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jiteshmeghwal9

  • 3 years ago

If \(\LARGE{x^2=\frac{1}{x^2}+1}\), then \(\LARGE{x^2+\frac{1}{x^2}=?}\)

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  1. jiteshmeghwal9
    • 3 years ago
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    @hartnn @ParthKohli @ajprincess plz help :)

  2. ParthKohli
    • 3 years ago
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    \[\dfrac{1}{x^2} = x^2 - 1 \ \ \ \ \]So,\[x^2 + x^2 - 1 = 2x^2 -1\]

  3. ParthKohli
    • 3 years ago
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    Should it be a number value?

  4. jiteshmeghwal9
    • 3 years ago
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    \[\Huge{\color{red}{Ans.-\sqrt{5}}}\]

  5. Yahoo!
    • 3 years ago
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    x^4 = x^2 + 1 x^4 - x^2 - 1 = 0 put y = x^2

  6. jiteshmeghwal9
    • 3 years ago
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    wait..

  7. ParthKohli
    • 3 years ago
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    That's a quartic equation in \(x\). So multiply through \(x^2\).\[x^4 = x^2 + 1\]

  8. jiteshmeghwal9
    • 3 years ago
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    from where do u gt x^4?

  9. ParthKohli
    • 3 years ago
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    I had typed a long explanation, but Yahoo! spilled the beans =P

  10. Yahoo!
    • 3 years ago
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    Solve the given Equation @jiteshmeghwal9 x^2 = 1 + 1/x^2

  11. ParthKohli
    • 3 years ago
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    \(x^2 = 1 + \dfrac{1}{x^2}\) is what you have. Multiply \(x^2\) to both sides.

  12. jiteshmeghwal9
    • 3 years ago
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    x^4=x^2+1

  13. ParthKohli
    • 3 years ago
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    That...

  14. jiteshmeghwal9
    • 3 years ago
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    Oops sorry i misread it

  15. jiteshmeghwal9
    • 3 years ago
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    AArgh.... I'm going to be crazy :(

  16. jiteshmeghwal9
    • 3 years ago
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    Sorry dudes but i'm nt getting the right thing :(

  17. jiteshmeghwal9
    • 3 years ago
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    Need help

  18. rahul91
    • 3 years ago
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    x^4-x^2-1=0 put x^2=t t^2-t-1=0 solve it

  19. jiteshmeghwal9
    • 3 years ago
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    using quadratic formula ?

  20. rahul91
    • 3 years ago
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    yep

  21. jiteshmeghwal9
    • 3 years ago
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    \[{1 \pm \sqrt{1+4}\over2}={1 \pm \sqrt{5}\over2}\]

  22. jiteshmeghwal9
    • 3 years ago
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    No.. i'm nt getting answer

  23. rahul91
    • 3 years ago
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    substitute for x^2 in the expression needed

  24. Yahoo!
    • 3 years ago
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    Hm..Hm...After Solving i got sqrt5

  25. jiteshmeghwal9
    • 3 years ago
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    can anybody give me steps to the solution, i'm getting confused :((((

  26. rahul91
    • 3 years ago
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    hey substitute x^2= (1-sqrt(5))/2 in the expression u get the answer -sqrt(5)

  27. jiteshmeghwal9
    • 3 years ago
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    ok

  28. Yahoo!
    • 3 years ago
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    @jiteshmeghwal9 ..Can u Check...ur Solution

  29. Yahoo!
    • 3 years ago
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    Is it sqrt5 or -sqrt5

  30. Yahoo!
    • 3 years ago
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    x^2 + 1 / x^2 = (x^4 + 1) / x^2

  31. ParthKohli
    • 3 years ago
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    \[x^4 -x^2 - 1 = 0 \]We can let \(t = x^2\).\[t^2 - t - 1 = 0 \iff \boxed{t = \dfrac{1\pm \sqrt 5}{2}}{}\]

  32. Yahoo!
    • 3 years ago
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    x^4=x^2+1 (x^2+2) / x^2 = 1 + 2/x^2

  33. sirm3d
    • 3 years ago
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    \[x^2=1/x^2+1\\x^4=1+x^2\\x^4-x^2-1=0\]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!
    • 3 years ago
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    nw put the value of x^2

  35. jiteshmeghwal9
    • 3 years ago
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    well \(\sqrt{5}=\pm\sqrt{5}\) but in my book it is only given \(\sqrt{5}\) as answer :)

  36. Yahoo!
    • 3 years ago
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    Lol....Then...Go By My method

  37. jiteshmeghwal9
    • 3 years ago
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    ohh! k

  38. Yahoo!
    • 3 years ago
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    1 + 2/x^2 1 + 4 / (1 + sqrt5) (5 + sqrt5)/ (1 + sqrt5) rationalize... u will get (4 sqrt 5)/4 = sqrt5

  39. Yahoo!
    • 3 years ago
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    Nw u Got it...:)

  40. jiteshmeghwal9
    • 3 years ago
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    A little doubt here, just last doubt

  41. Yahoo!
    • 3 years ago
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    Wat?@jiteshmeghwal9

  42. jiteshmeghwal9
    • 3 years ago
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    (x^2+2) / x^2 = 1 + 2/x^2 how did u gt this ?

  43. sirm3d
    • 3 years ago
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    \[x^2=\frac{1+\sqrt 5}{2}\\\frac{1}{x^2}=\frac{2}{1+\sqrt 5}\cdot\frac{1-\sqrt 5}{1-\sqrt 5}=-\frac{1}{2}+\frac{\sqrt 5}{2}\\x^2+\frac{1}{x^2}=(1/2+\sqrt 5/2)+(-1/2+\sqrt 5/2)=\sqrt {5}\]

  44. ParthKohli
    • 3 years ago
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    \[\dfrac{x^2 + 2}{x^2} = \dfrac{x^2}{x^2} + \dfrac{2}{x^2}\cdots\]

  45. jiteshmeghwal9
    • 3 years ago
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    I m just gt confused just like my bro @maheshmeghwal9 :(

  46. Yahoo!
    • 3 years ago
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    Oh...i just Did to make calculation easy....if u dont get it...Go My The method used by sirm

  47. jiteshmeghwal9
    • 3 years ago
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    Well...now i gt it, thanx all of u a lot :) Mainly @Yahoo! thanx buddy ;)

  48. Yahoo!
    • 3 years ago
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    Well..)) .welcome

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