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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[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 $
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Now it should be perfectly clear the two solutions are $\displaystyle x=\pm 2~.$
Hello, sciencepal!
$\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.