How Do You Solve This Complex Irrational Equation?

[anemone]

Solve the irrational equation $x^4-9x^3+16x^2+15x+26=\dfrac{7}{\sqrt{x^2-10x+26}+\sqrt{x^2-10x+29}+\sqrt{x^2-10x+41}}$.

 

[MarkFL]

My solution:

Spoiler

With a bit of factoring, we obtain:

\(\displaystyle (x-5)^4+11(x-5)^3+31(x-5)^2+1=\frac{7}{\sqrt{(x-5)^2+1}+\sqrt{(x-5)^2+4}+\sqrt{(x-5)^2+16}}\)

So, let $u=x-5$, and we have:

\(\displaystyle u^4+11u^3+31u^2+1=\frac{7}{\sqrt{u^2+1}+\sqrt{u^2+4}+\sqrt{u^2+16}}\)

Define:

\(\displaystyle f(u)=u^4+11u^3+31u^2+1\)

Hence:

\(\displaystyle f'(0)=4u^3+33u^2+62u=u\left(4u^2+33u+62\right)\)

Using the first derivative test, we then find a local minimum at:

\(\displaystyle u=-\frac{33+\sqrt{97}}{8}\)

A local maximum at:

\(\displaystyle u=-\frac{33-\sqrt{97}}{8}\)

And a local minimum at:

\(\displaystyle u=0\)

From this, we obtain:

\(\displaystyle f_{\min}=f(0)=1\)

Now, we we define:

\(\displaystyle g(u)=\frac{7}{\sqrt{u^2+1}+\sqrt{u^2+4}+\sqrt{u^2+16}}\)

It is easy to see that:

\(\displaystyle g_{\max}=g(0)=1\)

Hence, the only solution is:

\(\displaystyle u=0\implies x=5\)

 

[anemone]

Aww...that solution is superb! Well done MarkFL!

But I suspect it was Santa who told you how to do it, since you solved it so fast, hehehe...ho ho ho!:p