 Questions & Answers    Question Answers

# Solve the following equation: ${x^4} + \dfrac{8}{9}{x^2} + 1 = 3{x^3} + 3x$  Answer Verified
Hint- Replace the value of x as its inverse in the given equation and proceed with sum and product of roots.

Given equation is ${x^4} + \dfrac{8}{9}{x^2} + 1 = 3{x^3} + 3x \Rightarrow {x^4} + 3{x^3} + \dfrac{{8{x^2}}}{9} + 3x + 1 = 0{\text{ }} \to {\text{(1)}}$
Let us replace $x$ by $\dfrac{1}{x}$ in the given equation, we get
${\left( {\dfrac{1}{x}} \right)^4} + \dfrac{8}{9}{\left( {\dfrac{1}{x}} \right)^2} + 1 = 3{\left( {\dfrac{1}{x}} \right)^3} + \dfrac{3}{x} \Rightarrow \dfrac{1}{{{x^4}}} + \dfrac{8}{{9{x^2}}} + 1 = \dfrac{3}{{{x^3}}} + \dfrac{3}{x}$
Taking ${\text{9}}{x^4}$ as the LCM on the LHS and ${x^3}$ as the LCM on the RHS of the above equation
$\Rightarrow \dfrac{{9 + 8{x^2} + 9{x^4}}}{{9{x^4}}} = \dfrac{{3 + 3{x^2}}}{{{x^3}}} \Rightarrow \dfrac{{9 + 8{x^2} + 9{x^4}}}{{9x}} = 3 + 3{x^2} \Rightarrow \dfrac{{9 + 8{x^2} + 9{x^4}}}{9} = x\left( {3 + 3{x^2}} \right) \\ \Rightarrow 1 + \dfrac{{8{x^2}}}{9} + {x^4} = 3x + 3{x^3} \Rightarrow {x^4} + 3{x^3} + \dfrac{{8{x^2}}}{9} + 3x + 1 = 0 \\$
Clearly, the above equation which is obtained by replacing $x$ by $\dfrac{1}{x}$ in the given equation is the same as the given equation.
As, we know that in case of four degree polynomial (having two roots as $\alpha$, $\beta$) if $x$ is replaced by $\dfrac{1}{x}$ and the fourth degree polynomial comes out to be same as the previous one then the other two roots of that polynomial will be $\dfrac{1}{\alpha }$ and $\dfrac{1}{\beta }$.
Also, for any general fourth degree polynomial $a{x^4} + b{x^3} + c{x^2} + dx + e = 0$
Sum of all the roots$= - \dfrac{b}{a}$
Sum of product of different roots taken two at a time$= \dfrac{c}{a}$
According to the given equation (1), we can say $a = 1,{\text{ }}b = 3,{\text{ }}c = \dfrac{8}{9},{\text{ }}d = 3$ and $e = 1$.
Therefore, Sum of all the roots of the given equation (1) is given by $\alpha + \beta + \dfrac{1}{\alpha } + \dfrac{1}{\beta } = - \dfrac{3}{1} = - 3{\text{ }} \to {\text{(2)}}$
Also, sum of product of different roots taken two at a time of the given equation (1) is given by
$\alpha \beta {\text{ + }}\alpha {\text{.}}\dfrac{1}{\alpha }{\text{ + }}\alpha {\text{.}}\dfrac{1}{\beta }{\text{ + }}\beta {\text{.}}\dfrac{1}{\alpha }{\text{ + }}\beta {\text{.}}\dfrac{1}{\beta }{\text{ + }}\dfrac{1}{\alpha }{\text{.}}\dfrac{1}{\beta }{\text{ }} = \dfrac{{\dfrac{8}{9}}}{1} = \dfrac{8}{9}$
$\Rightarrow \alpha \beta {\text{ + }}1{\text{ + }}\alpha {\text{.}}\dfrac{1}{\beta }{\text{ + }}\beta {\text{.}}\dfrac{1}{\alpha }{\text{ + }}1{\text{ + }}\dfrac{1}{\alpha }{\text{.}}\dfrac{1}{\beta }{\text{ }} = \dfrac{8}{9} \Rightarrow \alpha \beta {\text{ + }}\alpha {\text{.}}\dfrac{1}{\beta }{\text{ + }}\beta {\text{.}}\dfrac{1}{\alpha }{\text{ + }}\dfrac{1}{\alpha }{\text{.}}\dfrac{1}{\beta } = \dfrac{8}{9} - 2 = - \dfrac{{10}}{9} \\ \Rightarrow \alpha \left( {\beta {\text{ + }}\dfrac{1}{\beta }} \right){\text{ + }}\dfrac{1}{\alpha }\left( {\beta {\text{ + }}\dfrac{1}{\beta }} \right) = - \dfrac{{10}}{9} \Rightarrow \left( {\alpha + \dfrac{1}{\alpha }} \right)\left( {\beta {\text{ + }}\dfrac{1}{\beta }} \right) = - \dfrac{{10}}{9}{\text{ }} \to {\text{(3)}} \\$
Equation (2) can be rearranged as $\left( {\alpha + \dfrac{1}{\alpha }} \right) = - 3 - \left( {\beta + \dfrac{1}{\beta }} \right)$
Put the value of $\left( {\alpha + \dfrac{1}{\alpha }} \right)$ in equation (3), we get
$\left[ { - 3 - \left( {\beta + \dfrac{1}{\beta }} \right)} \right]\left( {\beta {\text{ + }}\dfrac{1}{\beta }} \right) = - \dfrac{{10}}{9}$
Let $\left( {\beta + \dfrac{1}{\beta }} \right) = t$
$\Rightarrow \left[ { - 3 - t} \right]t = - \dfrac{{10}}{9} \Rightarrow - \left( {3 + t} \right)t = - \dfrac{{10}}{9} \Rightarrow \left( {3 + t} \right)t = \dfrac{{10}}{9} \Rightarrow 9{t^2} + 27t - 10 = 0 \\ \Rightarrow 9{t^2} + 27t - 10 = 0 \Rightarrow 9{t^2} - 3t + 30t - 10 = 0 \Rightarrow 3t\left( {3t - 1} \right) + 10\left( {3t - 1} \right) = 0 \\ \Rightarrow \left( {3t - 1} \right)\left( {3t + 10} \right) = 0 \\$
i.e., Either $3t - 1 = 0$ or $3t + 10 = 0$
$\Rightarrow t = \dfrac{1}{3} \Rightarrow \beta + \dfrac{1}{\beta } = \dfrac{1}{3} \Rightarrow \dfrac{{{\beta ^2} + 1}}{\beta } = \dfrac{1}{3} \Rightarrow 3{\beta ^2} + 3 = \beta \Rightarrow 3{\beta ^2} - \beta + 3 = 0 \Rightarrow \beta = \dfrac{{ - \left( { - 1} \right) \pm \sqrt {{{\left( { - 1} \right)}^2} - 4 \times 3 \times 3} }}{{2 \times 3}} \\ \Rightarrow \beta = \dfrac{{1 \pm \sqrt {1 - 36} }}{6} \Rightarrow \beta = \dfrac{{1 \pm \sqrt { - 35} }}{6} \Rightarrow \beta = \dfrac{{1 \pm i\sqrt {35} }}{6} \\$
or $t = - \dfrac{{10}}{3} \Rightarrow \beta + \dfrac{1}{\beta } = - \dfrac{{10}}{3} \Rightarrow \dfrac{{{\beta ^2} + 1}}{\beta } = - \dfrac{{10}}{3} \Rightarrow 3{\beta ^2} + 3 = - 10\beta \Rightarrow 3{\beta ^2} + 10\beta + 3 = 0 \Rightarrow 3{\beta ^2} + 9\beta + \beta + 3 = 0 \\ \Rightarrow 3\beta \left( {\beta + 3} \right) + 1\left( {\beta + 3} \right) = 0 \Rightarrow \left( {3\beta + 1} \right)\left( {\beta + 3} \right) = 0 \Rightarrow \beta = - \dfrac{1}{3}, - 3 \\$
Using equation (2) put the value of $\beta$, we will get the value of $\alpha$
$\Rightarrow \alpha = - \dfrac{1}{3}, - 3$ or $\alpha = \dfrac{{1 \pm i\sqrt {35} }}{6}$
Therefore, all the roots of the given equation are $- \dfrac{1}{3}, - 3,\dfrac{{1 \pm i\sqrt {35} }}{6}$.

Note- These types of problems can be solved by somehow checking for some properties regarding roots of a polynomial and then finding out an appropriate relation between the roots and hence solving further to get the values of these roots.
Bookmark added to your notes.
View Notes
Solve the Pair of Linear Equation  Roots of Polynomial Equation  How to Solve Linear Differential Equation?  Polynomial Equation Formula  ICSE Full Form  Polynomial for Class 10  Log Values From 1 to 10  Intercept Form of the Equation of a Plane  Equation of a Plane  Equation of a Line  