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

1. TeemoTheTerific

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

2. TeemoTheTerific

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

3. Azteck

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

4. enka

$x=\frac{1+\sqrt{5}}{2}\\ \frac{1}{x}=\frac{2}{1+\sqrt{5}}\times\frac{\sqrt{5}-1}{\sqrt{5}-1}=\frac{2\sqrt{5}-1}{4}=\frac{\sqrt{5}-1}{2}$ 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. Azteck

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

6. TeemoTheTerific

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

7. TeemoTheTerific

yes ok

8. enka

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

9. Azteck

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

10. Azteck

And you prove that the LHS=RHS.

11. TeemoTheTerific

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

12. Azteck

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

13. Azteck

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

14. Azteck

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

15. Azteck

numerator*

16. TeemoTheTerific

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

17. Azteck

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

18. Azteck

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

19. Azteck

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

20. enka

21. enka

or this

22. TeemoTheTerific

thank you enka

23. enka

you're welcome :)