2006-09-24 18:42:53 · 13 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
x^2+(1/x^2)-2=6-2
(x-1/x)^2=4
x-1/x=+/-2
2006-09-24 20:19:53 · answer #1 · answered by Anonymous · 1⤊ 0⤋
2
2006-09-25 01:53:49 · answer #2 · answered by Hari 1 · 1⤊ 0⤋
(x - 1/x)^2 = x^2 + 1/x^2 -2 = 6-2 =4
==> (x - 1/x) = +/- 2
2006-09-25 06:39:55 · answer #3 · answered by cosmic_ashim 2 · 0⤊ 0⤋
x² + 1/x² = 6 (x X²)
x^4 + 1 = 6x²
x^4 - 6x² + 1 = 0
Let z = x²
z² - 6z + 1 = 0 (Now use quadratic formula).
z = 5â828 427 125 (or) z = 0â171 572 875
But z = x²
x = âz
â x = â(5â828 427 125) (or) x = â(0â171 572 875)
â x = ± 2â414 213 562 (or) x = ± 0â585 786 437
Now check values for x in the original equation.
It will be seen that x = ± 2â414 213 562
x² + 1/x² = 6
(+ 2â414 213 562)² + 1/(+ 2â414 213 562)² = 6
(5â828 427 125...) + 1/(5â828 427 125...) = 6
(5â828 427 125...) + 0â171 572 875... = 6
6 = 6 Values for x confirmed.
The square root of a minus number is positive,
so - 2â414 213 562 is also a value for x.
(x - 1/x) =
(+ 2â414 213 562...) - 1/(+ 2â414 213 562...) =
(+ 2â414 213 562...) - 0â414 213 562 = 2
(x - 1/x) = 2
(x - 1/x) =
(- 2â414 213 562...) - 1/(- 2â414 213 562...) =
(- 2â414 213 562...) + 0â414 213 562 = - 2
(x - 1/x) = - 2
(x - 1/x) = ± 2
2006-09-25 04:03:17 · answer #4 · answered by Brenmore 5 · 0⤊ 0⤋
(x-1/x)^2=(x^2+1/x^2)-2*x*1/x
=6-2*1
=4
hence,
x-1/x=4^1/2
=+2 or -2
2006-09-25 06:12:20 · answer #5 · answered by raj 1 · 0⤊ 0⤋
3.2360684 or -1.2360684
(x^2)+1/(x^2)=6
(x^2)+1=6(x^2)
1=5(x^2)
x^2=1/5
x= (1/5)^((2)^(-1)) =+0.4472135 or -0.4472135
when x=+0.4472135
x-1/x=0.4472135-1/0.4472135
= -(0.5527865/0.4472135)
= -(1.2360684)
when x=-0.4472135
x-1/x= (-(0.4472135+1)) / (-(0.4472135))
= -(1.4472135) / (- 0.4472135)
= 3.2360684
2006-09-25 02:07:38 · answer #6 · answered by geniusboy 2 · 0⤊ 1⤋
x^2 = 6 - x^-2
x = (6 - x^-2)^(1/2)
x - x^-1 = (6 - x^-2)^(1/2) - (6 - x^-2)^(-1/2)
[(6 - x^-2)^(1/2)/(6 - x^-2)^(1/2)] = 1
[(6 - x^-2)^(1/2)/(6 - x^-2)^(1/2)][(6 - x^-2)^(1/2)] - 1/[(6 - x^-2)^(1/2)]
= (6 - x^-2)/[(6 - x^-2)^(1/2)] - 1/[(6 - x^-2)^(1/2)]
= (6 - x^-2 - 1)/[(6-x^-2)^(1/2)]
= (5 - x^-2)/[(6-x^-2)^(1/2)]
2006-09-25 02:05:59 · answer #7 · answered by johnny m 2 · 0⤊ 1⤋
(x ^2) + 1/(x ^2) =6
Re-arranging
x^4-6x^2+1=0
let z=x^2
z^2-6z+1=0
Solve using quadratic formula
z= (6+32^.5) /2 or (6-32^.5) /2
since z=x^2
x= z^.5
x= ((6+32^.5) /2)^.5 or ((6-32^.5) /2).5
substitute this into
(x - 1/x)
and u will get ur answer
which is 0.585786
2006-09-25 01:44:33 · answer #8 · answered by Anonymous · 0⤊ 1⤋
(x-1/x)^2=(x^2)+1/(x^2)-2(standard formula)
substituting the above value in this formula,
(x-1/x)^2=6-2=4;
(x-1/x)=sq.root of 4=2;
so,(x-1/x)=2
2006-09-25 02:02:40 · answer #9 · answered by Anonymous · 1⤊ 0⤋
(x - 1/x)^2 =
... = x^2 - 2 + 1/x^2
... = 6 - 2 = 4
so the answer is -2 or +2.
2006-09-25 02:30:06 · answer #10 · answered by dutch_prof 4 · 1⤊ 0⤋