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Math Algebra?

6x^4-16x^2-2=0

5 Answers

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  • 1 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.

  • 1 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))

  • 1 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 7
    1 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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  • 1 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)

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