1. Polynomial Func. and Eq.'s

If anyone could provide the answer so i can compare with mine, or tell me whats wrong with my answer that would be great.

1. Solve Algebraically

A) $\displaystyle 2x^2+5x-63>0$

My Answer: $\displaystyle {-7>x<9/2}$

B) $\displaystyle 20x^3-3x^2-17x+6 \leq 0$

My Answer: $\displaystyle {-1>x,2/5<x<3/4}$

C) $\displaystyle \frac{x+1}{x-3} - \frac{2x^2+3x}{4x^2-9} = \frac{5}{2x^2-9x+9}$

My Answer: $\displaystyle {x=2,-4}$

2. Sketch the graph of $\displaystyle f(x)=\frac{x+4}{x-5}$, Determine all intercepts, the domain and the range, and asymptotes.
Sketch the graph of $\displaystyle g(x)=3x-8$ on the same graph as f(x). Determine where $\displaystyle f(x) \geq g(x)$.

Intercepts: $\displaystyle y=-4/5$, $\displaystyle x=-4$
Domain and Range: Don't know how to write it on here
Vertical Asymptote: $\displaystyle x=5$
Horizontal Asymptote: $\displaystyle y=1$

I was able to graph it, I just need help on determining where $\displaystyle f(x) \geq g(x)$.

2. 2x^2 + 5x - 63

solving this, you get x= -7 and 9/2

2x^2 + 5x - 63 > 0

check if it holds true when x is smaller than -7, between -7 and 9/2, and greater than 9/2.

2(-9999)^2 + 5(-9999) - 63 > 0
huge positive number - smaller numbers > 0
positive > 0 (check!)

2(0)^2 + 5(0) - 63 > 0
-63 > 0 (.. nope!)

2(9999)^2 + 5(9999) - 63 > 0
huge positive number + smaller number - even smaller number > 0
positive > 0 (check!)

Therefore, x holds true when x < -7 and x > 9/2
(-infinite, -7) U (9/2, infinite)
(Note: not inclusive since it isn't 2x^2 + 5x - 63 >= 0)

--same concept for b-- btw, this is just one way to solve for x.

3. Originally Posted by Raj
My Answer: $\displaystyle {-7>x<9/2}$
Nice try on the notation shorthand, but never write anything like that again.

Trust me on this. Very ugly. If I were grading it, I would mark it wrong, at least partially.

4. Hello, Raj!

1. Solve algebraically

$\displaystyle A)\;\;2x^2+5x-63\:>\:0$

The graph of $\displaystyle y \:=\:2x^2+5x-63$ is an up-opening parabola.
. . We want to know when the graph is above the x-axis.
The function is positive to the left and the right of its x-intercepts.

x-intercepts: .$\displaystyle 2x^2+5x-63 \:=\:0\quad\Rightarrow\quad (x + 7)(2x - 9)\:=\:0\quad\Rightarrow\quad x \:=\:-7,\:\frac{9}{2}$
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Answer: .$\displaystyle \boxed{x < -7\:\text{ or }\:x > \frac{9}{2}}$

$\displaystyle B)\;\;20x^3-3x^2-17x+6 \:\leq \:0$

We have the cubic function: .$\displaystyle y \:=\:20x^3 - 3x^2 - 17x + 6$
. . We want know when this cubic is below the x-axis.

It factors: .$\displaystyle (x + 1)(5x-2)(4x-3) \:=\;0\quad\Rightarrow\quad x \:=\:-1,\:\frac{2}{5},\:\frac{3}{4}$
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Answer: .$\displaystyle \boxed{x < -1,\;\frac{2}{5} < x < \frac{3}{4}}$

$\displaystyle C)\;\;\frac{x+1}{x-3} - \frac{2x^2+3x}{4x^2-9} \:= \:\frac{5}{2x^2-9x+9}$

We have: .$\displaystyle \frac{x+1}{x-3} - \frac{x(2x+3)}{(2x -3)(2x+3)} \:=\;\frac{5}{(x-3)(2x-3)}$

. . $\displaystyle = \;\frac{x+1}{x-3} - \frac{x}{2x-3} \;=\;\frac{5}{(x-3)(2x-3)}$

Multiply by the LCD, $\displaystyle (x-3)(2x-3)$
. . $\displaystyle (x+1)(2x-3) - x(x-3) \:=\:5\quad\Rightarrow\quad x^2 + 2x - 8 \:=\:0\quad\Rightarrow\quad(x+4)(x-2) \:=\:0$

Therefore: .$\displaystyle \boxed{x \;=\;-4,\:2}$

You were right!