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Observe that:

\(\displaystyle 10=(5+2\sqrt{6})^1+(5-2\sqrt{6})^1\)

Take it away!!!

\(\displaystyle 10=(5+2\sqrt{6})^1+(5-2\sqrt{6})^1\)

Take it away!!!

How did you get a power of 1?Observe that:

\(\displaystyle 10=(5+2\sqrt{6})^1+(5-2\sqrt{6})^1\)

Take it away!!!

\(\displaystyle x^2-5x+5=1\)

\(\displaystyle x^2-5x+4=0\)

\(\displaystyle (x-4)(x-1)=0\)

\(\displaystyle x\in\{1,4\}\)

Observe that:

\(\displaystyle 10=(5+2\sqrt{6})^1+(5-2\sqrt{6})^1\)

Take it away!!!

\(\displaystyle \frac{1}{5\pm2\sqrt{6}}\cdot\frac{5\mp2\sqrt{6}}{5\mp2\sqrt{6}}=5\mp2\sqrt{6}\)

So, we see the two bases are each others' reciprocals. Let's begin by writing:

\(\displaystyle (5+2\sqrt{6})^u+(5-2\sqrt{6})^u=10\)

Using our discovery above, we can now write:

\(\displaystyle (5+2\sqrt{6})^u+(5+2\sqrt{6})^{-u}=10\)

If we multiply through by \(v=(5+2\sqrt{6})^u\) then we get (after arranging in standard form):

\(\displaystyle v^2-10v+1=0\)

\(\displaystyle v=5\pm2\sqrt{6}\)

This means:

\(\displaystyle (5+2\sqrt{6})^u=5\pm2\sqrt{6}\)

And using our initial result above, this means:

\(\displaystyle u=x^2-5x+5=\pm1\)

We've previously solved the positive case, so now the negative:

\(\displaystyle x^2-5x+5=-1\)

\(\displaystyle x^2-5x+6=0\)

\(\displaystyle (x-2)(x-3)=0\)

And so the complete set of real solutions is:

\(\displaystyle x\in\{1,2,3,4\}\)

Nicely done as always. Are you saying that my simply picture answer is not even close to proving that the LHS = RHS?

\(\displaystyle \frac{1}{5\pm2\sqrt{6}}\cdot\frac{5\mp2\sqrt{6}}{5\mp2\sqrt{6}}=5\mp2\sqrt{6}\)

So, we see the two bases are each others' reciprocals. Let's begin by writing:

\(\displaystyle (5+2\sqrt{6})^u+(5-2\sqrt{6})^u=10\)

Using our discovery above, we can now write:

\(\displaystyle (5+2\sqrt{6})^u+(5+2\sqrt{6})^{-u}=10\)

If we multiply through by \(v=(5+2\sqrt{6})^u\) then we get (after arranging in standard form):

\(\displaystyle v^2-10v+1=0\)

\(\displaystyle v=5\pm2\sqrt{6}\)

This means:

\(\displaystyle (5+2\sqrt{6})^u=5\pm2\sqrt{6}\)

And using our initial result above, this means:

\(\displaystyle u=x^2-5x+5=\pm1\)

We've previously solved the positive case, so now the negative:

\(\displaystyle x^2-5x+5=-1\)

\(\displaystyle x^2-5x+6=0\)

\(\displaystyle (x-2)(x-3)=0\)

And so the complete set of real solutions is:

\(\displaystyle x\in\{1,2,3,4\}\)

I do not feel so bad when you said "we missed the fact" in your reply.

Now it is solved for all to see.