## anonymous 3 years ago help me on a question please... x= 1/2 +1/2sqrt5 show that...

1. anonymous

$x=\frac{ 1 }{ 2 } + \frac{ 1 }{ 2 }\sqrt{5}$

2. anonymous

$\frac{ x^{2}+x ^{-2} }{ x-x ^{-1} }=3$

3. anonymous

$x=\frac{1}{2}+\frac{\sqrt{5}}{2}$ $x=\dfrac{1+\sqrt{5}}{2}$

4. anonymous

$x=\frac{1+\sqrt{5}}{2}\\ \frac{1}{x}=\frac{2}{1+\sqrt{5}}\times\frac{\sqrt{5}-1}{\sqrt{5}-1}=\frac{2\sqrt{5}-1}{4}=\frac{\sqrt{5}-1}{2}$ then $x-\frac{1}{x}=1\qquad(1)$ square both side to get: $x^2+\frac{1}{x^2}-2=1\\ x^2+\frac{1}{x^2}=3\qquad(3)$ now divide (3) by (1) to get the result

5. anonymous

$x^2+x^{-2}=3x-3x^{-1}$ Multiply both sides by x^2. $x^4+1=3x^3-3x$ $x^4-3x^3+3x+1=0$

6. anonymous

i dont get it... both of you :P

7. anonymous

yes ok

8. anonymous

first, we find the value of $x-\frac{1}{x}$

9. anonymous

You just sub in $\frac{1+\sqrt{5}}{2}$ into the original equation....

10. anonymous

And you prove that the LHS=RHS.

11. anonymous

but no matter how manytimes i try i can never get it to equal 3

12. anonymous

Use a calculator if you can't do it....Use a calculator to solve the denominator first. You get one as the denominator...

13. anonymous

Jeez, you have a tool that can help you but you don't want to use it.

14. anonymous

You get 3 as the denominator when you plug the values into the numerator...

15. anonymous

numerator*

16. anonymous

we need to show our working our using our skill of indices laws... But i dont have much

17. anonymous

There's no indice work when you learn that $x^{-1}=\frac{1}{x^1}$ $x^{-2}=\frac{1}{x^2}$

18. anonymous

Dude, now once you see the letter x, put that big fat fraction into your calculator and get your values...

19. anonymous

I can solve that by hand but that will take me ages, to write it on the computer, so no thank you...

20. anonymous

21. anonymous

or this

22. anonymous

thank you enka

23. anonymous

you're welcome :)