find all real roots:
x^4-6x^2+8=0
and the other polynomial is:
8x^3+1=0
2 answers
for the first one, all real roots are 2, -2, and the square root of 2. i'm not sure of the second one. hope this helps.
For the first one :
You put X = x²
Then you have (variable change) :
X² - 6X + 8 = 0
And you use common tools to solve that (use delta). You find X = 2 or X = 4.
You have x² = X, so x = sqrt(X) or x = -sqrt(X).
Here, the solutions are sqrt(2), -sqrt(2), 2, -2.
For the second one, you simply have :
x^3 = -1/8
x = sqrt3(-1/8) = -1/2
You put X = x²
Then you have (variable change) :
X² - 6X + 8 = 0
And you use common tools to solve that (use delta). You find X = 2 or X = 4.
You have x² = X, so x = sqrt(X) or x = -sqrt(X).
Here, the solutions are sqrt(2), -sqrt(2), 2, -2.
For the second one, you simply have :
x^3 = -1/8
x = sqrt3(-1/8) = -1/2