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

I square the both sides and get $(5+2\sqrt{6})^{x}+(5-2\sqrt{6})^{x}=98$. But I don't know how to carry on. Please help. Thank you.

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Let $t=( \sqrt{5 + 2\sqrt{6}})^{x}\implies (\sqrt{5-2\sqrt{6}})^{x}=\frac{1}{t} $

Thus the equation becomes,

$ t+\frac{1}{t}= 10\implies t^2-10t+1=0$ which is a quadratic equation and have roots $t=5+2\sqrt 6,5-2\sqrt 6$

If $t=5+2\sqrt 6\implies ( \sqrt{5 + 2\sqrt{6}})^{x}=5+2\sqrt 6\implies (5+2\sqrt 6)^{1-x/2}=1\implies 1-x/2=0\implies x=2$

If $t=5-2\sqrt 6 \implies (\sqrt{5+2\sqrt{6}})^{x}=5-2\sqrt 6=\frac{1}{5+2\sqrt 6}\implies (5+2\sqrt 6)^{1+x/2}=1\implies x=-2$

Thus, $x=2,-2$ are the two solutions.

Verification:

Putting $x=2$ in the original equation gives L.H.S=$5+2\sqrt 6+5-2\sqrt 6=10=$R.H.S

Similarly, putting $x=-2$ also gives L.H.S=$5-2\sqrt 6+5+2\sqrt 6=10=$R.H.S

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  • $\begingroup$ To be perfectly rigorous, you need to check (it's trivial here) that the two possibilities you get are indeed solutions; you only wrote implications ($\Rightarrow$), not equivalences ($\Leftrightarrow$). $\endgroup$ – Najib Idrissi Sep 25 '12 at 9:39
  • $\begingroup$ @nik: i already checked that by putting $x=2$ and $-2$ in the original equation and they both satisfy it. $\endgroup$ – Aang Sep 25 '12 at 9:41
  • $\begingroup$ But you did not write it. $\endgroup$ – xavierm02 Sep 25 '12 at 13:44
  • $\begingroup$ I included it now. $\endgroup$ – Aang Sep 25 '12 at 13:49
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Nice solution, just for clarification

$\left(5+2\sqrt{6}\right)^{\frac{x}{2}}\cdot \left(5-2\sqrt{6}\right)^{\frac{x}{2}}=\left[\left(5+2\sqrt{6}\right)\cdot \left(5-2\sqrt{6}\right)\right]^{\frac{x}{2}}=\left[5^2-\left(2\sqrt{6}\right)^2\right]^{\frac{x}{2}}=\left(25-24\right)^{\frac{x}{2}}=1$

and so for $t=\left(\sqrt{5+2\sqrt{6}}\right)^x$......

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  • $\begingroup$ Interesting. Firefox is rendering the first set of brackets as Floor brackets. Checked in IE and it's fine. Also, CTRL + and CTRL - (change font size) renders correctly. $\endgroup$ – Chris Cudmore Sep 25 '12 at 14:03
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Hint $\ $ Put $\rm\ b = 5 + 2\sqrt{6},\,\ a = b^{\,x/2}\ $ in

$$\rm a+ a^{-1} =\, b + b^{-1}\, \Rightarrow\ \{a,a^{-1}\} = \{b,b^{-1}\}\ \ since\ \ (x-a)(x-a^{-1})\, =\, (x-b)(x-b^{-1})$$

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.

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Let $t_1=(\sqrt{5+2\sqrt6})^x$, and $t_2=(\sqrt{5-2\sqrt6})^x$.

Now the given equation is:

$$\tag 1 t_1+t_2=10$$

But

$$\displaylines{ {t_1}\cdot{t_2}&=& {\left( {\sqrt {5 + 2\sqrt 6 } } \right)^x}{\left( {\sqrt {5 - 2\sqrt 6 } } \right)^x} \cr &=& {\left[ {\sqrt {\left( {5 + 2\sqrt 6 } \right)\left( {5 - 2\sqrt 6 } \right)} } \right]^x} \cr &=& {\left[ {\sqrt {{5^2} - {{\left( {2\sqrt 6 } \right)}^2}} } \right]^x} \cr &=& {\left( {\sqrt {25 - 24} } \right)^x} \cr &=& {1^x} = 1 \cr} $$

Thus

$$\begin{cases} t_1+t_2=10 \\t_1\cdot t_2=1\end{cases}$$

$\Rightarrow$ $t^2-10t+1=0$

Since $$\frac{1}{{{t_1}}} = {t_2}$$

we

have $${t_1} + \frac{1}{{{t_1}}} = 10$$ or

$$t_1^2 - 10{t_1} + 1 = 0$$

Note the system is symmetric on the unknowns.

For this quadratic equation we have:

$t_{1,2}=\frac{10\pm\sqrt{100-4}}{2}$

$t_{1,2}=\frac{10\pm 4\sqrt{6}}{2}$

$t_1=5+2\sqrt 6$, $t_2=5-2\sqrt 6$

Now return the inital substition:

$t_1=(\sqrt{5+2\sqrt6})^x$

$5+2\sqrt 6=(\sqrt{5+2\sqrt6})^x$

$(\sqrt{5+2\sqrt 6})^2=(\sqrt{5+2\sqrt6})^x$ $\Rightarrow$ $x=2$, and

$(\sqrt{5+2\sqrt6})^x=5-2\sqrt 6$

$(\sqrt{5+2\sqrt6})^x=(5+2\sqrt 6)^{-2}$ $\Rightarrow$ $x=-2$

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

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  • 2
    $\begingroup$ Please could you tell me the reason why they are negative points $\endgroup$ – Madrit Zhaku Sep 25 '12 at 19:23

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