# Solve the equation: $2{x^4} + {x^3} - 6{x^2} + x + 2 = 0$

Answer

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Hint- Here, we will be proceeding by diving the given equation by ${x^2}$ in order to convert this equation of degree four into a quadratic equation.

Given equation is $2{x^4} + {x^3} - 6{x^2} + x + 2 = 0$

Now let us the divide both sides by ${x^2}$, we get

$

\Rightarrow \dfrac{{2{x^4}}}{{{x^2}}} + \dfrac{{{x^3}}}{{{x^2}}} - \dfrac{{6{x^2}}}{{{x^2}}} + \dfrac{x}{{{x^2}}} + \dfrac{2}{{{x^2}}} = 0 \Rightarrow 2{x^2} + x - 6 + \dfrac{1}{x} + \dfrac{2}{{{x^2}}} = 0 \Rightarrow 2{x^2} + \dfrac{2}{{{x^2}}} + x + \dfrac{1}{x} - 6 = 0 \\

\Rightarrow 2\left( {{x^2} + \dfrac{1}{{{x^2}}}} \right) + \left( {x + \dfrac{1}{x}} \right) - 6 = 0 \\

$

Now, proceeding further by using method of completing the square

\[

\Rightarrow 2\left[ {\left( {{x^2} + \dfrac{1}{{{x^2}}} + 2\left( {{x^2}} \right)\left( {\dfrac{1}{{{x^2}}}} \right)} \right) - 2\left( {{x^2}} \right)\left( {\dfrac{1}{{{x^2}}}} \right)} \right] + \left( {x + \dfrac{1}{x}} \right) - 6 = 0 \Rightarrow 2\left[ {{{\left( {x + \dfrac{1}{x}} \right)}^2} - 2} \right] + \left( {x + \dfrac{1}{x}} \right) - 6 = 0 \\

\Rightarrow 2{\left( {x + \dfrac{1}{x}} \right)^2} - 4 + \left( {x + \dfrac{1}{x}} \right) - 6 = 0 \\

\Rightarrow 2{\left( {x + \dfrac{1}{x}} \right)^2} + \left( {x + \dfrac{1}{x}} \right) - 10 = 0 \\

\]

Put \[\left( {x + \dfrac{1}{x}} \right) = t\] in the above equation, we get

\[

2{t^2} + t - 10 = 0 \Rightarrow 2{t^2} - 4t + 5t - 10 = 0 \Rightarrow 2t\left( {t - 2} \right) + 5\left( {t - 2} \right) = 0 \Rightarrow \left( {t - 2} \right)\left( {2t + 5} \right) = 0 \\

\\

\]

Either $t = 2$ or $t = - \dfrac{5}{2}$

For $t = 2$ $ \Rightarrow \left( {x + \dfrac{1}{x}} \right) = t = 2 \Rightarrow \left( {\dfrac{{{x^2} + 1}}{x}} \right) = 2 \Rightarrow {x^2} + 1 = 2x \Rightarrow {x^2} - 2x + 1 = 0 \Rightarrow {\left( {x - 1} \right)^2} = 0 \Rightarrow x = 1$

Corresponding to $t = 2$, there exists two equal roots $x = 1$ of the given equation.

For $t = - \dfrac{5}{2}$ $

\Rightarrow \left( {x + \dfrac{1}{x}} \right) = t = - \dfrac{5}{2} \Rightarrow \left( {\dfrac{{{x^2} + 1}}{x}} \right) = - \dfrac{5}{2} \Rightarrow 2\left( {{x^2} + 1} \right) = - 5x \Rightarrow 2{x^2} + 5x + 2 = 0 \Rightarrow 2{x^2} + 4x + x + 2 = 0 \\

\Rightarrow 2x\left( {x + 2} \right) + 1\left( {x + 2} \right) = 0 \Rightarrow \left( {x + 2} \right)\left( {2x + 1} \right) = 0 \\

$

Either $ \Rightarrow x = - 2$ or $x = - \dfrac{1}{2}$

Corresponding to $t = - \dfrac{5}{2}$, there exists two roots of the given equation which are $x = - 2$, $x = - \dfrac{1}{2}$.

Therefore, all the four roots of the given equation are $x = 1,1, - 2, - \dfrac{1}{2}$.

Note- The given equation consists of a polynomial of degree four so ultimately it will be resulting in a total four roots for the given equation.

Given equation is $2{x^4} + {x^3} - 6{x^2} + x + 2 = 0$

Now let us the divide both sides by ${x^2}$, we get

$

\Rightarrow \dfrac{{2{x^4}}}{{{x^2}}} + \dfrac{{{x^3}}}{{{x^2}}} - \dfrac{{6{x^2}}}{{{x^2}}} + \dfrac{x}{{{x^2}}} + \dfrac{2}{{{x^2}}} = 0 \Rightarrow 2{x^2} + x - 6 + \dfrac{1}{x} + \dfrac{2}{{{x^2}}} = 0 \Rightarrow 2{x^2} + \dfrac{2}{{{x^2}}} + x + \dfrac{1}{x} - 6 = 0 \\

\Rightarrow 2\left( {{x^2} + \dfrac{1}{{{x^2}}}} \right) + \left( {x + \dfrac{1}{x}} \right) - 6 = 0 \\

$

Now, proceeding further by using method of completing the square

\[

\Rightarrow 2\left[ {\left( {{x^2} + \dfrac{1}{{{x^2}}} + 2\left( {{x^2}} \right)\left( {\dfrac{1}{{{x^2}}}} \right)} \right) - 2\left( {{x^2}} \right)\left( {\dfrac{1}{{{x^2}}}} \right)} \right] + \left( {x + \dfrac{1}{x}} \right) - 6 = 0 \Rightarrow 2\left[ {{{\left( {x + \dfrac{1}{x}} \right)}^2} - 2} \right] + \left( {x + \dfrac{1}{x}} \right) - 6 = 0 \\

\Rightarrow 2{\left( {x + \dfrac{1}{x}} \right)^2} - 4 + \left( {x + \dfrac{1}{x}} \right) - 6 = 0 \\

\Rightarrow 2{\left( {x + \dfrac{1}{x}} \right)^2} + \left( {x + \dfrac{1}{x}} \right) - 10 = 0 \\

\]

Put \[\left( {x + \dfrac{1}{x}} \right) = t\] in the above equation, we get

\[

2{t^2} + t - 10 = 0 \Rightarrow 2{t^2} - 4t + 5t - 10 = 0 \Rightarrow 2t\left( {t - 2} \right) + 5\left( {t - 2} \right) = 0 \Rightarrow \left( {t - 2} \right)\left( {2t + 5} \right) = 0 \\

\\

\]

Either $t = 2$ or $t = - \dfrac{5}{2}$

For $t = 2$ $ \Rightarrow \left( {x + \dfrac{1}{x}} \right) = t = 2 \Rightarrow \left( {\dfrac{{{x^2} + 1}}{x}} \right) = 2 \Rightarrow {x^2} + 1 = 2x \Rightarrow {x^2} - 2x + 1 = 0 \Rightarrow {\left( {x - 1} \right)^2} = 0 \Rightarrow x = 1$

Corresponding to $t = 2$, there exists two equal roots $x = 1$ of the given equation.

For $t = - \dfrac{5}{2}$ $

\Rightarrow \left( {x + \dfrac{1}{x}} \right) = t = - \dfrac{5}{2} \Rightarrow \left( {\dfrac{{{x^2} + 1}}{x}} \right) = - \dfrac{5}{2} \Rightarrow 2\left( {{x^2} + 1} \right) = - 5x \Rightarrow 2{x^2} + 5x + 2 = 0 \Rightarrow 2{x^2} + 4x + x + 2 = 0 \\

\Rightarrow 2x\left( {x + 2} \right) + 1\left( {x + 2} \right) = 0 \Rightarrow \left( {x + 2} \right)\left( {2x + 1} \right) = 0 \\

$

Either $ \Rightarrow x = - 2$ or $x = - \dfrac{1}{2}$

Corresponding to $t = - \dfrac{5}{2}$, there exists two roots of the given equation which are $x = - 2$, $x = - \dfrac{1}{2}$.

Therefore, all the four roots of the given equation are $x = 1,1, - 2, - \dfrac{1}{2}$.

Note- The given equation consists of a polynomial of degree four so ultimately it will be resulting in a total four roots for the given equation.

Last updated date: 28th May 2023

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