Find sum of the roots of $\frac{1}{x^2}+\frac{1}{(2-x)^2}=\frac{40}{9}$

Question

Find the sum of the roots of $$\frac{1}{x^2}+\frac{1}{(2-x)^2}=\frac{40}{9}$$ my attempt:

$\dfrac{1}{x^2}+\dfrac{1}{(2-x)^2}=\left[\dfrac{1}{x}-\dfrac{1}{(x-2)}\right]^2\!\!+\dfrac{2}{x(x-2)}=\dfrac{40}{9}$

Now let $\frac{1}{x(x-2)}=t$, so we have $$4t^2+2t=\frac{40}{9}$$ If we consider $\alpha, \beta$ to be the root of this equation, we have $$x^2-2x=\frac{1}{\alpha}$$ from here sum of the root is $2$ and $$x^2-2x=\frac{1}{\beta}$$ from here sum of the root is $2.$

Thus the sum of the roots of $$\frac{1}{x^2}+\frac{1}{(2-x)^2}=\frac{40}{9}$$ is $4$

I don't know if this approach is ok. Any help is appreciated.

Answer 1

Clearly by symmetry, if $a$ is a root of the given equation, so is $2-a$

So, the sum of roots will be $a_1+2-a_1+a_2+2-a_2=?$

Answer 2

Make your problem symmetric by letting

$$u=1-x$$

In this way, your expression is transformed into :

$$\frac{1}{(1-u)^2}+\frac{1}{(1+u)^2}=\frac{40}{9}$$

Reducing the LHS to the common denominator $(1-u^2)^2$, you get equation :

$$20 u^4-49 u^2 +11=0\tag{1}$$

As the coefficient of $u^3$ is zero, according to one of the Viète's formulas, the sum of the roots of (1) is zero :

$$u_1+u_2+u_3+u_4=0 \iff (1-x_1)+(1-x_2)+(1-x_3)+(1-x_4)=0$$

Otherwise said, the answer is

$$x_1+x_2+x_3+x_4=4$$

Answer 3

$$\frac{1}{x^2}+\frac{1}{(2-x)^2}=\frac{40}{9}$$

Multiplying by $x^2(2-x)^2$ and expanding:

$$x^4 - 4 x^3 + \frac{71}{20} x^2 + \frac{9}{10} x - \frac{9}{10}=0$$

$(x-z_1)(x-z_2)(x-z_3)(x-z_4)=x^4-(z_1+z_2+z_3+z_4)x^3+...$

$$-(z_1+z_2+z_3+z_4)x^3=-4x^3$$ $$z_1+z_2+z_3+z_4=4$$

So the sum of the roots is $4$

These are known results from Vieta's formulas