text: 1. ## 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.
  
  If that picture doesn't work, then this should:
  
  [IMG][/IMG]
  
  2. 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:
  
  $
  x^3-ax^2-3kx+m=(x-2)(x+1)(x-3)=0
  $
  
  expanding the brackets gives:
  
  $
  x^3-ax^2-3kx+m=x^3 - 4x^2 + x + 6=0
  $
  
  so $a=4,\ k=-1/3,\ m=6$
  
  RonL
  
  3. 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
  
  4. $
  \frac{2x-1}{x-4}>1,\ \ x \ne 4
  $
  
  Suppose $x>4$ so $x-4>0$, now multiply through by $x-4$ to get:
  
  $
  2x-1>x-4
  $
  ,
  
  or rearranging:
  
  $
  x>-3
  $
  ,
  
  which tells us nothing extra, as we had assumed that $x-4>0$, or $x>4$.
  
  Instead suppose $x-4<0$, then:
  
  $
  2x-1$
  ,
  
  or rearranging:
  
  $
  x<-3
  $
  .
  
  So our solution is $x>4$ or $x<-3$.
  
  RonL
  
  5. 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! )
  
  6. 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:
  
  $
  x=\frac{-k \pm \sqrt{k^2-4k}}{2}
  $
  ,
  
  and these are not real if $k^2-4k<0$, which is equivalent to $k<4$ and $k>0$.
  
  RonL
  
  7. 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: . $-2 \:<\:k-2 \:<\:2\quad\Rightarrow\quad \boxed{0 \:< \:k\:<\:4}$
  
  8. Terrific replies. I really appreciate it and I can do other problems like these now so thanks again.
explanation: The document contains multiple math questions and their corresponding answers. The questions cover topics such as finding the roots of a polynomial, solving inequalities, and determining the values of constants for which a quadratic function is always positive. Each question is followed by a detailed solution.
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distilabel_metadata.content: 1. ## 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.
  
  If that picture doesn't work, then this should:
  
  [IMG][/IMG]
  
  2. 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:
  
  $
  x^3-ax^2-3kx+m=(x-2)(x+1)(x-3)=0
  $
  
  expanding the brackets gives:
  
  $
  x^3-ax^2-3kx+m=x^3 - 4x^2 + x + 6=0
  $
  
  so $a=4,\ k=-1/3,\ m=6$
  
  RonL
  
  3. 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
  
  4. $
  \frac{2x-1}{x-4}>1,\ \ x \ne 4
  $
  
  Suppose $x>4$ so $x-4>0$, now multiply through by $x-4$ to get:
  
  $
  2x-1>x-4
  $
  ,
  
  or rearranging:
  
  $
  x>-3
  $
  ,
  
  which tells us nothing extra, as we had assumed that $x-4>0$, or $x>4$.
  
  Instead suppose $x-4<0$, then:
  
  $
  2x-1$
  ,
  
  or rearranging:
  
  $
  x<-3
  $
  .
  
  So our solution is $x>4$ or $x<-3$.
  
  RonL
  
  5. 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! )
  
  6. 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:
  
  $
  x=\frac{-k \pm \sqrt{k^2-4k}}{2}
  $
  ,
  
  and these are not real if $k^2-4k<0$, which is equivalent to $k<4$ and $k>0$.
  
  RonL
  
  7. 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: . $-2 \:<\:k-2 \:<\:2\quad\Rightarrow\quad \boxed{0 \:< \:k\:<\:4}$
  
  8. Terrific replies. I really appreciate it and I can do other problems like these now so thanks again.
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document contains multiple math questions and their corresponding answers. The questions cover topics such as finding the roots of a polynomial, solving inequalities, and determining the values of constants for which a quadratic function is always positive. Each question is followed by a detailed solution.
  
  Label: multiple-question-and-answer