Question 11
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4( x - 1/x )^2 -4(x + 1/x) +1 =0
(x - 1/x)^2 = ( x + 1/x)^2 -4
use this ,
4( x + 1/x )^2 -16 + 4(x + 1/x ) +1 =0
=> 4( x +1/x )^2 -4( x + 1/x ) -15 =0
let (x +1/x) = r
then,
4r^2 -4r -15 =0
4r ^2 -10r + 6r -15 =0
2r( 2r -5) +3( 2r -5) =0
( 2r +3)(2r -5) =0
r = -3/2 and 5/2
now,
x + 1/x =5/2
2x^2 +2 =5x
2x^2 -5x +2 =0
2x ^2-4x -x + 2=0
(x -2)(2x-1) =0
x =2 , 1/2
again ,
x + 1/x = -3/2
2x^2 +3x +2 =0
D =(3)^2 -4(4) <0
so no real solution of this
hence,
x = 2, 1/2
(x - 1/x)^2 = ( x + 1/x)^2 -4
use this ,
4( x + 1/x )^2 -16 + 4(x + 1/x ) +1 =0
=> 4( x +1/x )^2 -4( x + 1/x ) -15 =0
let (x +1/x) = r
then,
4r^2 -4r -15 =0
4r ^2 -10r + 6r -15 =0
2r( 2r -5) +3( 2r -5) =0
( 2r +3)(2r -5) =0
r = -3/2 and 5/2
now,
x + 1/x =5/2
2x^2 +2 =5x
2x^2 -5x +2 =0
2x ^2-4x -x + 2=0
(x -2)(2x-1) =0
x =2 , 1/2
again ,
x + 1/x = -3/2
2x^2 +3x +2 =0
D =(3)^2 -4(4) <0
so no real solution of this
hence,
x = 2, 1/2
abhi178:
(x-1/x)^2 =(x+1/x)^2-4
how
4{(x+1/x)^2-4} =4(x+1/x)^2-16
why -4
use formula (a-b)^2 =(a+b)^2-4ab
put in place of a =x and b =1/x
you see 4.x.1/x =4
In which class are u in?
hey
ABHI178
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