## anonymous 4 years ago x=sqrt(7+4sqrt3)+sqrt(7-4sqrt3)

1. anonymous

2. anonymous

You may see the equation here.

3. ash2326

We have $x=\sqrt{7+4\sqrt 3}+\sqrt{7-4\sqrt 3}$ Let's multiply and divide this by $$\sqrt{7+4\sqrt 3}-\sqrt{7-4\sqrt 3}$$ we have $x=\sqrt{7+4\sqrt 3}+\sqrt{7-4\sqrt 3} \times \frac{\sqrt{7+4\sqrt 3}-\sqrt{7-4\sqrt 3}}{\sqrt{7+4\sqrt 3}-\sqrt{7-4\sqrt 3}}$ We get $x=\frac{{(\sqrt{7+4\sqrt 3})^2 }-(\sqrt{(7-4\sqrt 3})^2 }{\sqrt{7+4\sqrt 3}-\sqrt{7-4\sqrt 3}}$ we get $x=\frac{7+4\sqrt 3-(7-4 \sqrt 3)}{\sqrt{7+4\sqrt 3}-\sqrt{7-4\sqrt 3}}$ so we get $x=\frac{8\sqrt 3}{\sqrt{7+4\sqrt 3}-\sqrt{7-4\sqrt 3}}$

4. anonymous

5. ash2326

Yeah can't be simplified more

6. anonymous

7. anonymous

8. anonymous

Ai Se Eu Te Pego -

9. anonymous

159

10. anonymous

rrsrsrs

11. anonymous

good

12. anonymous

De onde vc é Elodi?

13. anonymous

quoi

14. anonymous

Vc falou português

15. anonymous

i m from Brazil

16. anonymous

okk

17. anonymous

18. ash2326

I got it, Let me show you how this 4

19. anonymous

yes]

20. ash2326

21. anonymous

ok

22. anonymous

Bye Elodi See u around i ve got go

23. anonymous

y

24. ash2326

We have $x=\sqrt {7+4 \sqrt 3}+\sqrt {7-4 \sqrt 3}$ Let's square both the sides we get $x^2=(\sqrt {7+4 \sqrt 3})^2+(\sqrt {7-4 \sqrt 3})^2+2\times (\sqrt {7+4 \sqrt 3})\times (\sqrt {7-4 \sqrt 3})$ we get $x^2=7+4\sqrt 3+7-4\sqrt 3+2 \sqrt{(7^2-(4\sqrt 3)^2}$ we get $x^2=14+2\times \sqrt {49-16\times 3}$ we get $x^2=14+2\times \sqrt 1$ we get $x^2=14+2=16$ so $x=4$

25. anonymous

square on both sides|dw:1330781973743:dw|

26. anonymous

Good.