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? asked in Science & MathematicsMathematics · 2 months ago

Math: Algebra?

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

5 Answers

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  • 2 months ago
    Favorite Answer

    2x⁴ - 16x² - 2 = 0

    2.(x⁴ - 8x² - 1) = 0

    x⁴ - 8x² - 1 = 0

    x⁴ - 8x² = 1

    x⁴ - 8x² + 16 = 1 + 16

    (x² - 4)² = 17

    x² - 4 = ± √17

    x² = 4 ± √17 → a square cannot be negative

    x² = 4 + √17

    x = ± √(4 + √17)

  • 2 months ago

    First, remove the common factors

    ==> x^4 - 8x^2 - 1= 0

    If you look closely at this, you will see that it could be considered a quadratic equation using x^2 and the base, rather than just x

    so, substitute u for x^2

    ==> u^2 - 8u - 1 = 0

    now solve for u

    u = (8 +/- sqrt(64 + 4)/2

    ==> u = (8 +/- sqrt(68)/2

    ==> u = 4 + sqrt(17) or 4 - sqrt(17)

    x = sqrt(u)

    ==> x = +/- sqrt(4 + sqrt(17)) or +/- sqrt(4 - sqrt(17))

    Note that sqrt(4 -sqrt(17)) is a complex number (sqrt(17)> 4)

    so +/- sqrt(sqrt(17)-4)i where i = sqrt(-1)

  • ?
    Lv 7
    2 months ago

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

    2 (x^2 - 4)^2 - 34 = 0

    Real solutions:

    x = -sqrt(4 + sqrt(17))

    x = sqrt(4 + sqrt(17))

    Complex solutions:

    x = -i sqrt(sqrt(17) - 4)

    x = i sqrt(sqrt(17) - 4)

  • 2 months ago

    Presuming you want to solve for x:

    2x⁴ - 16x² - 2 = 0

    If we make this substitution:

    z = x²

    we get:

    2z² - 16z - 2 = 0

    We now have a quadratic that we can solve.  Let's simplify this by dividing both sides by 2, then I'll complete the square:

    z² - 8z - 1 = 0

    z² - 8z = 1

    z² - 8z + 16 = 1 + 16

    (z - 4)² = 17

    z - 4 = ± √17

    z = 4 ± √17

    Now we can substitute back the expression in terms of x and solve for x:

    x² = 4 ± √17

    x = ± √(4 ± √17)

    So we have four total roots.  Two of them are complex as 4 - √17 is negative.  The other two are real.

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  • 2 months ago

    what is the question?

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