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Math Help - Translation difficulties, Basic Integer Problem

  1. #1
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    Arrow Translation difficulties, Basic Integer Problem

    Hi MHF!

    I've been having some problems recently with word problems. I'm still not used to English, I guess.
    So:

    Consider the Integral expression in x.
    P=x^3+x^2+ax+1

    Where a=?? and the value of P is a rational number for any x which satisfies the equation x^2+2x-2=0, and in this case the value of P is:


    I'm sorry if there's any typos in there, I've copied this some time ago.

    What "the integral expression in x" means?

    \int[x^3+x^2+ax+1] or maybe d/dx [x^3+x^2+ax+1] ?

    Not having a clue on how to start, I've tried to resolve x^2+2x-2=0.
    Still with -1+(3)^(1/2) and -1-(3)^(1/2), no luck.

    P represents many numbers, right? But what number does it actually ask for?
    Is P=-1+(3)^(1/2) and -1-(3)^(1/2) ?

    (3)^(1/2) is supposed to be a root

    *completely lost*

    Thanks :3
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  2. #2
    Senior Member Sambit's Avatar
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    Quote Originally Posted by Zellator View Post
    What "the integral expression in x" means?

    \int[x^3+x^2+ax+1] or maybe d/dx [x^3+x^2+ax+1] ?
    The first one is integration (the second one is differentiation; not integration)
    Quote Originally Posted by Zellator View Post
    P represents many numbers, right? But what number does it actually ask for?
    Is P=-1+(3)^(1/2) and -1-(3)^(1/2) ?
    The roots of the quadratic equation are \frac{-2\pm\sqrt{12}}{2}
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  3. #3
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    Quote Originally Posted by Zellator View Post
    Hi MHF!

    I've been having some problems recently with word problems. I'm still not used to English, I guess.
    So:

    Consider the Integral expression in x.
    P=x^3+x^2+ax+1

    Where a=?? and the value of P is a rational number for any x which satisfies the equation x^2+2x-2=0, and in this case the value of P is:


    I'm sorry if there's any typos in there, I've copied this some time ago.

    What "the integral expression in x" means?
    The word "integral" has two completely different meanings in English. It can mean something to do with integration, or it can mean something to do with integers.

    My guess is that this is an algebra problem, not a calculus problem, and that "the integral expression in x" means "a function of x whose value is an integer". So forget about integration, and think about integers. As you correctly say, the solutions of the equation x^2+2x-2=0 are x = -1\pm\sqrt3. Plug those values into the expression P, and find the value of  a for which the answer is an integer. In other words, choose  a so that terms involving \sqrt3 cancel out and you are left with an "integral" answer.
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  4. #4
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    Thumbs up

    I'll have to revive this thread since I'm still having difficulties.

    Opalg suggested me to plug the values x = -1\pm\sqrt3 in the formula.
    I'm kind of confused, since the question says this is a value of P and not of x

    So what would be right?
    x = -1\pm\sqrt3=x^3+x^2+ax+1

    or probably
    P=(-1\pm\sqrt3)^3+(-1\pm\sqrt3)^2+a(-1\pm\sqrt3)+1
    In this case, what would happen to P?
    Would it equal zero? Would it equal -1\pm\sqrt3?

    Both cases, I still get a racional number that is nowhere near a=-4

    This sqrt3 is really annoying.
    Thanks!
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  5. #5
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    Quote Originally Posted by Zellator View Post
    I'll have to revive this thread since I'm still having difficulties.

    Opalg suggested me to plug the values x = -1\pm\sqrt3 in the formula.
    I'm kind of confused, since the question says this is a value of P and not of x

    So what would be right?
    x = -1\pm\sqrt3=x^3+x^2+ax+1

    or probably
    P=(-1\pm\sqrt3)^3+(-1\pm\sqrt3)^2+a(-1\pm\sqrt3)+1
    This second one.

    In this case, what would happen to P?
    Would it equal zero? Would it equal -1\pm\sqrt3?
    It must be equal to an integer. It might be 0 but certainly not -1\pm\sqrt{3}

    Both cases, I still get a racional number that is nowhere near a=-4
    ??? No one has said anything about a being equal to -4.

    This sqrt3 is really annoying.
    Thanks!
    If x= -1+ \sqrt{3} what is x^3? What is x^2? What is x^3+ x^2+ ax+ 1? What value of a will make that an integer?
    Last edited by HallsofIvy; May 27th 2011 at 09:46 AM.
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  6. #6
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    Thumbs up

    Hi HallsofIvy!
    hahah the a=-4 part is the answer!
    Sorry If I haven't made myself clear.

    Quote Originally Posted by HallsofIvy View Post
    If x= -1+ \sqrt{3} what is x^3? What is x^2? What is x^3+ x^2+ ax+ 1? What value of a will make that an integer?
    Right! x^3= -1+2\sqrt{3}-3 +\sqrt{3}-6+3\sqrt{3} when x= -1+ \sqrt{3}
    x^2= 1-2\sqrt{3}+3

    I'm not sure where you want to get here, sorry.
    I'm still not sure about the end of that equation.
    It also says that a is a rational number.
    This has been omited so far, I'm sorry.

    Working on the question we get.
    2\sqrt{3}-5+a(1+\sqrt{3})=??

    Ok, so if we equal this to zero and square both sides we can get
    a^2=73/4
    This gives something close to 4.23...
    Close enough, but is this conclusive?
    We still have to find the value of P when a is in our desired value.


    Edit: Can some moderator please move this topic to it's right place?
    This looks like but is not a calculus problem.
    Last edited by Zellator; May 27th 2011 at 08:58 AM. Reason: Ask for the movement of the topic
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  7. #7
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    Quote Originally Posted by Zellator View Post
    Hi HallsofIvy!
    hahah the a=-4 part is the answer!
    Sorry If I haven't made myself clear.



    Right! x^3= -1+2\sqrt{3}-3 +\sqrt{3}-6+3\sqrt{3} when x= -1+ \sqrt{3}
    x^2= 1-2\sqrt{3}+3
    Okay- x^2= 4- 2\sqrt{3} but that is not what I get for x^3. For x^3 you should have (-1+ \sqrt{3})(4- 2\sqrt{3})= -4+ 4\sqrt{3}+ 2\sqrt{3}- 6, not what you have.

    I'm not sure where you want to get here, sorry.
    I'm still not sure about the end of that equation.
    It also says that a is a rational number.
    This has been omited so far, I'm sorry.

    Working on the question we get.
    2\sqrt{3}-5+a(1+\sqrt{3})=??

    Ok, so if we equal this to zero and square both sides we can get
    a^2=73/4
    This gives something close to 4.23...
    Close enough, but is this conclusive?
    We still have to find the value of P when a is in our desired value.
    No, it does not equal 0- only the coefficient of \sqrt{3} has to be 0 in order that P be an integer.

    Edit: Can some moderator please move this topic to it's right place?
    This looks like but is not a calculus problem.
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  8. #8
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    Quote Originally Posted by HallsofIvy View Post
    Okay- x^2= 4- 2\sqrt{3} but that is not what I get for x^3. For x^3 you should have (-1+ \sqrt{3})(4- 2\sqrt{3})= -4+ 4\sqrt{3}+ 2\sqrt{3}- 6, not what you have.
    It is, actually!
    I left it this way because it is easier to make it disappear in the main question, this is about taste hahah
    -1-3-6+2\sqrt{3}+\sqrt{3}+3\sqrt{3}=-4+6\sqrt{3}-6

    Later on you have
    4-2\sqrt{3}

    Which gets to the final
    4\sqrt{3}-5+a(-1+\sqrt{3})

    not 2\sqrt{3} as I have mentioned earlier.


    No, it does not equal 0- only the coefficient of \sqrt{3} has to be 0 in order that P be an integer.
    4\sqrt{3}-5+a(-1+\sqrt{3})
    OK! I think I get it!

    4\sqrt{3}-5-a+a\sqrt{3}=P
    If a=-4
    4\sqrt{3}-5+4-4\sqrt{3}=P
    -1=P

    This is correct!
    I made some trivial mistakes here and there probably because I was lunching when I did this.
    Thanks for your help!
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