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Thread: to find the x value... i just wonder why the two exponents are not the same.

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    rcs
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    to find the x value... i just wonder why the two exponents are not the same.

    to find the x value... i just wonder how to find the x since the two exponents are not the same.
    to find the x value... i just wonder why the two exponents are not the same.-picture1.png
    5x^7+3x^4=242
    5x^7+3x^4-242 = 0
    i could not factor it ...

    have some help.
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    Re: to find the x value... i just wonder why the two exponents are not the same.

    Is your teacher a joker!

    Try x=1, then x=2:
    at least you'll see there's a solution >1 and <2
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    Re: to find the x value... i just wonder why the two exponents are not the same.

    Thank you Sir Denis... How did you solve it sir... please can you show the solution sir?

    5(1)^7+ 3(2)^4
    5(1)+3(16)
    5+48
    =53... it isnot 242.... ������
    Last edited by rcs; Mar 6th 2017 at 12:01 AM.
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    Re: to find the x value... i just wonder why the two exponents are not the same.

    Quote Originally Posted by rcs View Post
    Thank you Sir Denis... How did you solve it sir... please can you show the solution sir?

    5(1)^7+ 3(2)^4
    5(1)+3(16)
    5+48
    =53... it isnot 242.... ������
    this has to be solved using numeric methods.

    The only real solution is

    $x \approx 1.71275$

    There are 3 complex conjugate pair solutions as well.
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    rcs
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    Re: to find the x value... i just wonder why the two exponents are not the same.

    How is it done sir?
    Is it done by simply pluggig in any values of x element of N or Z?
    That takes time sir
    Last edited by rcs; Mar 6th 2017 at 12:51 AM.
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    Re: to find the x value... i just wonder why the two exponents are not the same.

    Oh My God. Is there any help i can have in here?
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    rcs
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    Re: to find the x value... i just wonder why the two exponents are not the same.

    Quote Originally Posted by romsek View Post
    this has to be solved using numeric methods.

    The only real solution is

    $x \approx 1.71275$

    There are 3 complex conjugate pair solutions as well.
    How come sir? How is it done :-(
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    Re: to find the x value... i just wonder why the two exponents are not the same.

    Quote Originally Posted by rcs View Post
    How come sir? How is it done :-(
    https://en.wikibooks.org/wiki/Numeri...uation_Solving
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    Re: to find the x value... i just wonder why the two exponents are not the same.

    Quote Originally Posted by rcs View Post
    Oh My God. Is there any help i can have in here?
    That is totally uncalled for. It is rude and off-putting.
    You have posted here many many times. Are you too lazy to use available web-based calculators.?
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    Re: to find the x value... i just wonder why the two exponents are not the same.

    I suspect the OP is frustrated because his follow up question is not understood. He wants to know analytically how Denis found that a solution lies between 1 and 2.

    There is no mechanical trick. Even if Denis did not graph the function, he undoubtedly observed that for non-negative values of x

    $f(x) = 5x^7 + + 3x^4$ increases rapidly and monotonically from a value of zero at x = 0.

    It is also obvious by inspection that f(1) = 8 and f(3) >> 400. So if there is an integer solution, it must be 2. But f(2) is too large.

    Thus, there is a positive solution between 1 and 2.
    Thanks from DenisB
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    rcs
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    Re: to find the x value... i just wonder why the two exponents are not the same.

    Quote Originally Posted by Plato View Post
    That is totally uncalled for. It is rude and off-putting.
    You have posted here many many times. Are you too lazy to use available web-based calculators.?
    Sorry Sir Plato... i dont understand you ...just Chill

    I just wanted to see whether this is doable or solvable algebraially... for me it is seemed not...
    Last edited by rcs; Mar 6th 2017 at 01:46 PM.
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