## jiteshmeghwal9 3 years ago If $$\LARGE{x^2=\frac{1}{x^2}+1}$$, then $$\LARGE{x^2+\frac{1}{x^2}=?}$$

1. jiteshmeghwal9

@hartnn @ParthKohli @ajprincess plz help :)

2. ParthKohli

$\dfrac{1}{x^2} = x^2 - 1 \ \ \ \$So,$x^2 + x^2 - 1 = 2x^2 -1$

3. ParthKohli

Should it be a number value?

4. jiteshmeghwal9

$\Huge{\color{red}{Ans.-\sqrt{5}}}$

5. Yahoo!

x^4 = x^2 + 1 x^4 - x^2 - 1 = 0 put y = x^2

6. jiteshmeghwal9

wait..

7. ParthKohli

That's a quartic equation in $$x$$. So multiply through $$x^2$$.$x^4 = x^2 + 1$

8. jiteshmeghwal9

from where do u gt x^4?

9. ParthKohli

I had typed a long explanation, but Yahoo! spilled the beans =P

10. Yahoo!

Solve the given Equation @jiteshmeghwal9 x^2 = 1 + 1/x^2

11. ParthKohli

$$x^2 = 1 + \dfrac{1}{x^2}$$ is what you have. Multiply $$x^2$$ to both sides.

12. jiteshmeghwal9

x^4=x^2+1

13. ParthKohli

That...

14. jiteshmeghwal9

15. jiteshmeghwal9

AArgh.... I'm going to be crazy :(

16. jiteshmeghwal9

Sorry dudes but i'm nt getting the right thing :(

17. jiteshmeghwal9

Need help

18. rahul91

x^4-x^2-1=0 put x^2=t t^2-t-1=0 solve it

19. jiteshmeghwal9

20. rahul91

yep

21. jiteshmeghwal9

${1 \pm \sqrt{1+4}\over2}={1 \pm \sqrt{5}\over2}$

22. jiteshmeghwal9

23. rahul91

substitute for x^2 in the expression needed

24. Yahoo!

Hm..Hm...After Solving i got sqrt5

25. jiteshmeghwal9

can anybody give me steps to the solution, i'm getting confused :((((

26. rahul91

hey substitute x^2= (1-sqrt(5))/2 in the expression u get the answer -sqrt(5)

27. jiteshmeghwal9

ok

28. Yahoo!

@jiteshmeghwal9 ..Can u Check...ur Solution

29. Yahoo!

Is it sqrt5 or -sqrt5

30. Yahoo!

x^2 + 1 / x^2 = (x^4 + 1) / x^2

31. ParthKohli

$x^4 -x^2 - 1 = 0$We can let $$t = x^2$$.$t^2 - t - 1 = 0 \iff \boxed{t = \dfrac{1\pm \sqrt 5}{2}}{}$

32. Yahoo!

x^4=x^2+1 (x^2+2) / x^2 = 1 + 2/x^2

33. sirm3d

$x^2=1/x^2+1\\x^4=1+x^2\\x^4-x^2-1=0$by quadratic formula,$x^2=\frac{1\pm \sqrt{5}}{2}$disregard $$\displaystyle x^2= \frac{1-\sqrt{5}}{2}$$ since this is not a real number

34. Yahoo!

nw put the value of x^2

35. jiteshmeghwal9

well $$\sqrt{5}=\pm\sqrt{5}$$ but in my book it is only given $$\sqrt{5}$$ as answer :)

36. Yahoo!

Lol....Then...Go By My method

37. jiteshmeghwal9

ohh! k

38. Yahoo!

1 + 2/x^2 1 + 4 / (1 + sqrt5) (5 + sqrt5)/ (1 + sqrt5) rationalize... u will get (4 sqrt 5)/4 = sqrt5

39. Yahoo!

Nw u Got it...:)

40. jiteshmeghwal9

A little doubt here, just last doubt

41. Yahoo!

Wat?@jiteshmeghwal9

42. jiteshmeghwal9

(x^2+2) / x^2 = 1 + 2/x^2 how did u gt this ?

43. sirm3d

$x^2=\frac{1+\sqrt 5}{2}\\\frac{1}{x^2}=\frac{2}{1+\sqrt 5}\cdot\frac{1-\sqrt 5}{1-\sqrt 5}=-\frac{1}{2}+\frac{\sqrt 5}{2}\\x^2+\frac{1}{x^2}=(1/2+\sqrt 5/2)+(-1/2+\sqrt 5/2)=\sqrt {5}$

44. ParthKohli

$\dfrac{x^2 + 2}{x^2} = \dfrac{x^2}{x^2} + \dfrac{2}{x^2}\cdots$

45. jiteshmeghwal9

I m just gt confused just like my bro @maheshmeghwal9 :(

46. Yahoo!

Oh...i just Did to make calculation easy....if u dont get it...Go My The method used by sirm

47. jiteshmeghwal9

Well...now i gt it, thanx all of u a lot :) Mainly @Yahoo! thanx buddy ;)

48. Yahoo!

Well..)) .welcome