# Solve for x in the given equation! [closed]

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 :/

## closed as off-topic by kingW3, Shailesh, projectilemotion, Arnaldo, DaveAug 18 '17 at 17:18

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – kingW3, projectilemotion, Arnaldo, Dave
If this question can be reworded to fit the rules in the help center, please edit the question.

• – lab bhattacharjee Aug 18 '17 at 8:44
• In order to get the best possible answers, it is helpful to state what your thoughts and attempts on the problem are; this will prevent people from telling you things you already know, and help them give their answers at the right level. – projectilemotion Aug 18 '17 at 8:45
• @projectilemotion sure – MathDude3013 Aug 18 '17 at 8:55

## 4 Answers

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: $$\{2,-2,\sqrt2,-\sqrt2\}$$

Hint:

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

Method$\#1$:

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

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

Method$\#2$:

$$a+\dfrac1a=5+2\sqrt6+\dfrac1{5+2\sqrt6}$$

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

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

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

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

and

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

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

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 $x = \pm 2, \pm \sqrt{2}$.