ّFind $x$ such that $ \frac{1}{x^2} + \frac{1}{(3-x)^2} = \frac{104}{25}$.

Question

ّFind $x$ such that $$ \frac{1}{x^2} + \frac{1}{(3-x)^2} = \frac{104}{25}\,.$$

My attempt:

After clearing the denominators, I obtain this quartic equation $$104 x^{4} -624 x^{3} +886 x^{2} +150x-225=0.$$ I don't know how to proceed from here.

Answer 1

Let $u=x$ and $v=3-x$. Then we have $$ u+v = 3, \qquad \frac{1}{u^2} + \frac{1}{v^2} = \frac{104}{25} $$ But $$ \frac{1}{u^2} + \frac{1}{v^2} = \frac{u^2+v^2}{(uv)^2} = \frac{(u+v)^2-2uv}{(uv)^2} = \frac{9-2uv}{(uv)^2} $$ Thus, $uv$ is a root of $$ \frac{9-2z}{z^2} = \frac{104}{25} $$ a quadratic equation. Once you know $uv$ and $u+v$, you know $u$ and $v$ by solving another quadratic equation.

Answer 2

Now you use the rational root theorem, Horner's algorithm and a little work to find the roots of the polynomial.

The rational candidates for the root are $\frac{p}{q}$ where $p$ is a factor of $225$ (i.e., $3^2\cdot 5^2$) and $q$ is a factor of $104$ (i.e., $2^3\cdot 13$).

Answer 3

Let $y:=x-\dfrac{3}{2}$. The equation becomes $$\frac{1}{\left(y+\frac{3}{2}\right)^2}+\frac{1}{\left(y-\frac{3}{2}\right)^2}=\frac{104}{25}\,.$$ This is equivalent to $$\frac{y^2+\frac{9}{4}}{\left(y^2-\frac{9}{4}\right)^2}=\frac{52}{25}\,.$$ Let $z:=\dfrac{1}{y^2-\frac{9}{4}}$, we have $$\frac{9}{2}z^2+z=z^2\left(\frac{1}{z}+\frac{9}{2}\right)=\frac{52}{25}\,.$$ That is, $$\frac{9}{2}\left(z+\frac{4}{5}\right)\left(z-\frac{26}{45}\right)=0\,.$$ Thus, $z=-\dfrac45$ or $z=\dfrac{26}{45}$.

In the case $z=-\dfrac{4}{5}$, we have $$y^2-\frac{9}{4}=\frac{1}{z}=-\frac{5}{4}\,,$$ so $y^2=1$, or $y=\pm1$. In this case, $x=\dfrac{1}{2}$ or $x=\dfrac{5}{2}$.

In the case $z=\dfrac{26}{45}$, we have $$y^2-\frac{9}{4}=\frac{1}{z}=\frac{45}{26}\,.$$ That is, $y^2=\dfrac{207}{52}$, so $y=\pm\dfrac{3\sqrt{299}}{26}$. Hence, $$x=\frac{39\pm3\sqrt{299}}{26}\,.$$

Answer 4

One can see that $1/2$ is a solution of $\frac{1}{x^2} + \frac{1}{(3-x)^2} = \frac{104}{25}$. Then it is easy to see that also $5/2$ is a solution.

Can you proceed ?

Answer 5

Now, you can make the following.

Easy to see that $\frac{1}{2}$ and $\frac{5}{2}$ they are roots of the equation,

which gives a factor $$(2x-1)(2x-5)=4x^2-12x+5$$ and $$104x^4-624x^3+886x^2+150x-225=$$ $$=104x^2-312x^3+130x^2-312x^3+936x^2-390x-180x^2+540x-225=$$ $$=26x^2(4x^2-12x+5)-78(4x^2-12x+5)-45(4x^2-12x+5)=$$ $$=(4x^2-12x+5)(26x^2-78x-45),$$ which gives the answer: $$\left\{\frac{1}{2},\frac{5}{2},\frac{3}{2}\left(1+\sqrt{\frac{23}{13}}\right),\frac{3}{2}\left(1-\sqrt{\frac{23}{13}}\right)\right\}$$

Answer 6

Due to the symmetry at $x=3/2$, it makes sense to let $x=(3-u)/2$, which simplifies things initially to

$${1\over(3-u)^2}+{1\over(3+u)^2}={26\over25}$$

and then to

$${9+u^2\over(9-u^2)^2}={13\over25}$$

This can now be turned into a quadratic in $U=u^2$: $25(9+U)=13(9-U)^2$, or, once you do the algebra, $13U^2-259U+828=0$, which factors into

$$(U-4)(13U-207)=0$$

On noting that $207=9\cdot29$, we see from this that $u=\sqrt U=\pm2$ and $\pm3\sqrt{29/13}$, so that from $x=(3-u)/2$ we get

$$x=1/2,\quad 5/2,\quad3(1-\sqrt{29/13})/2,\quad\text{and}\quad3(1+\sqrt{29/13})/2$$

Remarks: Knowing that the symmetry between $x$ and $3-x$ makes a substitution worth trying is mostly a matter of experience. Figuring out exactly what substitution makes life easiest is a combination of experience and fiddling around. The factorization of the quadratic is easiest if you notice that $25(9+4)=13(9-4)^2$; if you don't, then the quadratic formula (and a calculator to compute $\sqrt{259^2-4\cdot13\cdot828}$) will do the job.