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

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- May 1st 2013, 05:57 AMsciencepalsurds
find the number of solutions in $\displaystyle (5+2\sqrt{6})^{x^2-3}+(5-2\sqrt{6})^{x^2-3}=10 $

- May 1st 2013, 06:32 AMPlatoRe: surds
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.

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[TEX](5+2\sqrt{6})^{x^2-3}+(5-2\sqrt{6})^{x^2-3}=10 [/TEX] gives $\displaystyle (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 $\displaystyle \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 $\displaystyle x=\pm 2~.$ - May 1st 2013, 06:47 AMsciencepalRe: surds
Thanks for your advice but there are 4 solutions

$\displaystyle \pm\sqrt{2}$

$\displaystyle \pm2$

Could you please solve the equation? - May 1st 2013, 06:52 AMPlatoRe: surds
- May 1st 2013, 06:59 AMsciencepalRe: surds
That's fine but how do I exactly know how many solutions can the equation have?

- May 1st 2013, 07:06 AMPlatoRe: surds
- May 1st 2013, 07:19 AMsciencepalRe: surds
Well now I got the idea. This equation holds true when the powers are 1 or -1 and $\displaystyle a^{2}-b=1$. In this case $\displaystyle 5^{2}-(2\sqrt{6})^{2}=1$. Thanks.

- May 1st 2013, 04:48 PMSorobanRe: surds
Hello, sciencepal!

Quote:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

The $\displaystyle \pm$ gives us 2 solutions.

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