#### Question

Solve the equation

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

Collected in the board: Algebraic equation

Steven Zheng posted 3 months ago

Let

a=\dfrac{1}{x}
(1)
b=\dfrac{1}{x+2}
(2)

Substitute to original equation to get rid of fraction. We get a new equation.

a^2+b^2=\dfrac{10}{9}
(3)

In the meantime, take reciprocal of (1) and (2)

\dfrac{1}{a}=x

\dfrac{1}{b}=x+2

Subtraction of them gives

\dfrac{1}{a}-\dfrac{1}{b} = -2

Simplifying the equation gives

a-b=2ab
(4)

From (3), the following equation is valid after subtraction of 2ab from both sides

a^2+b^2-2ab=\dfrac{10}{9} -2ab

(a-b)^2 = \dfrac{10}{9} -2ab

Replacing ab with (4), we get a quadratic equation in terms of a-b

(a-b)^2 = \dfrac{10}{9} -(a-b)

Let p = a-b. The equation is simplified as

p^2+p- \dfrac{10}{9}=0

Solving the quadratic equation, we get two solutions

p_1=-\dfrac{5}{3}, p_2=\dfrac{2}{3}

When p = -\dfrac{5}{3}, let's add 2ab in both sides of (3)

a^2+b^2+2ab=\dfrac{10}{9} +2ab

(a+b)^2=\dfrac{10}{9} +2ab

Replace ab with the result of (4)

(a+b)^2=\dfrac{10}{9} +(a-b)
(5)

Since a-b=p= -\dfrac{5}{3}

(a+b)^2=\dfrac{10}{9} -\dfrac{5}{3}=-\dfrac{5}{9}<0

which is impossible. Therefore, p = -\dfrac{5}{3} is cancelled

When p = \dfrac{2}{3}, that is a-b = \dfrac{2}{3} substitute to (5)

(a+b)^2=\dfrac{10}{9} + \dfrac{2}{3} =\dfrac{16}{9}

Taking square root of both sides gives

a+b = \pm\dfrac{4}{3}

Plus a-b = \dfrac{2}{3}, we get two equation systems.

\begin{cases} a + b =\dfrac{4}{3} \\ a - b = \dfrac{2}{3} \end{cases}

and

\begin{cases} a + b =-\dfrac{4}{3} \\ a - b = \dfrac{2}{3} \end{cases}

Solving the first equation system, we get a = 1, and then x = 1

Solving the second equation system, we get a = -\dfrac{1}{3} , and then x =-3

The graph of y=\dfrac{1}{x^2}+\dfrac{1}{(x+2)^2}-\dfrac{10}{9} illustrates the points when y =0

Steven Zheng posted 3 months ago

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