Methods 3/4 Questions

[#1procrastinator]

(hopefully the last I'll ask about quadratic inequalities lol)

When you solve a quadratic inequality like x^2-x-6<0, you split it into two cases and
then combine the solution sets right? If one of the solution sets (for the above) is
x < -2 and x > 3, why doesn't it make sense to say that the set is (-∞, -2) U (3, ∞)
(aside from the fact that simple inspection will tell you it's wrong)?

----

And here's what kind of led me back to ^

Solve for |2x+6| - |x+3| = |x| for x

I think I've got the gist of how to solve these...
(=> is equal to greater than, lol, sorry don't have my symbols handy)

For |2x+6|
2x + 6 if x ≥ -3
-2x - 6 if x < -3

|x+3|
x+3 if x ≥ -3
-x-3 if x<-3

|x|
x if x ≥ 0
-x if x < 0

So if what I've gathered is right, you're supposed to find an x that fits in the intervals and then
solve the equation? e.g. I can't choose 2x+ 6, x+3 and x because if x can't be
greater than or equal to both 3 and 0

But what about x is less than -3 and x is less than 0? If it's less than -3 then it's also less than 0
but the equation has no solution.

Thanks

[TrueTears]

(hopefully the last I'll ask about quadratic inequalities lol)

When you solve a quadratic inequality like x^2-x-6, you split it into two cases and
then combine the solution sets right? If one of the solution sets (for the above) is
x < -2 and x > 3, why doesn't it make sense to say that the set is (-∞, -2) U (3, ∞)
(aside from the fact that simple inspection will tell you it's wrong)?

----

And here's what kind of led me back to ^

Solve for |2x+6| - |x+3| = |x| for x

I think I've got the gist of how to solve these...
(=> is equal to greater than, lol, sorry don't have my symbols handy)

For |2x+6|
2x + 6 if x ≥ -3
-2x - 6 if x < -3

|x+3|
x+3 if x ≥ -3
-x-3 if x<-3

|x|
x if x ≥ 0
-x if x < 0

So if what I've gathered is right, you're supposed to find an x that fits in the intervals and then
solve the equation? e.g. I can't choose 2x+ 6, x+3 and x because if x can't be
greater than or equal to both 3 and 0

But what about x is less than -3 and x is less than 0? If it's less than -3 then it's also less than 0
but the equation has no solution.

Thanks

for the quadratic question, after you factorise you will get 2 factors, call them A and B

eg, A*B >0

case 1. A>0 AND B>0, so you take the intersection of the solutions from the 2 inequalities

OR

case 2. A<0 AND B<0, again you take intersection for obvious reasons ('AND')

then you take the union of case 1 and 2, since it's either case 1 OR case 2

[Phy124]

For |2x+6|
2x + 6 if x ≥ -3
-2x - 6 if x < -3

|x+3|
x+3 if x ≥ -3
-x-3 if x<-3

|x|
x if x ≥ 0
-x if x < 0

(Ceebz with latex, soz )

Following on from what you had done... therefore;

|2x + 6| - |x + 3| equals;
 
(2x + 6) - (x + 3) if x ≥ -3
(-2x - 6)-(-x - 3) if x < -3

Which consequently equals;

x + 3 if x ≥ -3
-x - 3 if x < -3

So now we're trying to find when

x + 3 if x ≥ -3
-x - 3 if x < -3

is equal to

x if x ≥ 0
-x if x < 0

We can see by the gradients that x + 3 and x will be parallel, as is the same with -x - 3 and -x, so disregard those (they won't overlap).

Now we are left with x + 3 and -x, along with -x - 3 and x.

x + 3 and -x occur for x ≥ -3 and x < 0 respectively, so they will overlap, therefore;

x + 3 = -x

2x = -3
x= -3/2

You can check the answer by |2(-3/2)+6|  -|(-3/2)+3|= 3/2 = |(-3/2)| = |x|

Sorry about my convoluted answer and probability most likely lies with an easier method for working these out, but hey, this'll do for now

[#1procrastinator]

@TrueTears: sorry, I forgot to add <0 to the inequality but I guess it doesn't change my question. For case 2, 'you take the intersection for obvious reasons' - that's what's not obvious to me lol

@MyLittlePony - haha, no problem. That working makes sense to me, I'll try to applying it to other problems (that was one I just made up whilst trying to understand damn absolute values, suppose I'll have to look at their graphs as well)

[TrueTears]

because it's 'and' (http://en.wikipedia.org/wiki/Intersection_%28set_theory%29#Basic_definition)

and

http://en.wikipedia.org/wiki/Logical_conjunction

take note of

    x ∈ A ∩ B if and only if

        * x ∈ A and
        * x ∈ B.

[#1procrastinator]

^ Thanks

[#1procrastinator]

How do you simplify this? It should evaluate to either root 3 or 0 according to my calculator (it should be plus or minus between root 3 and the big root

[yawho]

How do you simplify this? It should evaluate to either root 3 or 0 according to my calculator (it should be plus or minus between root 3 and the big root

simplified to
.:

[#1procrastinator]

How'd you simplify it to 1+root3?

[yawho]

express it as a + sqrt(b), square both sides to find a and b

[#1procrastinator]

sorry, you're dealing with a moron here, could you give me a bit more? you're allowed to square it?

[TrueTears]

and


a = 1 and b = 3

or and b = 1

In either we are left with

i think that's what he means lol

[#1procrastinator]

^ aah, thanks a lot! where do you guys pick these little tricks up from?

geniuses lol

[#1procrastinator]

Why can't we have a negative base for logs? If the number on the bottom is negative, than can't the base be negative (for certain integers). Like log(-2)(-8) = 3

And what does it mean to differentiate with respect to a certain variable? It's coming up in implicit differentiation and it's got me confused but I'm asking here cause it's a mcnoob question

[TrueTears]

Why can't we have a negative base for logs? If the number on the bottom is negative, than can't the base be negative (for certain integers). Like log(-2)(-8) = 3

And what does it mean to differentiate with respect to a certain variable? It's coming up in implicit differentiation and it's got me confused but I'm asking here cause it's a mcnoob question

Well you can have a negative base for logs, it's just that it's not defined.

Think about it, if we have some function y = a^x where a<0, ie, say y = (-2)^x

then clearly for some values of x, y will be undefined in the real space, you could at least think of 1 very obvious one, ie, when x = 1/2

Thus when we take the inverse, x = (-2)^y -> y = log_(-2)(x) this would not make a very useful or nice function, it's discontinuous, because it's inverse is discontinuous, and it's simply a plot of different points rather than smooth joining curve, there's just simply not much use defining a negative base for a log. Other than that, what you have written isn't wrong and is intuitively reasonable, it's just that it's not well-defined.