How Do You Solve This Complex Irrational Equation?

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The forum discussion centers on solving the complex irrational equation \(x^4-9x^3+16x^2+15x+26=\dfrac{7}{\sqrt{x^2-10x+26}+\sqrt{x^2-10x+29}+\sqrt{x^2-10x+41}}\). User MarkFL provided a solution that was praised for its clarity and efficiency. The conversation highlights the collaborative nature of problem-solving in mathematics, with a humorous nod to Santa as a potential source of insight.

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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
 

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