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xddxogm3
09-26-2004, 04:43 PM
I'm going to attempt to ask an educated algebra question. Please forgive a bad translation.

Inequalities and Powers/Roots.
Legend:
() <-Power
[] <-Square Root

Solve:
the square of x is less than 4
x(2)<4

would this translate to
x is less then the positive/negative of the square root of 4.
x<+/>-[4]
or
x<2 and x>-2
;)
-2<x<2


??????????????????????????
My book doesn't address this type of inequality, and would really like to know this.

PJYelton
09-26-2004, 04:44 PM
Yes, that is correct.

xddxogm3
09-26-2004, 04:46 PM
note the change of the greater than and less than symbol on the
x<2 and x>-2
is that change of symbol accurate as well?

Perspective
09-26-2004, 04:49 PM
its usually written this way


-2 < x < 2

xddxogm3
09-26-2004, 04:54 PM
yes, I understand that, but when you have an inequality like the one I was working, does it follow the muliplication/division rule when working with negatives? Do we always switch the > or < upon noticing that the square root will result in a negative value? again I appologize if i'm not to clear. I'm new to saying math. I think math, but never have to explain it much. I'm sorry if I'm repeating myself. Did you put the -2<x<2 to state that we do follow the above mentioned multiplication/devision rule of negatives and inequalities?

Sang-drax
09-26-2004, 05:00 PM
We usually write it using absolute value:

x^2 < 4 <=>

|x| < 2

The inverse of x^2 is sqrt(x) when x >= 0 and
sqrt( -x ) when x < 0


As for switching signs, remember that:
-x < y <=> x > -y

xddxogm3
09-26-2004, 05:13 PM
Thank you all for the help.
One last attempt to clarify.
Please do not think I'm an idiot.
I have an exam and want to make sure all concepts are know front and back.

x^2 < 4

{x|-2<x<2}

Correct? Yes or No

Sang-drax
09-26-2004, 05:15 PM
Yes, that is correct.

Here's how to derive it:

x^2 < 4
|x| < 2
+- x < 2
(i) x < 2
(ii) -x < 2 <=> x > -2

(i) & (ii) <=> -2 < x < 2 <=> |x| < 2

xddxogm3
09-26-2004, 05:18 PM
Have I said how much you people on this board rock lately?
You all kick some major @$$.

Perspective
09-26-2004, 05:53 PM
On an exam be sure to follow Sang-drax's solution. If you leave out the absolute value profs will mash your marks down like gravy soaked potatoes.....*drool* im sooo hungry.

okinrus
09-26-2004, 10:12 PM
Using the squareroot on both sides of the inequality looks correct, but unless if you have theorem in your book that tells you it is correct, you should not rely on it without first proving it. Ie., you must first show that if x^2 < a, then sqrt(x^2) < sqrt(a).