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

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### 4 Answers

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. **

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}**

To find x . if x^2+(1/x^2)+x+(1/x)=4 ?

x^2+1/x^2 +x+1/x = 4.

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

t^2+t -2 = 4, where t = x+1/x.

t^2+t-6 = 0.

(t+3)(t-2) = 0.

t = -3 or t = 2.

t-2 = 0 gives x+1/x = 2.

Or x+1/x-2 = 0.

Multiply by x:

x^2-2x+1 = 0.

(x-1)^2 = 0

x-1 = 0.

x=1.

t+3 = 0 gives x+1/x+3 = 0.

Multiply by x:

x^2+3x+1 = 0.

We get the roots by using the quadratic formula.

x = (-3+sqrt(9^2-4*1*1)}/ 2 = (-3+sqrt5)/2.

Or x= {-3-sqrt5}/2.

To solve the equation, we'll substitute the sum of terms x + (1/x) = t.

To get x^2 + (1/x^2) = (x + 1/x)^2 - 2

x^2 + (1/x^2) = t^2 - 2

We'll re-write the equation in t:

t^2 - 2 + t = 4

We'll subtract 4 both sides:

t^2 - 2 + t - 4 = 0

We'll combine like terms:

t^2 + t - 6 = 0

We'll apply quadratic formula:

t1 = [-1 + sqrt(1 + 24)]/2

t1 = (-1+5)/2

t1 = 2

t2 = -3

But x + 1/x = t1

We'll substitute t1:

x + 1/x = 2

We'll multiply by x:

x^2 - 2x + 1 = 0

(x-1)^2 = 0

x1 = x2 =1

We'll put x + 1/x = t2:

x + 1/x = -3

We'll multiply by x:

x^2 + 3x + 1 = 0

x3 = [-3+sqrt(9-4)]/2

x3 = (-3+sqrt5)/2

x4 = (-3-sqrt5)/2

**The 4 solutions of the equation are: {(-3-sqrt5)/2 ; (-3+sqrt5)/2; 1}.**