# Math Help - surds

1. ## surds

find the number of solutions in $(5+2\sqrt{6})^{x^2-3}+(5-2\sqrt{6})^{x^2-3}=10$

2. ## Re: surds

Originally Posted by sciencepal
find the number of solutions in (5 + 2*6^0.5)^(x^2-3) + (5 - 2*6^0.5)^(x^2-3) = 10
Why don’t you take some time to learn to post symbols?
It really is so easy. And it makes most of us more willing to find out how to help.
This subforum will help you with the code. Once you begin, you quickly learn the code.
LaTeX Help

[TEX](5+2\sqrt{6})^{x^2-3}+(5-2\sqrt{6})^{x^2-3}=10 [/TEX] gives $(5+2\sqrt{6})^{x^2-3}+(5-2\sqrt{6})^{x^2-3}=10$
If you click on the “go advanced tab” you should see $\boxed{\Sigma}$ on the tool-bar. That gives the [TEX]..[/TEX] wrap. Your LaTeX code goes between them.

Now it should be perfectly clear the two solutions are $x=\pm 2~.$

3. ## Re: surds

$\pm\sqrt{2}$
$\pm2$
Could you please solve the equation?

4. ## Re: surds

Originally Posted by sciencepal

$\pm\sqrt{2}$
$\pm2$
Could you please solve the equation?
Show that $\pm\sqrt2$ are solutions.

What does ${\left( {5 + 2\sqrt 6 } \right)^{ - 1}} = ~?$

5. ## Re: surds

That's fine but how do I exactly know how many solutions can the equation have?

6. ## Re: surds

Originally Posted by sciencepal
That's fine but how do I exactly know how many solutions can the equation have?
As far as I know there is no elementary solution to this question.

Drawing a graph would give you some idea.

7. ## Re: surds

Well now I got the idea. This equation holds true when the powers are 1 or -1 and $a^{2}-b=1$. In this case $5^{2}-(2\sqrt{6})^{2}=1$. Thanks.

8. ## Re: surds

Hello, sciencepal!

$\text{Find the number of solutions in: }\:(5+2\sqrt{6})^{x^2-3}+(5-2\sqrt{6})^{x^2-3}\:=\:10$

We note that: . $5\pm2\sqrt{6} \:=\:\left(\sqrt{3}\pm\sqrt{2}\right)^2$

The equation becomes: . $\left(\sqrt{3}+\sqrt{2}\right)^{2(x^2-3)} + \left(\sqrt{3}-\sqrt{2}\right)^{2(x^2-3)} \:=\:10$

We further note that: . $\sqrt{3}-\sqrt{2} \:=\:\frac{\sqrt{3}-\sqrt{2}}{1}\cdot\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3} +\sqrt{2}} \:=\: \frac{1}{\sqrt{3}+\sqrt{2}}$

The equation becomes: . $\left(\sqrt{3}+\sqrt{2}\right)^{2(x^2-3)} + \frac{1}{(\sqrt{3}+\sqrt{2})^{2(x^2-3)}} \:=\:1$

Multiply by $\left(\sqrt{3}+\sqrt{3}\right)^{2(x^2-3)}$

. . $\left(\sqrt{3}+\sqrt{2}\right)^{4(x^2-3)} + 1 \:=\:10\left(\sqrt{3}+\sqrt{2}\right)^{2(x^2-3)}$

. . $\left(\sqrt{3}+\sqrt{2}\right)^{4(x^2-3)} - 10\left(\sqrt{3}+\sqrt{2}\right)^{2(x^2-3)} + 1 \;=\;0$

Let $u \:=\:\left(\sqrt{3}+\sqrt{2}\right)^{2(x^2-30}$

We have: . $u^2 - 10u + 1 \:=\:0$

Quadratic Formula: . $u \;=\;5 \pm2\sqrt{3} \;=\;\left(\sqrt{3}\pm\sqrt{2}\right)^2$

Back-substitute: . $\left(\sqrt{3}+\sqrt{2}\right)^{2(x^2-3)} \;=\;\left(\sqrt{3}\pm\sqrt{2}\right)^2$

Take logs: . $\ln\left(\sqrt{3}+\sqrt{2}\right)^{2(x^2-3)} \;=\;\ln\left(\sqrt{3}\pm\sqrt{2}\right)^2$

. . $2(x^2-3)\ln\left(\sqrt{3}+\sqrt{2}\right) \;=\;2\ln\left(\sqrt{3}\pm\sqrt{2}\right)$

. . $x^2-3 \;=\;\frac{\ln(\sqrt{3}\pm\sqrt{2})}{\ln(\sqrt{3}+ \sqrt{2})}$

The $\pm$ gives us 2 solutions.

And the $x^2$ gives us $2\cdot 2 \:=\:4$ solutions.