college algebra

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Solve, what is the easiest way to solve this equation?

9x^4+4=13x^2 and

Thanks, in advance..

  • college algebra -

    I am solving the first one.....second one you can try yourself.

    so 9x^4+4=13x^2
    this can be written,

    as x^2 is always +ve
    so to be -ve,

    we see that
    it comes as -1.2<x<1.2

    next we will see if there is some integer solution.
    at x=1 and x=-1
    the function has its value 0
    so x=1 and x=-1 are solution to the equation.

    next we will see if there is some maxima or minima in this func,(9x^4+4-13x^2)

    after differentiating,
    and taking it =0,
    the func. has minima at +sqrt(13/18), -sqrt(13/18)

    and at x=+1, x=-1 it has value 0,
    so the another solutions are
    x=1-2*sqrt(13/18) and x=-1+2*sqrt(13/18)

    these are 4 solutions.
    Ask me if you have any doubt.

  • college algebra -

    There are two equations there.
    In the first one, let y = x^2 and solve the quadratic.
    9y^2 -13y +4 = 0
    (y -1)(9y-4) = 0
    y = 1 or 4/9
    x = sqrt(y) = +/-1 and +/-2/3

    In the second one, let y = x^3 and again solve the quadratic.

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