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Math Help - 3 Function/Pre-Calculus Questions

  1. #1
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    3 Function/Pre-Calculus Questions

    I can't solve these 3 solutions. Could someone post up solutions or tell me how to do them. Greatly appreciated.





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  2. #2
    Grand Panjandrum
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    1. The roots of x^3-ax^2-3kx+m=0 are 2, -1, 3 find the values of a, k, m.

    That the roots are 2,\ -1,\ 3 means that:

    <br />
x^3-ax^2-3kx+m=(x-2)(x+1)(x-3)=0<br />

    expanding the brackets gives:

    <br />
x^3-ax^2-3kx+m=x^3 - 4x^2 + x + 6=0<br />

    so a=4,\ k=-1/3,\ m=6

    RonL
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  3. #3
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    If x^3-ax^2-3kx+m=0 has the roots x = 2, -1, 3. Find a, k, m.

    The simplest way to do this is to construct the polynomial that has those roots. So:
    (x - 2)(x - (-1))(x - 3) = (x -2)(x+1)(x-3)

    =  (x^2 - x - 2)(x - 3) = x^3 - x^2 - 2x - 3x^2 + 3x + 6

    = x^3 - 4x^2 + x + 6

    So I get that
    -a = -4
    -3k = 1
    m = 6

    So a = 4, k = -1/3, m = 6.

    -Dan
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  4. #4
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    <br />
\frac{2x-1}{x-4}>1,\ \ x \ne 4<br />

    Suppose x>4 so x-4>0, now multiply through by x-4 to get:

    <br />
2x-1>x-4<br />
,

    or rearranging:

    <br />
x>-3<br />
,

    which tells us nothing extra, as we had assumed that x-4>0, or x>4.

    Instead suppose x-4<0, then:

    <br />
2x-1<x-4<br />
,

    or rearranging:

    <br />
x<-3<br />
.

    So our solution is x>4 or x<-3.

    RonL
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  5. #5
    Forum Admin topsquark's Avatar
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    Solve \frac{2x-1}{x-4}>1

    First:

    \frac{2x-1}{x-4}-1>0

    \frac{2x-1}{x-4} - \frac{x-4}{x-4} > 0

    \frac{x+3}{x-4}>0

    There are critical points where the numerator is 0 and where the denominator is 0. So we have critical points at:
    x + 3 = 0 --> x = -3
    x - 4 = 0 --> x = 4

    So break the real line up into  (- \infty , -3) \cup (-3, 4) \cup (4, \infty) and test each interval:

    (-\infty, -3): \frac{x+3}{x-4}>0 (Check!)
    (-3, 4): \frac{x+3}{x-4}<0 (Nope!)
    (4, \infty): \frac{x+3}{x-4}>0 (Check!)

    So the solution set is x \in (-\infty, -3) \cup (4, \infty)

    -Dan

    (Hi CaptainBlack! )
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  6. #6
    Grand Panjandrum
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    For f(x)=x^2+kx+k determine all values of k such that f(x)>0.

    Now for large |x|\ f(x)>0 as the leading term is dominant. So the question is realy asking us to find the values of k such that f(x)=0 has no real roots.

    From the quadratic formula we know the roots of f(x)=0 are:

    <br />
x=\frac{-k \pm \sqrt{k^2-4k}}{2}<br />
,

    and these are not real if k^2-4k<0, which is equivalent to k<4 and k>0.

    RonL
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  7. #7
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    Hello, SportfreundeKeaneKent!

    If the roots of the equation x^3 - ax^2 - 3kx + m\:=\:0 are 2,\:-1, and 3,
    find the values of a,\:k, and m.

    If 2,\:-1, and 3 are roots of the cubic,
    . . then (x - 2),\:(x + 1),\;(x - 3) are factors of the cubic.

    Hence, the cubic is: . (x - 2)(x + 1)(x - 3)\:=\:x^3- 4x^2+x - 6

    Therefore: . \begin{Bmatrix}\text{-}a = \text{-}4 \\ \text{-}3k = 1 \\ m = 6\end{Bmatrix}\;\Rightarrow\;\begin{Bmatrix}a = 4\\ k = \text{-}\frac{1}{3} \\ m = 6\end{Bmatrix}


    Solve: . \frac{2x-1}{x-4}\:>\:1 . where x \neq 4.

    Multiply both sides by (x - 4).

    If x - 4 \,>\,0 or x \,<\,4,
    . . we have: . 2x - 1 \,>\,x - 4\quad\Rightarrow\quad x > -3
    The "stronger" inequality is: . x > 4

    If x - 4 \,<\,0 or x \,<\,4,
    . . we have: . 2x - 1 \,< \,x - 4\quad\Rightarrow\quad x < -3
    The "stronger" inequality is: . x < -3

    Solution: . \boxed{x < -3\;\text{ or }\;x > 4}



    For f(x)\:=\:x^2 + kx + k, determine all the values of k such that f(x) > 0.

    If f(x) > 0, the parabola is completely above the x-axis.

    If it had x-intercepts, they would occur where f(x) = 0.
    . . That is: . x^2 + kx + k\:=\:0

    Quadratic Formula: . x \:=\:\frac{-k \pm \sqrt{k^2 - 4k}}{2}

    To have no x-intercepts, the discriminant must be negative.
    . . That is: . k^2 - 4k \:< \:0

    Complete the square: . k^2 - 4k + 4 \:< \:4

    And we have: . (k - 2)^2 \:< \:4\quad\Rightarrow\quad |k-2|\:<\:2

    Hence: . -2 \:<\:k-2 \:<\:2\quad\Rightarrow\quad \boxed{0 \:< \:k\:<\:4}

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  8. #8
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    Terrific replies. I really appreciate it and I can do other problems like these now so thanks again.
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