jiteshmeghwal9
If \(\LARGE{x^2=\frac{1}{x^2}+1}\), then \(\LARGE{x^2+\frac{1}{x^2}=?}\)
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jiteshmeghwal9
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@hartnn @ParthKohli @ajprincess plz help :)
ParthKohli
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\[\dfrac{1}{x^2} = x^2 - 1 \ \ \ \ \]So,\[x^2 + x^2 - 1 = 2x^2 -1\]
ParthKohli
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Should it be a number value?
jiteshmeghwal9
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\[\Huge{\color{red}{Ans.-\sqrt{5}}}\]
Yahoo!
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x^4 = x^2 + 1
x^4 - x^2 - 1 = 0
put y = x^2
jiteshmeghwal9
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wait..
ParthKohli
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That's a quartic equation in \(x\). So multiply through \(x^2\).\[x^4 = x^2 + 1\]
jiteshmeghwal9
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from where do u gt x^4?
ParthKohli
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I had typed a long explanation, but Yahoo! spilled the beans =P
Yahoo!
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Solve the given Equation @jiteshmeghwal9 x^2 = 1 + 1/x^2
ParthKohli
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\(x^2 = 1 + \dfrac{1}{x^2}\) is what you have. Multiply \(x^2\) to both sides.
jiteshmeghwal9
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x^4=x^2+1
ParthKohli
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That...
jiteshmeghwal9
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Oops sorry i misread it
jiteshmeghwal9
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AArgh.... I'm going to be crazy :(
jiteshmeghwal9
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Sorry dudes but i'm nt getting the right thing :(
jiteshmeghwal9
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Need help
rahul91
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x^4-x^2-1=0
put x^2=t
t^2-t-1=0 solve it
jiteshmeghwal9
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using quadratic formula ?
rahul91
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yep
jiteshmeghwal9
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\[{1 \pm \sqrt{1+4}\over2}={1 \pm \sqrt{5}\over2}\]
jiteshmeghwal9
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No.. i'm nt getting answer
rahul91
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substitute for x^2 in the expression needed
Yahoo!
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Hm..Hm...After Solving i got sqrt5
jiteshmeghwal9
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can anybody give me steps to the solution, i'm getting confused :((((
rahul91
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hey substitute x^2= (1-sqrt(5))/2 in the expression u get the answer -sqrt(5)
jiteshmeghwal9
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ok
Yahoo!
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@jiteshmeghwal9 ..Can u Check...ur Solution
Yahoo!
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Is it sqrt5 or -sqrt5
Yahoo!
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x^2 + 1 / x^2 = (x^4 + 1) / x^2
ParthKohli
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\[x^4 -x^2 - 1 = 0 \]We can let \(t = x^2\).\[t^2 - t - 1 = 0 \iff \boxed{t = \dfrac{1\pm \sqrt 5}{2}}{}\]
Yahoo!
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x^4=x^2+1
(x^2+2) / x^2 = 1 + 2/x^2
sirm3d
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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
Yahoo!
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nw put the value of x^2
jiteshmeghwal9
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well \(\sqrt{5}=\pm\sqrt{5}\) but in my book it is only given \(\sqrt{5}\) as answer :)
Yahoo!
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Lol....Then...Go By My method
jiteshmeghwal9
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ohh! k
Yahoo!
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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
Yahoo!
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Nw u Got it...:)
jiteshmeghwal9
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A little doubt here, just last doubt
Yahoo!
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Wat?@jiteshmeghwal9
jiteshmeghwal9
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(x^2+2) / x^2 = 1 + 2/x^2 how did u gt this ?
sirm3d
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\[x^2=\frac{1+\sqrt 5}{2}\\\frac{1}{x^2}=\frac{2}{1+\sqrt 5}\cdot\frac{1-\sqrt 5}{1-\sqrt 5}=-\frac{1}{2}+\frac{\sqrt 5}{2}\\x^2+\frac{1}{x^2}=(1/2+\sqrt 5/2)+(-1/2+\sqrt 5/2)=\sqrt {5}\]
ParthKohli
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\[\dfrac{x^2 + 2}{x^2} = \dfrac{x^2}{x^2} + \dfrac{2}{x^2}\cdots\]
jiteshmeghwal9
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I m just gt confused just like my bro @maheshmeghwal9 :(
Yahoo!
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Oh...i just Did to make calculation easy....if u dont get it...Go My The method used by sirm
jiteshmeghwal9
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Well...now i gt it, thanx all of u a lot :)
Mainly @Yahoo! thanx buddy ;)
Yahoo!
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Well..)) .welcome