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# Solve the following equation: ${x^4} + \dfrac{8}{9}{x^2} + 1 = 3{x^3} + 3x$ Verified
366k+ views
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.
Last updated date: 27th Sep 2023
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