# Quadratic Equation using surds property

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

Using property of surd can we simplify the above expression like:

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

Am I right here please guide....

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Yes we can. But is it easier to solve? –  Berci Feb 19 '13 at 16:31

Absolutely. Since you know that $$\sqrt{2\pm\sqrt{3}}=\frac{\sqrt{3}\pm1}{\sqrt{2}},$$ you can always make such a substitution. All you're doing is rewriting the same thing in a different-looking form, and there's no problem with doing that.
I'd take a different route, though. Since $\sqrt{\alpha}=\alpha^{1/2}$ for any positive $\alpha$, we can rewrite the equation using exponential properties as $$\left(2+\sqrt{3}\right)^{x/2}+\left(2-\sqrt{3}\right)^{x/2}=4^{x/2},$$ then make the substitution $y=x/2$ to get $$\left(2+\sqrt{3}\right)^y+\left(2-\sqrt{3}\right)^y=4^y.$$ It should be clear just by looking that $y=1$ ($x=2$) provides a solution to this equation. In fact, it is the only solution, which can be proved in various ways.
How to do substitution here because of R.H.S = $2^x$ –  Sachin Sharmaa Feb 19 '13 at 16:10
And of course: $$\sqrt{2+\sqrt{3}}=\frac{1}{\sqrt{2-\sqrt{3}}}$$ –  Babak S. Feb 19 '13 at 18:13