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Thread: solving equations

  1. #1
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    solving equations

    Solve the following algebraically
    $\displaystyle a)x^3+5x^2=2x+6$
    $\displaystyle b)x^4-13x^2=-36$
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  2. #2
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    Quote Originally Posted by william View Post
    Solve the following algebraically
    $\displaystyle a)x^3+5x^2=2x+6$
    $\displaystyle b)x^4-13x^2=-36$
    a) $\displaystyle x^3 + 5x^2 - 2x - 6 = 0$

    Notice that $\displaystyle f(-1) = (-1)^3 + 5(-1)^2 - 2(-1) - 6 = -1 + 5 + 2 - 6 = 0$

    So by the factor and remainder theorems, $\displaystyle x + 1$ is a factor.

    Using long division, you should find that

    $\displaystyle f(x) = (x + 1)(x^2 + 4x - 6) = 0$

    By the null factor law

    $\displaystyle x + 1 = 0$ or $\displaystyle x^2 + 4x - 6 = 0$

    The first equation is easy to solve, the second is solved using the Quadratic Formula.

    So $\displaystyle x = -2 - \sqrt{10}, x = -1, x = -2 + \sqrt{10}$ are the solutions.
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  3. #3
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    Quote Originally Posted by william View Post
    Solve the following algebraically
    $\displaystyle a)x^3+5x^2=2x+6$
    $\displaystyle b)x^4-13x^2=-36$
    b) $\displaystyle x^4 - 13x^2 + 36 = 0$

    Notice that if I write $\displaystyle X = x^2$ the equation becomes

    $\displaystyle X^2 - 13X + 36 = 0$

    $\displaystyle (X - 4)(X - 9) = 0$

    $\displaystyle X = 4$ or $\displaystyle X = 9$.


    Since $\displaystyle X = x^2$

    $\displaystyle x^2 = 4$ or $\displaystyle x^2 = 9$

    So $\displaystyle x = -3, x = -2, x = 2, x = 3$ are the solutions.
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