Question

Solve ${{x}^{4}}+{{x}^{3}}-16{{x}^{2}}-4x+48=0$ given that the product of two roots is 6.

Answer 1

Hint: Here we have to find the roots of the polynomial equation. We know, the given equation is a polynomial equation, where the product of two roots is given. So we will use the relation of the roots with coefficients of the polynomial equation one by one to get the value of all the roots of the equation.

Complete step-by-step answer:
Here the given equation is a polynomial equation and it is also given that the product of two roots is 6.
Let the roots of the equation be $\alpha ,\beta ,\gamma \And \delta $
We have $\alpha \beta =6.........\left( a \right)$
We know the product of all four roots is equal to the ratio of constant term to the coefficient of${{x}^{4}}$.
Therefore,
$\Rightarrow$ $\alpha \beta \gamma \delta =48$
Now, we will put the value of$\alpha \beta $here
$\Rightarrow$ $6.\gamma \delta =48$
On further simplification, we get
$\gamma \delta =8$……………$\left( 1 \right)$
We will write all relations of the roots with the coefficient of the equation.
\[\begin{align}
  &\Rightarrow ~\alpha +\beta +\gamma +\delta =-1.............\left( 2 \right) \\
 &\Rightarrow \alpha \beta +\alpha \delta +\alpha \gamma +\beta \gamma +\beta \delta +\gamma \delta =-16~..............\left( 3 \right) \\
 &\Rightarrow \alpha \beta \gamma +\alpha \beta \delta +\alpha \gamma \delta +\beta \gamma \delta =4..............\left( 4 \right) \\
\end{align}\]
We will put the value of $\alpha \beta \And \gamma \delta $ in equation 3 now.
$\Rightarrow$ \[6+\alpha \delta +\alpha \gamma +\beta \gamma +\beta \delta +8=-16\]
On simplification, we get
$\Rightarrow$ \[\left( \alpha +\beta \right)\left( \gamma +\delta \right)=-30...........\left( 5 \right)\]
Let \[\left( \alpha +\beta \right)=t\]then \[\left( \gamma +\delta \right)=-1-t\]
On putting these values in equation 5, we get
$\Rightarrow$ \[t\left( 1+t \right)=30\]
Thus, the values of t are 5 and -6.
We will put the value of $\alpha \beta \And \gamma \delta $ in equation 4,
\[6\gamma +6\delta +8\alpha +8\beta =4.\]
We can write the equation as
$\Rightarrow$ \[6\left( \gamma +\delta \right)+8\left( \alpha +\beta \right)=4.\]
We know the value of \[\left( \alpha +\beta \right)=t\] and \[\left( \gamma +\delta \right)=-1-t\]
Therefore,
$\Rightarrow$ \[6\left( -1-t \right)+8t=4.\]
Only $t=5$ is satisfying the equation.
Therefore,
$\Rightarrow$ \[\left( \alpha +\beta \right)=5...........\left( 6 \right)\]
$\Rightarrow$ \[\left( \gamma +\delta \right)=-6.........\left( 7 \right)\]
On solving equation (a) and equation 6, we get
$\Rightarrow$ $\alpha =2\ and \beta =3$
On solving equation 1 and equation 7, we get
$\Rightarrow$ $\gamma =-2\ and \delta =-4$
Thus, the required roots are -4, -2, 2, 3.

Note: Here we have found the roots of the polynomial equation. Here, roots of a polynomial equation are defined as the value of the variable which when put in the equation makes the polynomial equation zero. We generally use the relation of roots with the coefficients of polynomial equations to calculate the roots.