3 Function/Pre-Calculus Questions

**SportfreundeKeaneKent** · October 27th 2006, 02:05 PM

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:

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**CaptainBlack** · October 27th 2006, 02:27 PM

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

**topsquark** · October 27th 2006, 02:31 PM

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

**CaptainBlack** · October 27th 2006, 02:38 PM

$ \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<x-4 $ ,

or rearranging:

$ x<-3 $ .

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

RonL

**topsquark** · October 27th 2006, 02:40 PM

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!

)

**CaptainBlack** · October 27th 2006, 02:47 PM

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

**Soroban** · October 27th 2006, 02:52 PM

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

**SportfreundeKeaneKent** · October 28th 2006, 02:37 PM

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

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