+2\alpha \beta +2\beta \gamma +2\gamma \alpha \right]-2\left( \alpha \beta +\beta \gamma +\gamma \alpha \right)=-b \\
\end{align}\]
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
$\begin{align}
  & \Rightarrow 2{{\left( 0 \right)}^{2}}-2\left( -7 \right)=-b \\
 & \Rightarrow 14=-b \\
 & \Rightarrow b=-14 \\
\end{align}$
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
\[\begin{align}
  & \alpha +\beta +\gamma =0 \\
 & \Rightarrow \beta +\gamma =-\alpha ..........\left( vi \right) \\
 & \Rightarrow \alpha +\gamma =-\beta ...........\left( vii \right) \\
 & \Rightarrow \alpha +\beta =-\gamma ............\left( viii \right) \\
\end{align}\]
Now using the values from equations (vi), (vii) and (viii) in equation (v), we get:
$\begin{align}
  & {{\left( -\gamma \right)}^{2}}{{\left( -\alpha \right)}^{2}}+{{\left( -\alpha \right)}^{2}}{{\left( -\beta \right)}^{2}}+{{\left( -\beta \right)}^{2}}{{\left( -\gamma \right)}^{2}}=c \\
 & {{\gamma }^{2}}{{\alpha }^{2}}+{{\alpha }^{2}}{{\beta }^{2}}+{{\beta }^{2}}{{\gamma }^{2}}=c..................\left( ix \right) \\
\end{align}$
From equation (ii), we have
$\alpha \beta +\beta \gamma +\gamma \alpha =-7$
On squaring both sides, we get:
$\begin{align}
  & {{\left( \alpha \beta +\beta \gamma +\gamma \alpha \right)}^{2}}={{\left( -7 \right)}^{2}} \\
 & \Rightarrow {{\alpha }^{2}}{{\beta }^{2}}+{{\beta }^{2}}{{\gamma }^{2}}+{{\gamma }^{2}}{{\alpha }^{2}}+2\alpha {{\beta }^{2}}\gamma +2\alpha \beta {{\gamma }^{2}}+2{{\alpha }^{2}}\beta \gamma =49 \\
 & \Rightarrow {{\alpha }^{2}}{{\beta }^{2}}+{{\beta }^{2}}{{\gamma }^{2}}+{{\gamma }^{2}}{{\alpha }^{2}}+2\alpha \beta \gamma \left( \alpha +\beta +\gamma \right)=49.................\left( x \right) \\
\end{align}$
From equation (i), (iii) and (x) we get
$\begin{align}
  & \Rightarrow {{\alpha }^{2}}{{\beta }^{2}}+{{\beta }^{2}}{{\gamma }^{2}}+{{\gamma }^{2}}{{\alpha }^{2}}+0=49 \\
 & \Rightarrow {{\alpha }^{2}}{{\beta }^{2}}+{{\beta }^{2}}{{\gamma }^{2}}+{{\gamma }^{2}}{{\alpha }^{2}}=49 \\
\end{align}$
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
$\begin{align}
  & \Rightarrow {{\left( -\gamma \right)}^{2}}{{\left( -\alpha \right)}^{2}}{{\left( -\gamma \right)}^{2}}=-d \\
 & \Rightarrow {{\alpha }^{2}}{{\beta }^{2}}{{\gamma }^{2}}=-d \\
 & \Rightarrow {{\left( \alpha \beta \gamma \right)}^{2}}=-d \\
 & \Rightarrow d=-{{\left( \alpha \beta \gamma \right)}^{2}} \\
 & \Rightarrow d=-{{\left( -6 \right)}^{2}} \\
 & \Rightarrow d=-36 \\
\end{align}$
So, the equation required is
${{x}^{3}}-14{{x}^{2}}+49x-36=0$
On simplifying this, we get
$\begin{align}
  & \Rightarrow x\left( {{x}^{2}}-14x+49 \right)=36 \\
 & \Rightarrow x{{\left( x-7 \right)}^{2}}=36\text{ }\therefore \text{Using identity }{{a}^{2}}+{{b}^{2}}+2ab={{\left( a+b \right)}^{2}} \\
\end{align}$
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.