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Math Help - Equations

  1. #1
    Member classicstrings's Avatar
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    Equations

    Hi! Thanks for helping me!

    Show that the equation x^3 - x^2 +x - 21 = 0 has only one solution.

    and

    Without dividing find the remainder when x^3 - 4x + 12x - 2 is divided by x+1.

    and one more

    For what values of k does the graph of y = 2x^2 - kx + 3 have a TP on the x-axis?

    Thanks for your great help!
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  2. #2
    Member classicstrings's Avatar
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    Ok i got number 3, i realised that delta = 0 when there is a TP on x-axis.
    So k = +-2 root 6.
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  3. #3
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    Quote Originally Posted by classicstrings
    Hi! Thanks for helping me!

    Show that the equation x^3 - x^2 +x - 21 = 0 has only one solution.
    You can easily show it by demonstrating that this function is increasing only, and hence intersects the x-axis exactly one. Thus, it has only one real solution.

    But you need to have the knowledge of calculus, I do not know if you studied it. Did you?
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  4. #4
    Member classicstrings's Avatar
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    Well i know a bit of calculus as we just started but there is no other way to answer question?
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  5. #5
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    Quote Originally Posted by classicstrings
    Well i know a bit of calculus as we just started but there is no other way to answer question?
    Hello,

    your equation T(x)=x^3 - x^2 +x - 21 = 0 consists of a term in x T(x) which is equal to zero. If there is a solution you can split this term into a linear factor (x-s) (s stands for solution) and the remaining term R(x). So you get T(x)=(x-s) \cdot R(x). When you multiply the RHS of this equation you can see, that s must be a divisor of the constant summand of your equation. All integer divisors of 21 are: +-1, +-3, +-7, +-21.
    Now do a little bit trial and error and you'll find pretty soon, that 3 must be a solution of your equation. That means you can divide the LHS of your equation by (x-3):
    \left( x^3 - x^2 +x - 21 \right)/\left(x-3 \right) = x^2+2x+7
    Because x^2+2x+7 > 0 for all x \in \mathbb{R} there are no other possible solutions.

    Greetings

    EB
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  6. #6
    Member classicstrings's Avatar
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    Oh i see, thanks! Yes you need to sub in factors of 21. Normally, I just sub in values starting from 1, then -1, to 2 to -2 and so on.
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