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Thread: Solving Equations

  1. #1
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    Red face Solving Equations

    Hi everyone,

    I was working on some solving equations in preparation for university, and came across a few that I just could not understand how to do them. I have the answers, but do not know how to get to them, how painful....

    1. Find a cubic function of x whose graph has x-intercepts equal to -1 and 2, and in which f(0)=6 and f(3)=12.

    Answer: 2x^3-5x^2-x+6

    2. The roots of the equation x^2+px+1=0 are a and b, and the roots of the equation x^2-9x+q=0 are a+2b and 2a+b. (a and b are greek symbols).

    Answer: p=-3, q=19

    3. If the roots of the equation x^3-15x^2+cx-105, are a, a+b, and a-b, determine the values of a,b, and c.

    Answer: a=5, b=+2,-2 c=71

    Thanks in advance!
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  2. #2
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    3. $\displaystyle x^3 - 15x^2 + cx - 105$ has roots $\displaystyle x = a, x = a+ b, x = a - b$.

    That means $\displaystyle f(a) = 0, f(a + b) = 0, f(a - b) = 0$.


    So $\displaystyle a^3 - 15a^2 + ac - 105 = 0$
    $\displaystyle (a+b)^3 - 15(a+b)^2 + (a+b)c - 105 = 0$
    $\displaystyle (a-b)^3 - 15(a-b)^2 + (a-b)c - 105 = 0$.


    Solve these three equations simultaneously for $\displaystyle a,b,c$.
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  3. #3
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    Quote Originally Posted by KelvinScale View Post
    Hi everyone,

    I was working on some solving equations in preparation for university, and came across a few that I just could not understand how to do them. I have the answers, but do not know how to get to them, how painful....

    1. Find a cubic function of x whose graph has x-intercepts equal to -1 and 2, and in which f(0)=6 and f(3)=12.

    Answer: 2x^3-5x^2-x+6

    [snip]
    Use the model $\displaystyle f(x) = a(x + 1)(x - 2)(x - b)$ (why?) and use the fact that f(0) = 6 and f(3) = 12 to get two simuatenous equations for a and b.

    Quote Originally Posted by KelvinScale View Post
    [snip]
    2. The roots of the equation x^2+px+1=0 are a and b, and the roots of the equation x^2-9x+q=0 are a+2b and 2a+b. (a and b are greek symbols).

    Answer: p=-3, q=19

    [snip]
    $\displaystyle (x - a)(x - b) = x^2 + px + 1$. Therefore:

    $\displaystyle ab = 1$ .... (1)

    $\displaystyle a + b = -p$ .... (2)

    In a similar way, $\displaystyle (x - a - 2b) (x - 2a - b) = x^2 - 9x + q$. Therefore:

    ...... (3)

    ....... (4)

    Solve equations (1), (2), (3) and (4) simultaneously to get p and q.


    If you need more help, please show all your work and say where you get stuck.
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  4. #4
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    Thank you so so much for both your answers!
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