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Math Help - Solve x for log (98+sqrt(x^3-x^2+12x+36))=2

  1. #16
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    Re: Solve x for log (98+sqrt(x^3-x^2+12x+36))=2

    Quote Originally Posted by SlipEternal View Post
    Suppose the original problem was

    \log_{10}(98+\sqrt{x^3-x^2-12x+36}) = 2

    (The coefficient of x is -12 instead of 12)

    Then the problem simplifies to

    x^3-x^2-12x+32 = 0

    This factors as:

    (x+4)(x^2-5x+8) = 0

    That has x=-4 as a solution.
    SSSSSSSSSS, I missed a small thing, I wrote +12x, instead of -12x.

    I think It is the reason .

    But tell me how did you take (x+4)(x^2-5x+8) = 0 to x= -4
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  2. #17
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    Re: Solve x for log (98+sqrt(x^3-x^2+12x+36))=2

    Quote Originally Posted by NameIsHidden View Post
    SSSSSSSSSS, I missed a small thing, I wrote +12x, instead of -12x.

    I think It is the reason .

    But tell me how did you take (x+4)(x^2-5x+8) = 0 to x= -4
    what value of x would make the (x+4) term equal 0?

    What is 0 times anything?
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  3. #18
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    Re: Solve x for log (98+sqrt(x^3-x^2+12x+36))=2

    Quote Originally Posted by romsek View Post
    what value of x would make the (x+4) term equal 0?

    What is 0 times anything?
    Sorry, I forgot that.

    Either x+4=0

    or

     x^2-5x+8=0


    If x+4=0, then we know x=-4.

    What if  x^2-5x+8=0. x is an imaginary number. How did it become -4.

    My maths teacher also did like this for one question.

    It's something like (An expression)X(Another expression)=0.

    He made the solution only if first expression=0. He did not make the solution if another equation is 0.
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  4. #19
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    Re: Solve x for log (98+sqrt(x^3-x^2+12x+36))=2

    A variable can only take one value at a time. If x=-4, the whole product becomes zero. That means it is a solution. The second equation cannot be solved over the reals, so the equation only has one real solution.
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  5. #20
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    Re: Solve x for log (98+sqrt(x^3-x^2+12x+36))=2

    Quote Originally Posted by NameIsHidden View Post
    Sorry, I forgot that.

    Either x+4=0

    or

     x^2-5x+8=0


    If x+4=0, then we know x=-4.

    What if  x^2-5x+8=0. x is an imaginary number. How did it become -4.

    My maths teacher also did like this for one question.

    It's something like (An expression)X(Another expression)=0.

    He made the solution only if first expression=0. He did not make the solution if another equation is 0.
    if the solution to $x^2-5x+8=0$ then clearly $x\neq -4$ as $(-4)^2-5(-4)+8=44$

    The solution is satisfied if either factor is 0, so in this case either

    $x=-4$ OR $x=\dfrac 1 2 \left(5 \pm \imath \sqrt 7\right)$

    your professor probably ignored the other factor when he/she did what they did because they were only interested in real solutions. I don't know, I wasn't there.
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