## Wednesday, March 2, 2016

### Solve for real solutions of the system below: [MATH]\small \frac{1}{x-1}+\frac{2}{x-2}+\frac{6}{x-6}+\frac{7}{x-7}=x^2-4x-4[/MATH]

Solve for real solutions of the system below:

[MATH]\frac{1}{x-1}+\frac{2}{x-2}+\frac{6}{x-6}+\frac{7}{x-7}=x^2-4x-4[/MATH]

My solution:

First of all, notice that if one wants to solve the given system by clearing the fraction on the left, one would definitely end up with having to solve the polynomial of degree 6, which is a giant PITA!

We therefore should plan a strategy to brilliantly simplify the given system as we go. Yes, hear hear, talk is cheap! Of course! But, we're not asking you to think of the way to go to the moon without riding a rocket! :D We're only asking you to observe and use what is already been given wisely to simplify things, and we assure you that if you have the patient, and is inclined, and is passionate and wanted to learn more to become a better problem solver, you would eagerly wanted to read the rest of this blog post or even try out the problem on paper your own, and then checking the answer with our when you have finished.

It is important to recognize how useful the $1$ can be, as it can take any of the quotient forms below, as long as he quotient has the same numerator and denominator:

[MATH]1=\frac{a}{a}[/MATH]

[MATH]1=\frac{a+b+c}{a+b+c}[/MATH]

[MATH]1=\frac{x-2}{x-2}[/MATH]

Thus, the given equality can be rewritten, if we're to add 4 on both sides of the equation:

[MATH]\frac{1}{x-1}+\frac{2}{x-2}+\frac{6}{x-6}+\frac{7}{x-7}+4=x^2-4x-4+4[/MATH]

We will distribute the 4 equally to the four fractions on the left, that is, each fraction gets a $1$:

[MATH]\frac{1}{x-1}+1+\frac{2}{x-2}+1+\frac{6}{x-6}+1+\frac{7}{x-7}+1=x(x-4)[/MATH]

Now, we transform the $1$ on the left as [MATH]\frac{x-1}{x-1},\,\frac{x-2}{x-2},\,\frac{x-6}{x-6},\,\frac{x-7}{x-7}[/MATH] respectively and get:

[MATH]\frac{1}{x-1}+\frac{x-1}{x-1}+\frac{2}{x-2}+\frac{x-2}{x-2}+\frac{6}{x-6}+\frac{x-6}{x-6}+\frac{7}{x-7}+\frac{x-7}{x-7}=x(x-4)[/MATH]

[MATH]\frac{1+x-1}{x-1}+\frac{2+x-2}{x-2}+\frac{6+x-6}{x-6}+\frac{7+x-7}{x-7}=x(x-4)[/MATH]

[MATH]\frac{x}{x-1}+\frac{x}{x-2}+\frac{x}{x-6}+\frac{x}{x-7}=x(x-4)[/MATH]

Factor out the common factor, which is an $x$ on the left, we have:

[MATH]x\left(\frac{1}{x-1}+\frac{1}{x-2}+\frac{1}{x-6}+\frac{1}{x-7}\right)=x(x-4)[/MATH](*)

Can you see it now that we've greatly simplified the expression on the left and it's not hard to tell by now that $x=0$ is one of the solutions of the system.

When $x\ne 0$, we can divide through the equation (*) above by $x$ and get:

[MATH]\frac{1}{x-1}+\frac{1}{x-2}+\frac{1}{x-6}+\frac{1}{x-7}=x-4[/MATH](**)

Now, if you're observant enough, you would see $x=4$ is actually another solution for the system as:

[MATH]\frac{1}{4-1}+\frac{1}{4-2}+\frac{1}{4-6}+\frac{1}{4-7}=\frac{1}{3}+\frac{1}{2}-\frac{1}{2}-\frac{1}{3}=0=4-4[/MATH]

What happens if you're not observant? We would still arrive at the same conclusion if we are to let $u=x-4$ to the equation (**):

[MATH]\frac{1}{u+3}+\frac{1}{u+2}+\frac{1}{u-2}+\frac{1}{u-3}=u[/MATH]

Next, collect [MATH]\frac{1}{u+3}+\frac{1}{u-3}[/MATH] and [MATH]\frac{1}{u+2}+\frac{1}{u-2}[/MATH] and add them up to get:

[MATH]\frac{1}{u+3}+\frac{1}{u-3}+\frac{1}{u+2}+\frac{1}{u-2}=u[/MATH]

[MATH]\frac{u-3+u+3}{(u+3)(u-3)}+\frac{u-2+u+2}{(u+2)(u-2)}=u[/MATH]

[MATH]\frac{2u}{u^2-9}+\frac{2u}{u^2-4}=u[/MATH]

[MATH]u\left(\frac{2}{u^2-9}+\frac{2}{u^2-4}\right)=u[/MATH]

Again, we reach to the conclusion where $u=x-4=0\implies x=4$ is another solution for the given system. See it now? Even if you didn't assume $x=4$ is one of the solutions of the system, we could still work out the problem and find out that $x=4$ is one of the solutions.

Are we done yet? Most definitely not! That system is a polynomial of sixth degree, so we should expect to obtain the 6 real roots. So far we only managed to locate the two roots, which are $x=0,\,x=4$.

Continue with what we have left off, we could divide through the equation [MATH]u\left(\frac{2}{u^2-9}+\frac{2}{u^2-4}\right)=u[/MATH] by $u$ when $u\ne 0$ to look for the rest of the four solutions of the system.

[MATH]\frac{2}{u^2-9}+\frac{2}{u^2-4}=1[/MATH]

[MATH]\frac{2(u^2-4)}{u^2-9}+\frac{2(u^2-9)}{u^2-4}=1[/MATH]

[MATH]2(u^2-4)+2(u^2-9)=(u^2-4)(u^2-9)[/MATH]

[MATH]4u^2-26=u^4-13u^2+36[/MATH]

[MATH]u^4-17u^2+62=0[/MATH]

Solving the equation for $u^2$ by using the quadratic formula, we see that:

[MATH]u^2=\frac{17\pm\sqrt{(-17)^2-4(1)(62)}}{2}=\frac{17\pm\sqrt{41}}{2}[/MATH]

[MATH]u=\pm \sqrt{\frac{17\pm\sqrt{41}}{2}}[/MATH]

Back substituting $u=x-4$ into the above equation we therefore get the rest of the solutions for the system:

[MATH]x-4=\pm \sqrt{\frac{17\pm\sqrt{41}}{2}}[/MATH]

[MATH]x=4\pm \sqrt{\frac{17\pm\sqrt{41}}{2}}[/MATH]

Thus,

[MATH]x=0,\,4-\sqrt{\frac{17-\sqrt{41}}{2}},\,4- \sqrt{\frac{17+\sqrt{41}}{2}},\,4,\,4+\sqrt{\frac{17-\sqrt{41}}{2}},\,4+ \sqrt{\frac{17+\sqrt{41}}{2}}[/MATH].