Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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....

share|improve this question
1  
Yes we can. But is it easier to solve? –  Berci Feb 19 '13 at 16:31

1 Answer 1

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.

share|improve this answer
1  
How to do substitution here because of R.H.S = $2^x$ –  Sachin Sharmaa Feb 19 '13 at 16:10
    
You can make a substitution with an equivalent expression anywhere in any equation. There is no need to make substitutions for every expression in an equation. Your substitutions may not do you a lot of good, but there's certainly nothing preventing you from making them. –  Cameron Buie Feb 19 '13 at 16:23
    
Let me suggest an alternative approach. I will edit my answer now. –  Cameron Buie Feb 19 '13 at 16:28
    
And of course: $$\sqrt{2+\sqrt{3}}=\frac{1}{\sqrt{2-\sqrt{3}}}$$ –  B. S. Feb 19 '13 at 18:13

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.