# What is x if x^2+(1/x^2)+x+(1/x)=4 ?

x^2 + ( 1/x^2 ) + x + 1/x = 4

First we will re-write into two parts.

==> ( x^2 + 1/x^2 ) + ( x+ 1/x) = 4

Let x + 1/x = y..............(1)

==> (x+ 1/x)^2 = y^2

=> x^2 + 2 + 1/x^2 = y^2

==> x^2 + 1/x^2 = y^2 - 2............(2)

Now we will substitute ( 1) and (2).

==> (y^2 - 2 ) + ( y) = 4

==> y^2 + y - 2 = 4

==> y^2 + y - 6  = 0

==> ( y + 3) ( y-2) = 0

==> y1= -3

==> x + 1/x  = -3

==> x^2 + 1 = -3x

==> x^2 + 3x + 1 = 0

==> x1 = ( -3 + sqrt(5) /2

==> x2= ( -3-sqrt5)/2

==> y2= 2

==> x+ 1/x = 2

==> x^2 + 1 = 2x

==> x^2  -2x + 1 =0

==> ( x-1)^2 = 0

==> x= 1

Then x values are:

x = { 1, (-3+sqrt5)/2 , (-3-sqrt5)/2}

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We have to solve for x given that x^2+(1/x^2)+x+(1/x) = 4

Now x^2+(1/x^2)+x+(1/x) = 4

Let x + 1/x = y

=> x^2 + 1/x^2 + 2 = y^2

=> x^2 + 1/x^2 = y^2 -2

Therefore x^2+(1/x^2)+x+(1/x) = 4

=> y^2 - 2 + y = 4

=> y^2 + y - 6 = 0

=> y^2 + 3y - 2y - 6 =0

=> y(y + 3) - 2(y + 3) =0

=> (y - 2)(y + 3) =0

So y = 2 and -3

As x + 1/x = y

=> x + 1/x = 2

=> x^2 - 2x + 1 = 0

=> (x - 1)^2 = 0

=> x - 1 = 0

=> x = 1

and x + 1/x = -3

=> x^2 + 3x + 1 = 0

=> x1 = [-3 + sqrt(9 - 4)]/2 = -3/2 + (sqrt 5) / 2

=> x2 = [-3 - sqrt(9 - 4)]/2 = -3/2 - (sqrt 5) / 2

Therefore the roots are 1 , -3/2+(sqrt 5)/2 and -3/2-(sqrt 5)/2.

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