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

Solve the following equation: x4+1−3(x3+x)=2x2{x^4} + 1 - 3\left( {{x^3} + x} \right) = 2{x^2}.

Explanation

Solution

Hint: Here, we will proceed by replacing xx by 1x\dfrac{1}{x}in the given equation in order to prove that it will come out as the same.

Given, equation is x4+1−3(x3+x)=2x2⇒x4−3x3−2x2−3x+1=0 →(1){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 xx by 1x\dfrac{1}{x} in equation (1), we get
⇒(1x)4−3(1x)3−2(1x)2−3x+1=0 ⇒1x4−3x3−2x2−3x+1=0 ⇒1−3x−2x2−3x3+x4x4=0 ⇒x4−3x3−2x2−3x+1=0  \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 xx by 1x\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 α\alpha and β\beta ) if xx is replaced by 1x\dfrac{1}{x} and the polynomial comes out to be same as previous one then the other two roots of that polynomial will be 1α\dfrac{1}{\alpha } and 1β\dfrac{1}{\beta }.
Also, for any general polynomial of degree four ax4+bx3+cx2+dx+e=0a{x^4} + b{x^3} + c{x^2} + dx + e = 0

Sum of all the roots=−ba Sum of product of different roots taken two at a time =ca  {\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=1a = 1,b=−3b = - 3,c=−2c = - 2,d=−3d = - 3and e=1e = 1.
Therefore, Sum of all the roots of the given equation (1) is given by α+β+1α+1β=−(−3)1=3 →(2)\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

αβ + α.1α + α.1β + β.1α + β.1β + 1α.1β =(−2)1=−2 ⇒αβ + 1 + α.1β + β.1α + 1 + 1α.1β =−2⇒αβ + α.1β + β.1α + 1α.1β=−2−2=−4  \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 \\\

⇒α(β + 1β) + 1α(β + 1β)=−4⇒(α+1α)(β + 1β)=−4 →(3) \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 (α+1α)=3−(β+1β)\left( {\alpha + \dfrac{1}{\alpha }} \right) = 3 - \left( {\beta + \dfrac{1}{\beta }} \right)
Put the value of (α+1α) \left( {\alpha + \dfrac{1}{\alpha }} \right){\text{ }}in equation (3), we get
[3−(β+1β)](β + 1β)=−4\left[ {3 - \left( {\beta + \dfrac{1}{\beta }} \right)} \right]\left( {\beta {\text{ + }}\dfrac{1}{\beta }} \right) = - 4
Let (β+1β)=t\left( {\beta + \dfrac{1}{\beta }} \right) = t

⇒[3−t]t=−4⇒t2−3t−4=0⇒t2+t−4t−4=0⇒t(t+1)−4(t+1)=0 ⇒(t+1)(t−4)=0  \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 \\\

⇒t=−1 \Rightarrow t = - 1 or t=4t = 4
β+1β=−1⇒β2+1β=−1⇒β2+β+1=0⇒β=−(1)±(1)2−4×1×12×1=−1±1−42=−1±i32\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 β+1β=4⇒β2+1β=4⇒β2−4β+1=0⇒β=−(−4)±(−4)2−4×1×12×1=4±16−42=4±232=2±3\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, β=−1±i32\beta = \dfrac{{ - 1 \pm i\sqrt 3 }}{2}or β=2±3\beta = 2 \pm \sqrt 3
Using equation (2) put the value of β\beta , we will get the value for α\alpha
α=2±3\alpha = 2 \pm \sqrt 3 or α=−1±i32\alpha = \dfrac{{ - 1 \pm i\sqrt 3 }}{2}
Therefore, all the roots of the given are 2±32 \pm \sqrt 3 , −1±i32\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.