Can Quadratics Help Solve Inequalities?

[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) &gt; 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...