SOLUTION: Solve (x^2-x)^2+(x^2-x)-2=0 in exact values

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Question 990252: Solve
(x^2-x)^2+(x^2-x)-2=0 in exact values
Found 2 solutions by ikleyn, MathTherapy:
Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
.
Your equation is

%28x%5E2-x%29%5E2%2B%28x%5E2-x%29-2 = 0.

The standard and universal method for solving such equations is introducing new variable.

In our case let  y = x%5E2-x.

Then the original equation takes the form

y%5E2+%2B+y+-+2 = 0.

It is a quadratic equation.  You can easily solve it by applying the quadratic formula or the Viete's theorem.  Its roots are

y%5B1%5D = -2,   y%5B2%5D = 1.

Now we should solve two quadratic equations to find  x.  They are

x%5E2+-+x = -2         (1)

and

x%5E2+-x = 1.         (2)

The first of these two equations is

x%5E2+-+x+%2B+2 = 0.

It has no real solutions  (the discriminant  d = b%5E2-4ac = 1 - 4*2 = -7  is negative).

The second of these two equations is

x%5E2-x-1 = 0.

It has two roots   x%5B1%2C2%5D = %281+%2B-+sqrt%285%29%29%2F2.

Answer.  The given equation has two roots:  %281%2Bsqrt%285%29%29%2F2  and  %281-sqrt%285%29%29%2F2.


Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!

Solve
(x^2-x)^2+(x^2-x)-2=0 in exact values
Exact values: highlight_green%28x+=+%281+%2B-sqrt%285%29%29%2F2%29