If x = (2 + root(5))^1/2 + (2 - root(5))^1/2 and y = (2+ root(5))^1/2 - (2 - root(5))^1/2 then evaluate x^2 + y^2
![Comparing spirals for the Nautilus shell, the Fibonacci numbers and the Equal Tempered Chromatic Music Scale Comparing spirals for the Nautilus shell, the Fibonacci numbers and the Equal Tempered Chromatic Music Scale](http://mathman.biz/html/colorspirals_files/fibsp.gif)
Comparing spirals for the Nautilus shell, the Fibonacci numbers and the Equal Tempered Chromatic Music Scale
![Calculating 1/(2+Sqrt(5)) + 1/(2-Sqrt(5)) should return -4 · Issue #125 · axkr/symja_android_library · GitHub Calculating 1/(2+Sqrt(5)) + 1/(2-Sqrt(5)) should return -4 · Issue #125 · axkr/symja_android_library · GitHub](https://user-images.githubusercontent.com/19369448/55780969-56100300-5ad3-11e9-87a4-613583a2b27f.png)
Calculating 1/(2+Sqrt(5)) + 1/(2-Sqrt(5)) should return -4 · Issue #125 · axkr/symja_android_library · GitHub
15. 1/( 1 + √()2) + 1/ (√()2 +√()3) + 1/( √()3 + √()4) + 1/ ( √()4 + √()5) + 1/ (√()5 + √()6) +1/ (√()6 + √()7) + 1/ (√()7 + √()8 ) + 1 / √()8 + √()9) is equal to . . A) 0 B) 1 C) 2 D) 4
![The Golden Mean or Ratio[(1]sqrt(5))/2] to 20,000 Places: Buy The Golden Mean or Ratio[(1]sqrt(5))/2] to 20,000 Places by unknown at Low Price in India | Flipkart.com The Golden Mean or Ratio[(1]sqrt(5))/2] to 20,000 Places: Buy The Golden Mean or Ratio[(1]sqrt(5))/2] to 20,000 Places by unknown at Low Price in India | Flipkart.com](https://rukminim1.flixcart.com/image/850/1000/jmqmpow0/book/6/6/7/the-golden-mean-or-ratio-1-sqrt-5-2-to-20-000-places-original-imaf9kk5cd3ffznh.jpeg?q=90)
The Golden Mean or Ratio[(1]sqrt(5))/2] to 20,000 Places: Buy The Golden Mean or Ratio[(1]sqrt(5))/2] to 20,000 Places by unknown at Low Price in India | Flipkart.com
![If [math]x = \sqrt{\frac{\sqrt5+1}{\sqrt5 -1}}[/math] then the value of [math]5x^2 -5x-1[/math] is? - Quora If [math]x = \sqrt{\frac{\sqrt5+1}{\sqrt5 -1}}[/math] then the value of [math]5x^2 -5x-1[/math] is? - Quora](https://qph.cf2.quoracdn.net/main-qimg-3aaca14f655ca6ab6d42885f86373f4d.webp)
If [math]x = \sqrt{\frac{\sqrt5+1}{\sqrt5 -1}}[/math] then the value of [math]5x^2 -5x-1[/math] is? - Quora
![abstract algebra - proving that $\mathbb{Q}(\sqrt{5}, \sqrt{6}) = \mathbb{Q}(\sqrt{5}+ \sqrt{6}) $ - Mathematics Stack Exchange abstract algebra - proving that $\mathbb{Q}(\sqrt{5}, \sqrt{6}) = \mathbb{Q}(\sqrt{5}+ \sqrt{6}) $ - Mathematics Stack Exchange](https://i.stack.imgur.com/u590Y.png)
abstract algebra - proving that $\mathbb{Q}(\sqrt{5}, \sqrt{6}) = \mathbb{Q}(\sqrt{5}+ \sqrt{6}) $ - Mathematics Stack Exchange
![Sum of `1/(sqrt(2)+sqrt(5))+1/(sqrt(5)+sqrt(8))+1/(sqrt(8)+sqrt(11))+1/(sqrt (11)+sqrt(14))+..to n` - YouTube Sum of `1/(sqrt(2)+sqrt(5))+1/(sqrt(5)+sqrt(8))+1/(sqrt(8)+sqrt(11))+1/(sqrt (11)+sqrt(14))+..to n` - YouTube](https://i.ytimg.com/vi/8ciPBDdOSfg/maxresdefault.jpg)