Solve the following equation. $$(5+2\sqrt6)^{x^2-3}+(5-2\sqrt6)^{x^2-3} = 10$$

Solve for $x$ in this equation! I am stuck :/


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Let $(5+2\sqrt6)^{x^2-3}=t$. Hence, $t+\frac{1}{t}=10$ and we have $t=5\pm2\sqrt{6}$.

Thus, $x^2-3=1$ or $x^2-3=-1$, which gives the answer: $$\{2,-2,\sqrt2,-\sqrt2\}$$



As $(5+2\sqrt6)(5-2\sqrt6)=1$


set $(5+2\sqrt6)^{x^2-3}=a$ to find $$a+\dfrac1a=10$$

So, we have $$(5+2\sqrt6)^{x^2-3}=a=(5+2\sqrt6)^{\pm1}$$



  • $\begingroup$ What if their product wasn't equal to 1? $\endgroup$ – MathDude3013 Aug 18 '17 at 8:46
  • 2
    $\begingroup$ @LinuxGeek, If there is no relationship between two radicals, things will be horribly complex. $\endgroup$ – lab bhattacharjee Aug 18 '17 at 8:49

$\left(5-2 \sqrt{6}\right) \left(5+2 \sqrt{6}\right)=1$

So $5-2 \sqrt{6}=\dfrac{1}{5+2 \sqrt{6}}$

substitute $\left(2 \sqrt{6}+5\right)^{x^2-3}=t$ so that $\left(5-2 \sqrt{6}\right)^{x^2-3}=\dfrac{1}{t}$

and solve $t+\dfrac{1}{t}=10$ which gives $t_1= 5-2 \sqrt{6},\;t_2=5+2 \sqrt{6}$

Substitute back

$\left(2 \sqrt{6}+5\right)^{x^2-3}=5-2 \sqrt{6}$

$\left(2 \sqrt{6}+5\right)^{x^2-3}=\left(5+2 \sqrt{6}\right)^{-1}$

$x^2-3=-1\to x=\pm\sqrt 2$


$\left(2 \sqrt{6}+5\right)^{x^2-3}=5+2 \sqrt{6}$

$x^2-3=1\to x=\pm 2$


If you write $a = (5+2\sqrt6)^{x^2-3}$ then you have $a+1/a =10$ and thus $a_{1,2} = {5\pm 2\sqrt6}$.

So $x^2-3 = \pm 1$ and thus $x = \pm 2, \pm \sqrt{2}$.


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