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Find the number of real roots of x^6 - 2x^4 - x^2 + 2 = 0.

 Jun 8, 2020
 #1
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Solve for x:
x^6 - 2 x^4 - x^2 + 2 = 0

The left hand side factors into a product with four terms:
(x - 1) (x + 1) (x^2 - 2) (x^2 + 1) = 0

Split into four equations:
x - 1 = 0 or x + 1 = 0 or x^2 - 2 = 0 or x^2 + 1 = 0

Add 1 to both sides:
x = 1 or x + 1 = 0 or x^2 - 2 = 0 or x^2 + 1 = 0

Subtract 1 from both sides:
x = 1 or x = -1 or x^2 - 2 = 0 or x^2 + 1 = 0

Add 2 to both sides:
x = 1 or x = -1 or x^2 = 2 or x^2 + 1 = 0

Take the square root of both sides:
x = 1 or x = -1 or x = sqrt(2) or x = -sqrt(2) or x^2 + 1 = 0

Subtract 1 from both sides:
x = 1 or x = -1 or x = sqrt(2) or x = -sqrt(2) or x^2 = -1

Take the square root of both sides:

 x = 1     or      x = -1      or      x = sqrt(2)      or      x = -sqrt(2) 

 Jun 9, 2020

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