|
|
#11
06-04-2007, 07:33 PM
|
|||
|
|||
Re: quadratic inequality
[ QUOTE ]
[ QUOTE ] [ QUOTE ] sqrt(9) is |3| [/ QUOTE ] I don't know if this will cause more confusion or less, but sqrt(9) is just 3 There's nothing deep about it, it's just the way we have defined sqrt() [/ QUOTE ] This is definitely not true. 3*3 = 9 AND -3*-3=9. The negative answer just gets ignored most times because lots of times it is not relevant to real world solutions. [/ QUOTE ] While 3*3 =9 and -3*-3=9, the sqrt(9) is 3, it has to be +/-sqrt() for both answers 
|
|
#12
06-04-2007, 09:03 PM
|
|||
|
|||
|
Re: quadratic inequality
[ QUOTE ]
hi, dummy here. for whatever reason, i can't figure this out: x^2-9<0 i broke it down as such: (x-3)(x+3)<0 so: x=3, x=-3 where do i go from here? thanks, dummy out. [/ QUOTE ] (x-3)(x+3)<0 x-3<0 or x+3<0 x<3 or x<-3 -3<x<3
|
|
#13
06-05-2007, 02:49 AM
|
|||
|
|||
|
Re: quadratic inequality
[ QUOTE ]
[ QUOTE ] hi, dummy here. for whatever reason, i can't figure this out: x^2-9<0 i broke it down as such: (x-3)(x+3)<0 so: x=3, x=-3 where do i go from here? thanks, dummy out. [/ QUOTE ] (x-3)(x+3)<0 x-3<0 or x+3<0 x<3 or x<-3 -3<x<3 [/ QUOTE ] x<3 or x<-3 =/= -3<x<3 -3<x<3 means x is greater than -3 and less than 3
|
|
#14
06-05-2007, 03:46 AM
|
|||
|
|||
|
Re: quadratic inequality
[ QUOTE ]
(x-3)(x+3)<0 x-3<0 or x+3<0 x<3 or x<-3 -3<x<3 [/ QUOTE ] The logic of this should go as follows. How can the product of two numbers be less than zero? It can happen in two ways. The first is less than zero AND the second is greater than zero. The other way is, the first if greater than zero and the second is less than zero. So asking for (x-3)(x+3) <0 means either Case 1. (x-3)<0 AND (x+3)> 0 ie. x<3 AND x>-3 ........... or Case 2. (x-3)> 0 AND (x+3)<0 ie. x>3 AND x<-3 ie. this case is impossible. So we get no solutions from Case 2. But Case 1 means that all numbers x are solutions where x<3 AND x>-3 This is commonly denoted -3<x<3 PairTheBoard
|
|
#15
06-05-2007, 06:39 AM
|
|||
|
|||
|
Re: quadratic inequality
[ QUOTE ]
i guess i have a question about the conversion from |x| < 3 to x > -3 how did you get there? [/ QUOTE ] The *definition* of |x| is: (i) If x >= 0, then |x| = x (ii) If x < 0, then |x| = -x Thus applying this definition to the condition |x|<3: (1) If x >= 0, implies x < 3 by (i) (2) If x < 0, implies -x < 3 by (ii), which implies x > -3 Since both (1) and (2) must hold, -3<x<3. Any clearer? [img]/images/graemlins/smile.gif[/img]
|
|
#16
06-05-2007, 06:52 AM
|
|||
|
|||
|
Re: quadratic inequality
[ QUOTE ]
[ QUOTE ] [ QUOTE ] sqrt(9) is |3| [/ QUOTE ] I don't know if this will cause more confusion or less, but sqrt(9) is just 3 There's nothing deep about it, it's just the way we have defined sqrt() [/ QUOTE ] This is definitely not true. 3*3 = 9 AND -3*-3=9. The negative answer just gets ignored most times because lots of times it is not relevant to real world solutions. [/ QUOTE ] No. By *definition*, sqrt(x) >= 0, for all x >= 0. The solution to the *equation* x^2 = a for unknown x and real a is: x = sqrt(a) or x = -sqrt(a). You are confusing the definition of the sqrt() operation with finding roots of quadratics.
|
|
#18
06-05-2007, 05:32 PM
|
|||
|
|||
|
Re: quadratic inequality
[ QUOTE ]
[ QUOTE ] [ QUOTE ] hi, dummy here. for whatever reason, i can't figure this out: x^2-9<0 i broke it down as such: (x-3)(x+3)<0 so: x=3, x=-3 where do i go from here? thanks, dummy out. [/ QUOTE ] (x-3)(x+3)<0 x-3<0 or x+3<0 x<3 or x<-3 -3<x<3 [/ QUOTE ] x<3 or x<-3 =/= -3<x<3 -3<x<3 means x is greater than -3 and less than 3 [/ QUOTE ] You're right. By "or" I meant one or the other, but not both, and I was sloppy in my solution. PTB did what I was going for.
|
|
#19
06-05-2007, 07:37 PM
|
|||
|
|||
|
Re: quadratic inequality
[ QUOTE ]
[ QUOTE ] [ QUOTE ] [ QUOTE ] hi, dummy here. for whatever reason, i can't figure this out: x^2-9<0 i broke it down as such: (x-3)(x+3)<0 so: x=3, x=-3 where do i go from here? thanks, dummy out. [/ QUOTE ] (x-3)(x+3)<0 x-3<0 or x+3<0 x<3 or x<-3 -3<x<3 [/ QUOTE ] x<3 or x<-3 =/= -3<x<3 -3<x<3 means x is greater than -3 and less than 3 [/ QUOTE ] You're right. By "or" I meant one or the other, but not both, and I was sloppy in my solution. PTB did what I was going for. [/ QUOTE ] That's interesting. Isn't that logical connective denoted XOR? ie. A XOR B means A or B but not (A and B). I've always found the XOR confusing in practice. It almost works in this case but not quite because of the strict inequality. (x-3)(x+3)<0 does not quite translate to (x-3)<0 XOR (x+3)<0 The reason being that x<3 And x=-3 , ie. x = -3 satisfies the XOR statement but does not satisfy the original strict inequality. Maybe that's why the XOR is so seldom used. It's too tricky. PairTheBoard
|
|
#20
06-10-2007, 10:14 PM
|
|||
|
|||
|
Re: quadratic inequality
The correct way to do and to see how these types of problems are done is to realize all polynomials are continuous. That means the value of the polynomial does not jump from one value to another - the value moves from one value to another by hitting every value between the first value and the second value.
On this problem x^2-9<0, x^2-9 = (x+3)(x-3). x^2-9 = 0 when x=3 or x=-3. That means x^2-9 is either positive or negative on each of the intervals (-inf, -3), (-3, 3), (3, inf). Simply set x to any of the values in each of the intervals. If x^2-9 < 0 on that interval then x^2-9 < 0 for every value in that interval. The intervals where x^2-9<0 are the correct answer.
|
 |
|
|
Linear Mode
