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# Solve the following equation: ${x^4} + 1 - 3\left( {{x^3} + x} \right) = 2{x^2}$.

Last updated date: 13th Jul 2024
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Hint: Here, we will proceed by replacing $x$ by $\dfrac{1}{x}$in the given equation in order to prove that it will come out as the same.

Given, equation is ${x^4} + 1 - 3\left( {{x^3} + x} \right) = 2{x^2} \Rightarrow {x^4} - 3{x^3} - 2{x^2} - 3x + 1 = 0{\text{ }} \to {\text{(1)}}$
Let us replace $x$ by $\dfrac{1}{x}$ in equation (1), we get
$\Rightarrow {\left( {\dfrac{1}{x}} \right)^4} - 3{\left( {\dfrac{1}{x}} \right)^3} - 2{\left( {\dfrac{1}{x}} \right)^2} - \dfrac{3}{x} + 1 = 0{\text{ }} \Rightarrow \dfrac{1}{{{x^4}}} - \dfrac{3}{{{x^3}}} - \dfrac{2}{{{x^2}}} - \dfrac{3}{x} + 1 = 0{\text{ }} \Rightarrow \dfrac{{1 - 3x - 2{x^2} - 3{x^3} + {x^4}}}{{{x^4}}} = 0 \\ \Rightarrow {x^4} - 3{x^3} - 2{x^2} - 3x + 1 = 0 \\$
Clearly, the above equation which is obtained by replacing $x$ by $\dfrac{1}{x}$in the given equation (1) is the same as the given equation (1).
As, we know that when in a polynomial of degree four (having two roots as $\alpha$ and $\beta$ ) if $x$ is replaced by $\dfrac{1}{x}$ and the polynomial comes out to be same as previous one then the other two roots of that polynomial will be $\dfrac{1}{\alpha }$ and $\dfrac{1}{\beta }$.
Also, for any general polynomial of degree four $a{x^4} + b{x^3} + c{x^2} + dx + e = 0$
${\text{Sum of all the roots}} = - \dfrac{b}{a} \\ {\text{Sum of product of different roots taken two at a time }} = \dfrac{c}{a} \\$
According to the given equation (1), we can say $a = 1$,$b = - 3$,$c = - 2$,$d = - 3$and $e = 1$.
Therefore, Sum of all the roots of the given equation (1) is given by $\alpha + \beta + \dfrac{1}{\alpha } + \dfrac{1}{\beta } = - \dfrac{{\left( { - 3} \right)}}{1} = 3{\text{ }} \to {\text{(2)}}$
Sum of product of different roots taken two at a time of the given equation (1) is given by
$\alpha \beta {\text{ + }}\alpha {\text{.}}\dfrac{1}{\alpha }{\text{ + }}\alpha {\text{.}}\dfrac{1}{\beta }{\text{ + }}\beta {\text{.}}\dfrac{1}{\alpha }{\text{ + }}\beta {\text{.}}\dfrac{1}{\beta }{\text{ + }}\dfrac{1}{\alpha }{\text{.}}\dfrac{1}{\beta }{\text{ }} = \dfrac{{\left( { - 2} \right)}}{1} = - 2 \\ \Rightarrow \alpha \beta {\text{ + }}1{\text{ + }}\alpha {\text{.}}\dfrac{1}{\beta }{\text{ + }}\beta {\text{.}}\dfrac{1}{\alpha }{\text{ + }}1{\text{ + }}\dfrac{1}{\alpha }{\text{.}}\dfrac{1}{\beta }{\text{ }} = - 2 \Rightarrow \alpha \beta {\text{ + }}\alpha {\text{.}}\dfrac{1}{\beta }{\text{ + }}\beta {\text{.}}\dfrac{1}{\alpha }{\text{ + }}\dfrac{1}{\alpha }{\text{.}}\dfrac{1}{\beta } = - 2 - 2 = - 4 \\$
$\Rightarrow \alpha \left( {\beta {\text{ + }}\dfrac{1}{\beta }} \right){\text{ + }}\dfrac{1}{\alpha }\left( {\beta {\text{ + }}\dfrac{1}{\beta }} \right) = - 4 \Rightarrow \left( {\alpha + \dfrac{1}{\alpha }} \right)\left( {\beta {\text{ + }}\dfrac{1}{\beta }} \right) = - 4{\text{ }} \to {\text{(3)}}$
Now, equation (2) can be re-arranged as $\left( {\alpha + \dfrac{1}{\alpha }} \right) = 3 - \left( {\beta + \dfrac{1}{\beta }} \right)$
Put the value of $\left( {\alpha + \dfrac{1}{\alpha }} \right){\text{ }}$in equation (3), we get
$\left[ {3 - \left( {\beta + \dfrac{1}{\beta }} \right)} \right]\left( {\beta {\text{ + }}\dfrac{1}{\beta }} \right) = - 4$
Let $\left( {\beta + \dfrac{1}{\beta }} \right) = t$
$\Rightarrow \left[ {3 - t} \right]t = - 4 \Rightarrow {t^2} - 3t - 4 = 0 \Rightarrow {t^2} + t - 4t - 4 = 0 \Rightarrow t\left( {t + 1} \right) - 4\left( {t + 1} \right) = 0 \\ \Rightarrow \left( {t + 1} \right)\left( {t - 4} \right) = 0 \\$
$\Rightarrow t = - 1$ or $t = 4$
$\beta + \dfrac{1}{\beta } = - 1 \Rightarrow \dfrac{{{\beta ^2} + 1}}{\beta } = - 1 \Rightarrow {\beta ^2} + \beta + 1 = 0 \Rightarrow \beta = \dfrac{{ - \left( 1 \right) \pm \sqrt {{{\left( 1 \right)}^2} - 4 \times 1 \times 1} }}{{2 \times 1}} = \dfrac{{ - 1 \pm \sqrt {1 - 4} }}{2} = \dfrac{{ - 1 \pm i\sqrt 3 }}{2}$ or $\beta + \dfrac{1}{\beta } = 4 \Rightarrow \dfrac{{{\beta ^2} + 1}}{\beta } = 4 \Rightarrow {\beta ^2} - 4\beta + 1 = 0 \Rightarrow \beta = \dfrac{{ - \left( { - 4} \right) \pm \sqrt {{{\left( { - 4} \right)}^2} - 4 \times 1 \times 1} }}{{2 \times 1}} = \dfrac{{4 \pm \sqrt {16 - 4} }}{2} = \dfrac{{4 \pm 2\sqrt 3 }}{2} = 2 \pm \sqrt 3$
Hence, $\beta = \dfrac{{ - 1 \pm i\sqrt 3 }}{2}$or $\beta = 2 \pm \sqrt 3$
Using equation (2) put the value of $\beta$, we will get the value for $\alpha$
$\alpha = 2 \pm \sqrt 3$or $\alpha = \dfrac{{ - 1 \pm i\sqrt 3 }}{2}$
Therefore, all the roots of the given are $2 \pm \sqrt 3$, $\dfrac{{ - 1 \pm i\sqrt 3 }}{2}$.

Note: These types of problems are 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 them.