﻿ Solve \Big( \dfrac{x}{x-1} \Big) ^2+\Big( \dfrac{x}{x+1} \Big) ^2=\dfrac{10}{9}

Question

Solve \Big( \dfrac{x}{x-1} \Big) ^2+\Big( \dfrac{x}{x+1} \Big) ^2=\dfrac{10}{9}

Collected in the board: Quartic Equations

Steven Zheng posted 1 week ago

Using the identity a^2+b^2 = (a+b)^2-2ab to construct perfect square in the LHS

\Big( \dfrac{x}{x-1} \Big) ^2+\Big( \dfrac{x}{x+1} \Big) ^2

=\Big( \dfrac{x}{x-1}+ \dfrac{x}{x+1}\Big) ^2-2 \dfrac{x}{x-1}\dfrac{x}{x+1}

=\Big( \dfrac{4x^2}{(x^2-1)^2}\Big) ^2 -\dfrac{2x^2}{x^2-1}=\dfrac{10}{9}

Let

u =\dfrac{2x^2}{x^2-1}
(1)

The equation is simplified as

u^2-u-\dfrac{10}{9} =0

Multiply each term by 9

9u^2-9u-10=0

The root of teh quadratic equation could be determined by the formula

u = \dfrac{9\pm\sqrt{9^2+4\cdot 9\cdot 10}}{2\cdot 9}

=\dfrac{1}{2}\pm\dfrac{7}{6}

Therefore,

\begin{cases} u_1 = \dfrac{5}{3} \\ u_2 = -\dfrac{2}{3} \end{cases}

Substitute to (1), then we have 2 quadratic equations.

\dfrac{2x^2}{x^2-1}=\dfrac{5}{3}
(2)

and

\dfrac{2x^2}{x^2-1}=-\dfrac{2}{3}
(3)

Solving the equation (2)

x^2=-5

x_{1,2}= \pm\sqrt{5}i

Solving the equation (3)

6x^2 = -2x^2+2

x_{3,4} = \pm \dfrac{1}{2}

Steven Zheng posted 1 week ago

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