Quadratic substitution question: applying substitution $p=x+\frac1x$ to $2x^4+x^3-6x^2+x+2=0$

Like this,

as the equation is Reciprocal Equation of the First type with $x\ne0,$

divide either sides by $\displaystyle x^{\frac42}=x^2$ to reduce the degree of the equation by half

$$2x^2+x-6+\frac1x+\frac2{x^2}=0$$

$$\implies 2\left(x^2+\frac1{x^2}\right)+\left(x+\frac1x\right)-6=0$$

$$\implies 2\{\left(x+\frac1x\right)^2-2\}+\left(x+\frac1x\right)-6=0$$

Reference : Reciprocal Equation is explained here:

Chapter XI of Higher Algebra,Barnard & Child and

Article $568−570$ of Higher algebra, Hall & Knight


Hint: Try working backwards. Start with: $$ 2\left(x + \dfrac{1}{x}\right)^2 + \left(x + \dfrac{1}{x}\right) - 10 = 0 $$ then try to obtain the original equation by expanding then clearing the denominators.

Tags:

Algebra Precalculus

Quadratics

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