school help: math broblem plz help - Help.com

Sorry. calculus was never my thing, but have you tried Yahoo Answers? www.answers.yahoo.com
they seem to get a lot of math questions.

it looks similar to what i’m performing in pre-calculus. what’s the exercise topic? if it’s solving inequalities then i can tell you step by step how to solve it.

Yeah if it’s an inequality, this is how you work it:
x^2-x-2/(x+4)^2>0,
Multiply both sides by (x+4)^2 (keep in mind x=-4 would be undefined)
x^2-x-2 > 0
Factor the left side
(x-2)(x+1) > 0
So, we get the critical points, 2 and -1.
On the interval (-infinity, -1) the function is > 0 {but keep in mind it’s undefined at x=-4}
On the interval {-1,2} the function is =0
And on the interval from (2, infinity), the function is >0.

So the solution to this inequality is true when x is on the interval from:
(-inf, -4), (-4,-1), (2, inf)
Keep in mind the () mean do not include the stated number and {} means inlcude the stated number. You can not include inf or - inf.

Creal Default wrote:
Yeah if it’s an inequality, this is how you work it:
x^2-x-2/(x+4)^2>0,
Multiply both sides by (x+4)^2 (keep in mind x=-4 would be undefined)
x^2-x-2 > 0
Factor the left side
(x-2)(x+1) > 0
So, we get the critical points, 2 and -1.
On the interval (-infinity, -1) the function is > 0 {but keep in mind it’s undefined at x=-4}
On the interval {-1,2} the function is =0
And on the interval from (2, infinity), the function is >0.

So the solution to this inequality is true when x is on the interval from:
(-inf, -4), (-4,-1), (2, inf)
Keep in mind the () mean do not include the stated number and {} means inlcude the stated number. You can not include inf or - inf.

bravo

follow creal default, i believe he or she has the right idea.

x+4^2 is the common denominator for the equation which is the first step in solving. once you factor out x^2-x-2 then x-2, x+1 become 2 and -1 after you set them equal to zero and solve for x. x+4^2 becomes -4 once set equal to zero. so then you’re left with -1,2, and -4. usually, you would draw a number line with those three numbers on there and pick a number between those and plug it back into the equation. say you chose 0 being that 0 is between -1 and 2 and plug it back into the equation. if the result is greater than (>)0 then you have a true statement for the intervals of -1 and 2. but you have to plug in a number greater than -4, between -1 and -4, between -1 and 2, and greater than 2. by completing this concept you’ll achieve the solution of (-inf,-4)U(-4,-1)U(2,inf).

Yeah, I know it’s a little confusing. Thanks for clarifying mills.

As far as multiplying the (x+4)^2 to both sides, the left side will cancel with the denominator.

Look at this example, where I multiply 4 by both sides:
(x - 2)/4 = 0
4 * (x - 2)/4 = 4 * 0
(x - 2) = 0
x = 2

You could FOIL the (x+4)^2, but it won’t cancel or reduce anything.

my teacher explains it so quick, the only thing i can remember is how to solve the equation. i don’t take into account the terminology with numbers and signs or anything. it’s also easier to explain on pencil and paper rather than the net. but hopefully you understand the basic concepts with solving inequalities.

thank you for instilling and refreshing my memory creal default:)