(x^2 + 1/x^2) + (x + 1/x) - 4 = 0
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Answer:
Step-by-step explanation:
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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