Algebra: Math?

2x^4+20x^2−2=0

?2021-03-01T11:59:01Z

We can first divide through by 2 to get:

x⁴ + 10x² - 1 = 0

i.e. (x²)² + 10x² - 1 = 0

Using the quadratic formula we have:

x² = [-10 ± √(10² - 4(1)(-1))]/2

so, x² = [-10 ± √104]/2

or, x² = [-10 ± 2√26]/2

Then, x² = -5 ± √26

Now, x² is positive, so x² = -5 + √26

so, x² = 0.0990

Hence, x =  ±0.315

:)>

Nestor2021-03-01T10:16:31Z

Let X = x², then 2X² + 20X - 2 = 0

D = 20² - 4 * 2 * (-2) = 416 = (16 sqrt(26))²

then X1 = (-20 - 16sqrt(26)) / 4 impossible

X2 = (-20 + 16sqrt(26)) / 4 = -5 + 4sqrt(26)

and now x = sqrt(-5 + 4sqrt(26)) or x = -sqrt(-5 + 4sqrt(26))

la console2021-03-01T10:15:28Z

2x⁴ + 20x² - 2 = 0

2.(x⁴ + 10x² - 1) = 0

x⁴ + 10x² - 1 = 0

x⁴ + 10x² = 1

x⁴ + 10x² + 25 = 1 + 25

(x² + 5)² = 26

x² + 5 = ± √26

x² = - 5 ± √26 → a square cannot be negative

x² = - 5 + √26

x = ± √(- 5 + √26)

x ≈ 0.3146736

Bryce2021-03-01T09:43:39Z

x= -0.3147, 0.3147 (Solve graphically.)