Solving inequalities in precalculus

[brainygirl]

I need help on solving inequalities? Someone please help me. I 'm currently taking Pre-calculus.

 

[moose]

Like

$$x^2+2x-1\geq0$$
?

 

[brainygirl]

equations like the one implied ,

 

[moose]

Ok,
This is the way described through my math teachers, although it is not exactly the method I use now.
$$x^2+2x-3\geq$$
First, solve the problem as if it were a normal equality.
$$x^2+2x-3=0$$
and you get
$$x_1=1 x_2=-3$$
Now, write down all sets of numbers between those.
$$(-\infty,-3] <br /> [-3,1]<br /> [1,\infty)$$
Now set up a table like so
...set...|||||||||||sample point|||||||||||(x-1)|||||||||||(x+3)|||||||||||+ or - ?
$$(-\infty,-3]$$...[/color]|||||||||||...-4...|||||||||||..-..|||||||||||..-..|||||||||||...+...
$$[-3,1]$$...[/color]|||||||||||...0...|||||||||||..-..|||||||||||..+..|||||||||||...-...
$$[1,\infty)$$...[/color]|||||||||||...2...|||||||||||..+..|||||||||||..+..|||||||||||...+...


Now, since it was greater than or equal to, you know it has to be greater than zero, therefore the positive ones are the ones you want.

Therefore, the two sets $$(-\infty,-3]$$ and $$[1,\infty)$$ work.Now, you know that your answer is $$(-\infty,-3]\cup[1,\infty)$$now, try to solve this one on your own

$$x^2-5x+6\geq0$$

 

[hypermorphism]

You can do the familiar algebraic manipulation with inequalities, provided you remember to reverse the direction of the inequality whenever you multiply (or divide) by a negative quantity and practice simple logic. So by the example above,
$$x^2+2x-3 \geq 0$$
$$(x-1)(x+3) \geq 0$$
Now, if $ab \geq 0$, either ($a \geq 0$ and $b \geq 0$) or ($a \leq 0$ and $b \leq 0$) as you should easily justify. Let's evaluate the first set:
$x-1 \geq 0$ and $x+3 \geq 0$
implies that
$x \geq 1$ and $x \geq -3$
which is the set $\{x: x \geq 1\}$. Remember that x must satisfy both inequalities when using "and".
The second set evaluates to $\{x: x \leq -3\}$, so we have the set $\{x: x \geq 1$ or $x \leq -3\}$, or written another way $\{x: x \geq 1\} \cup \{x: x \leq -3\}$.
This is just the purely algebraic way. Choose whichever way you feel most comfortable with.

 

[ivybond]

My favorite word: VISUALIZE.

For generic inequality
$$f(x) > g(x)$$:
$$h(x) = f(x) - g(x)$$.
Find the intervals where a graph $$y=h(x)$$ is above the x-axis (you'll have to find/estimate the roots of the $$y=h(x)$$ and points where $$h(x)$$ is undetermined).

Try
$$\frac{x+2}{x}\leq \frac{1}{2-x}$$
Could you post your answer?

 

[QueenFisher]

gosh, that looks confusing!

i was just taught to regard the inequality as a quadratic, make it equal to 0, draw the graph and solve it from there.

that may be what ^^ was saying though...