Algebra: Math?
2x^4+20x^2−2=0
2x^4+20x^2−2=0
?
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
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Nestor
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 console
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
Bryce
x= -0.3147, 0.3147 (Solve graphically.)