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

Last updated date: 15th Mar 2023

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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.

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.

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