5x^4-26x^2+5=0/ x1=5,x2=0,2?

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To find the remaining roots, we need to factorize the given equation.

Starting from the original equation:
5x^4 - 26x^2 + 5 = 0

Let y = x^2, then we get:
5y^2 - 26y + 5 = 0

Using the quadratic formula to solve for y:
y = [26 ± sqrt((-26)^2 - 455)] / 2*5
y = [26 ± sqrt(676 - 100)] / 10
y = [26 ± sqrt(576)] / 10
y = [26 ± 24] / 10

So, y = 5 or y = 1/5

Now substitute back y = x^2:
For y = 5:
x^2 = 5
x = ±√5

For y = 1/5:
x^2 = 1/5
x = ±√(1/5)
x = ±1/√5
x = ±√5/5

Therefore, the roots of the given equation are: x = 5, x = -5, x = √5, x = -√5, x = √5/5, x = -√5/5.