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

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

    Solve x for \log (98+\sqrt{x^3-x^2+12x+36})=2

    I thought of taking to 10^2=98+\sqrt{x^3-x^2+12x+36}



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



    \Rightarrow 2=\sqrt{x^3-x^2+12x+36}



    \Rightarrow 4=x^3-x^2+12x+36



    \Rightarrow -32=x^3-x^2+12x





    I am just 9 th grader. I have no idea about solving a cubic equation.

    I also think that there should be a way to solve it with the help of logarithmic identities.
    Last edited by NameIsHidden; May 15th 2014 at 04:18 AM.
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    Re: Solve x for log (98+sqrt(x^3-x^2+12x+36))=2

    I agree up to $\displaystyle \begin{align*} 4 = x^3 - x^2 + 12x + 36 \end{align*}$. Then you need to make use of the null factor law, which states if the product of two terms is 0, then at least one of them is 0.

    So you need to set your equation equal to 0 and factorise it. Then you can set each factor equal to 0 and solve for x.
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    Re: Solve x for log (98+sqrt(x^3-x^2+12x+36))=2

    You can also use the rational root theorem: "If the rational number \frac{m}{n} is a root of the polynomial a_nx^n+ a_{n-1}x^{n-1}\cdot\cdot\cdot+ a_1x+ a_0= 0, then n evenly divides the leading coefficient, a_n, and m evenly divides the constant term, a_0."

    Of course, there is no guarantee that a polynomial equation has a rational root but if it does, that limits the possible roots. Here, your equation is [tex]x^3- x^2+ 12x+ 32= 0[tex]. The leading coefficient is 1 so the denominator of any rational root must evenly divide 1- i.e. must be \pm 1 so any rational root must be an integer. The constant term is 32 so if there is a rational root it must be \pm 1, [tex]\pm 2[tex], \pm 4, \pm 8, \pm 16, or \pm 32. Try those and see if any work.
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    Re: Solve x for log (98+sqrt(x^3-x^2+12x+36))=2

    Quote Originally Posted by Prove It View Post
    I agree up to $\displaystyle \begin{align*} 4 = x^3 - x^2 + 12x + 36 \end{align*}$. Then you need to make use of the null factor law, which states if the product of two terms is 0, then at least one of them is 0.

    So you need to set your equation equal to 0 and factorise it. Then you can set each factor equal to 0 and solve for x.
    Okay I take it to 0=x^3-x^2+12x+32. Factorising it,

    \\0=x^3-x^2+12x+32\\<br /> <br />
0=x^2(x-1)+4(3x+8)<br />

    What to do then?? How can I make it to satisfy null factor law.
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    Re: Solve x for log (98+sqrt(x^3-x^2+12x+36))=2

    Quote Originally Posted by HallsofIvy View Post
    You can also use the rational root theorem: "If the rational number \frac{m}{n} is a root of the polynomial a_nx^n+ a_{n-1}x^{n-1}\cdot\cdot\cdot+ a_1x+ a_0= 0, then n evenly divides the leading coefficient, a_n, and m evenly divides the constant term, a_0."

    Of course, there is no guarantee that a polynomial equation has a rational root but if it does, that limits the possible roots. Here, your equation is [tex]x^3- x^2+ 12x+ 32= 0[tex]. The leading coefficient is 1 so the denominator of any rational root must evenly divide 1- i.e. must be \pm 1 so any rational root must be an integer. The constant term is 32 so if there is a rational root it must be \pm 1, [tex]\pm 2[tex], \pm 4, \pm 8, \pm 16, or \pm 32. Try those and see if any work.
    The chapter is about surds, indices and logarithms. Polynomials are dealt in next chapter. So the solution has to be related to surds, indices and logarithms.
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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
    Okay I take it to 0=x^3-x^2+12x+32. Factorising it,

    \\0=x^3-x^2+12x+32\\<br /> <br />
0=x^2(x-1)+4(3x+8)<br />

    What to do then?? How can I make it to satisfy null factor law.
    That is not factorised. Like HallsofIvy advised, substitute numerical factors of 32 into the polynomial. If one of them (call it "a") gives the polynomial equalling 0, then that means (x - a) is a factor.
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    Re: Solve x for log (98+sqrt(x^3-x^2+12x+36))=2

    Quote Originally Posted by Prove It View Post
    That is not factorised. Like HallsofIvy advised, substitute numerical factors of 32 into the polynomial. If one of them (call it "a") gives the polynomial equalling 0, then that means (x - a) is a factor.
    I am not taught about rational root of a polynomial. I saw in my book that there is a thing called factor theorem. But I am not taught about it.

    I can neither understand what HallsofIvy told. Can anyone explain me what he told?
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    Re: Solve x for log (98+sqrt(x^3-x^2+12x+36))=2

    @prove it, please tell me am I right till x^3-x^2+12x+32=0.
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    Re: Solve x for log (98+sqrt(x^3-x^2+12x+36))=2

    For the cubic equation you are solving you should be able to factorize it so that it looks like
    (x-a)(x-b)(x-c)=0

    When you expand that you will see that -abc is the only term that is not multiplied by x or x2 or x3
    In your equation the only term not multiplied by x or x2 or x3 is 32 so -abc=32
    If a, b and c are whole numbers (which is not always true but usually is for grade 9 maths) then in order for them to multiply together to give 32 they have to be the factors of 32. The only factors of 32 are: 1, -1, 2, -2, 4, -4, 8, -8, 16, -16, 32, -32 so if you try putting those into the equation you will find some of them which make the equation equal to zero. If you tried 4, for example, and found that putting that into the equation gave an answer of 0 then you know that 4 is a solution of the equation and (x-4) is a factor.
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    Re: Solve x for log (98+sqrt(x^3-x^2+12x+36))=2

    Quote Originally Posted by Shakarri View Post
    For the cubic equation you are solving you should be able to factorize it so that it looks like
    (x-a)(x-b)(x-c)=0

    When you expand that you will see that -abc is the only term that is not multiplied by x or x2 or x3
    In your equation the only term not multiplied by x or x2 or x3 is 32 so -abc=32
    If a, b and c are whole numbers (which is not always true but usually is for grade 9 maths) then in order for them to multiply together to give 32 they have to be the factors of 32. The only factors of 32 are: 1, -1, 2, -2, 4, -4, 8, -8, 16, -16, 32, -32 so if you try putting those into the equation you will find some of them which make the equation equal to zero. If you tried 4, for example, and found that putting that into the equation gave an answer of 0 then you know that 4 is a solution of the equation and (x-4) is a factor.
    I will consider it. My portion includes upto quadratic equations only. So, there must be a way to solve it with logarithms instead of cubic equation.
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    Re: Solve x for log (98+sqrt(x^3-x^2+12x+36))=2

    That is the usual way which pre-university students use to solve cubic equations. I have never heard of using logarithms to solve a cubic equation. There is a way to exactly calculate the roots of a cubic equation but it is horribly complicated



    For a cubic equation in the form ax^3+ bx^2+ cx+d=0
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    Question Re: Solve x for log (98+sqrt(x^3-x^2+12x+36))=2

    Quote Originally Posted by Shakarri View Post
    That is the usual way which pre-university students use to solve cubic equations. I have never heard of using logarithms to solve a cubic equation. There is a way to exactly calculate the roots of a cubic equation but it is horribly complicated



    For a cubic equation in the form ax^3+ bx^2+ cx+d=0
    Thanks for the formula, I have been Google-ing this for several hours.

    I will now calculate it easily. I will work it out.

    1 doubt. Isn't \sqrt{x^2}=\pm x But you gave + in one term and - in another term.
    Last edited by NameIsHidden; May 15th 2014 at 08:04 AM.
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    Re: Solve x for log (98+sqrt(x^3-x^2+12x+36))=2

    Quote Originally Posted by Shakarri View Post
    That is the usual way which pre-university students use to solve cubic equations. I have never heard of using logarithms to solve a cubic equation. There is a way to exactly calculate the roots of a cubic equation but it is horribly complicated



    For a cubic equation in the form ax^3+ bx^2+ cx+d=0


    I should get 3 roots. Shouldn't I?

    But I get only one answer which says x \approx 5.2.......


    I should get x=-4.
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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
    I should get 3 roots. Shouldn't I?

    But I get only one answer which says x \approx 5.2.......


    I should get x=-4.
    A cubic with real coefficients may have 1, 2, or 3 distinct real roots. This particular cubic has only 1.

    And that root is not approximately 5.2

    $5.2^3 - 5.2^2 + 12 * 5.2 + 32 = 140.608 - 27.04 + 62.4 + 32 >> 0.$

    In fact, the solution is an irrational number between -1 and -2.

    $(-1)^3 - (-1)^2 + 12 * (-1) +32 = -1 - 1 - 12 + 32 = 18.$

    $(-2)^3 - (-2)^2 + 12 * (-2) +32 = -8 - 4 - 24 + 32 = - 4.$

    And so the answer is not - 4.

    This seems like a very strange problem to give to someone who has not studied polynomials. I wonder whether once again you have not given us the original problem.
    Last edited by JeffM; May 15th 2014 at 11:40 AM.
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    Re: Solve x for log (98+sqrt(x^3-x^2+12x+36))=2

    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.
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