x^4-10x^2+1=0?

x⁴ - 10x² + 1 = 0
x⁴ - 10x² = -1
x⁴ - 10x² + 25 = -1 + 25
(x² - 5)² = 24
x² - 5 = ±√24
x² = 5 ± 2√6
x = ±√(5 ± 2√6)

Let √(5 ± 2√6) = √a ± √b where a > b
(5 ± 2√6)]² = (√a ± √b)²
5 ± 2√6 = (a + b) ± 2√ab
Then, a = 3 and b = 2
√(5 ± 2√6) = √3 ± √2

Hence,
x = √3 + √2, -√3 - √2, √3 - √2, -√3 + √2

x⁴ - 10x² + 1 = 0

Since we have exponents of 4 and 2, we can make this substitution to turn it into a quadratic:

y = x²

So now we have:

y² - 10y + 1 = 0

Now we solve this quadratic. I'll complete the square:

y² - 10y = -1

Add 25 to both sides:

y² - 10y + 25 = 24

factor, then square root:

(y - 5)² = 24
y - 5 = ± √24

simplify:

y = 5 ± √24
y = 5 ± 2√6

But we aren't asked to solve for y, but for x, so let's put the x² back in and solve for x:

x² = 5 ± 2√6
x = ± √(5 ± 2√6)

Doing a sanity check on if 5 - 2√6 is negative (if it is we can take it out of the solution set if we are only interested in real values), but it is positive, barely: 2√6 ≈ 4.898979, so subtracting that from 5 you still have a positive value).

So we have 4 solutions with a square root of a square root:

x = ± √(5 ± 2√6)

Let u = x^2, then
u^2 - 10u + 1 = 0;
use the quadratic formula --
u = 5 +/- (1/2)sqrt(96)
= 5 +/- (1/2)sqrt(16*6)
= 5 +/- 2*sqrt(6).

So x = +/- sqrt[5 +/- 2*sqrt(6)].
Approximate decimal solutions are
+/- 3.1463 and +/- 0.31784.