MHB How Do You Solve This Complex Irrational Equation?

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The discussion centers around solving the complex irrational equation $x^4-9x^3+16x^2+15x+26=\dfrac{7}{\sqrt{x^2-10x+26}+\sqrt{x^2-10x+29}+\sqrt{x^2-10x+41}}$. A participant praises another's solution, highlighting its excellence and speed. There is a light-hearted comment suggesting that Santa may have assisted in finding the solution quickly. The overall tone is playful, yet the focus remains on the mathematical challenge presented. The thread showcases the collaborative spirit of problem-solving in mathematics.
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Solve the irrational equation $x^4-9x^3+16x^2+15x+26=\dfrac{7}{\sqrt{x^2-10x+26}+\sqrt{x^2-10x+29}+\sqrt{x^2-10x+41}}$.
 
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My solution:

With a bit of factoring, we obtain:

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

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

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

Define:

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

Hence:

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

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

$$u=-\frac{33+\sqrt{97}}{8}$$

A local maximum at:

$$u=-\frac{33-\sqrt{97}}{8}$$

And a local minimum at:

$$u=0$$

From this, we obtain:

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

Now, we we define:

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

It is easy to see that:

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

Hence, the only solution is:

$$u=0\implies x=5$$
 
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
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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