jiteshmeghwal9
  • jiteshmeghwal9
If \(\LARGE{x^2=\frac{1}{x^2}+1}\), then \(\LARGE{x^2+\frac{1}{x^2}=?}\)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
jiteshmeghwal9
  • jiteshmeghwal9
@hartnn @ParthKohli @ajprincess plz help :)
ParthKohli
  • ParthKohli
\[\dfrac{1}{x^2} = x^2 - 1 \ \ \ \ \]So,\[x^2 + x^2 - 1 = 2x^2 -1\]
ParthKohli
  • ParthKohli
Should it be a number value?

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More answers

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

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