# Find the solution of this equation $(\sqrt{2-\sqrt 3})^x+(\sqrt{2+\sqrt 3})^x=4$ [duplicate]

Possible Duplicate:
Solve $(\sqrt{5+2\sqrt{6}})^{x}+(\sqrt{5-2\sqrt{6}})^{x}=10$.

Help me solve this equation $(\sqrt{2-\sqrt 3})^x+(\sqrt{2+\sqrt 3})^x=4$

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$x=2$ it is one clear solution –  dato datuashvili Sep 26 '12 at 19:13
Functional duplicate of a very recent question. –  Did Sep 26 '12 at 19:18

## marked as duplicate by lhf, Pedro Tamaroff, William, Aang, Ｊ. Ｍ.Sep 27 '12 at 10:20

Let $t_1=(\sqrt{2+\sqrt3})^x$, and $t_2=(\sqrt{2-\sqrt3})^x$. Now the given equation is:

$t_1+t_2=4$..........(1)

Above it follows that:

$t_1\cdot t_2=(\sqrt{2+\sqrt 3})^x\cdot(\sqrt{2-\sqrt3})^x=(\sqrt{({2+\sqrt3})({2-\sqrt3}})^x=(\sqrt{2^2-(\sqrt3)^2})^x=(\sqrt{4-\sqrt 3^2})^x=(\sqrt{4-3})^x=1^x=1$ $\Rightarrow$ $t_1\cdot t_2=1$...........................(2)

For (1) and (2) we have:

$t_1+t_2=4$

$t_1\cdot t_2=1$

$\Rightarrow$ $t^2-4t+1=0$

$t_{1,2}=\frac{4\pm\sqrt{16-4}}{2}$

$t_{1,2}=\frac{4\pm 2\sqrt{3}}{2}$

$t_1=2+\sqrt 3$, $t_2=2-\sqrt 3$

Now return the inital substition:

$t_1=(\sqrt{2+\sqrt3})^x$

$2+\sqrt 3=(\sqrt{2+\sqrt3})^x$

$(\sqrt{2+\sqrt 3})^2=(\sqrt{2+\sqrt3})^x$ $\Rightarrow$ $x=2$, and

$(\sqrt{2+\sqrt3})^x=2-\sqrt 3$

$(\sqrt{2+\sqrt3})^x=\sqrt{(2+\sqrt 3)}^{-2}$ $\Rightarrow$ $x=-2$

Definitly $x=2$, and $x=-2$ is solve.

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HINT: Note that $2-\sqrt{3} = \dfrac1{2+\sqrt{3}}$.

Denote $\left(\sqrt{2-\sqrt{3}} \right)^x$ as $t$ and proceed to solve the quadratic in $t$ and hence solve for $x$.

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would not be it quadratic equation with two unknown variable? –  dato datuashvili Sep 26 '12 at 19:16
@dato You will get $t+ \dfrac1t = 4$ since $\left(\sqrt{2+\sqrt{3}} \right)^x = \dfrac1{\left(2-\sqrt{3} \right)^x}$ –  user17762 Sep 26 '12 at 19:17
@mixedmath I received two down-votes within the last one hour for correct answers. Could you let me know if they are from the same person? –  user17762 Sep 26 '12 at 20:01
Hint $\$ First, recall that a pair of reals $\rm\:a,a'\:$ is determined uniquely by their sum $\rm\,s\,$ and product $\rm\,p.\,$ Indeed, $\rm\:a,a'\:$ are the unique roots of $\rm\:(x-a)(x-a') = x^2 - s\, x + p = 0.\:$ Any other pair $\rm\,b,b'\,$ with the same sum $\rm\,s\,$ and product $\rm\,p\,$ are roots of the same polynomial, so are the same up to order.
Your problem is a special case. Let $\rm\, b = 2 + \sqrt{3},\,$ so $\rm\,b^{-1} = 2-\sqrt{3}.\,$ Let $\rm\, a = b^{\,x/2}.\,$ Then your equation is $\rm\:a+a^{-1} = b+b^{-1}.\:$ Hence the pairs $\rm\,a,a^{-1}$ and $\rm\,b,b^{-1}$ have the same sum, and the same product $(=1).\,$ Hence, as above, the pairs are equal up to order. Therefore we conclude that$\rm\: a = b^{\pm1},\:$ i.e. $\rm\:(2+\sqrt{3})^{\,x/2} = 2\pm\sqrt{3}.\:$ The rest is easy.
Note $\,$ Generally every pair of numbers in a ring is uniquely determined by their sum and product iff the ring is a domain. Above is the special case of this uniqueness result where the product $= 1.\:$ As I often emphasize, uniqueness theorems provide powerful tools for proving equalities.