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### 5 Answers

- PinkgreenLv 71 month ago
6x^4-16x^2-2=0 has 2 distinct real & 2 non-real

roots which are

x=1.669221

x=-1.669221

x=+/- 0.34588i

approximately.

Check:

6(x^2)^2-16(x^2)-2=0

=>

3(x^2)^2-8(x^2)-1=0

=>

x^2=[8+/-sqr(64+12)]/6

=>

x^2=2.786299648

or

x^2=-0.11963298

=>

x=+/-1.66922127

or

x=+/-0.345880008i

valid.

- KrishnamurthyLv 71 month ago
6x^4 - 16x^2 - 2 = 0

6 (x^2 - 4/3)^2 - 38/3 = 0

Real solutions:

x = -sqrt(4/3 + sqrt(19)/3)

x = sqrt(4/3 + sqrt(19)/3)

Complex solutions:

x = -i sqrt(1/3 (sqrt(19) - 4))

x = i sqrt(1/3 (sqrt(19) - 4))

- la consoleLv 71 month ago
6x⁴ - 16x² - 2 = 0

6.[x⁴ - (8/3).x² - (1/3)] = 0

x⁴ - (8/3).x² - (1/3) = 0

x⁴ - (8/3).x² = (1/3)

x⁴ - (8/3).x² + (4/3)² = (1/3) + (4/3)²

x⁴ - (8/3).x² + (4/3)² = 19/9

[x² - (4/3)]² = 19/9

x² - (4/3) = ± √(19/9)

x² = (4/3) ± [√(19/9)]

x² = (4/3) ± [(1/3).√19]

x² = (4 ± √19)/3 → a square cannot be negative

x² = (4 + √19)/3

x = ± √[(4 + √19)/3]

- ?Lv 71 month ago
Hint: realize that is this is a second-degree polynomial. A quadratic.

Indeed, let x² = t.

What conclude from that?

Answer that then we will take it from there if need be.

Don't forget to vote me best answer for being the first to correctly walk you through without spoiling the answer. That way it gives you a chance to work at it and forget good at it.

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- Daniel HLv 51 month ago
quadratic but

x^2 = (-b^2 +/- sqrt(b^2 - 4ac)/(2a)

x^2 = (16 +/- sqrt(256 + 48)/(12)

x^2 = 4/3 +/- sqrt(19)/3

x = +/- sqrt(4/3 - sqrt(19)/3) is invalid

x = +/- sqrt(4/3 + sqrt(19)/3)