Inequality difficulties D= (1 Viewer)

[Slidey]

I don't get the answer to Q1. Wouldn't it be |x| > |a|, which then would have 4 solutions (each depending on whether x and/or a are negative)? I haven't done the graphing stuff yet.

Typically 'solve' means find x=

So you don't want any functions attached to it, including absolute value.

 

[lolokay]

that's why I added 'which then would have 4 solutions'. I just couldn't be bothered typing them out.

 

[Finx]

Guysssss ._.

How do I do Q1 in algebraic form?

 

[iCpurplehippos]

I don't get it. What do you mean by algebraic form? If you use the testing method, is that doing it in algebraic form?

 

[lolokay]

Yeah, I edited my post when I realised it was true =P

I still don't know how to express x^2 > a^2 in algebraic form. I even asked some of my classmates today how to do it, and they also told me to draw a graph. I doubt the question will ask me to prove my answer by drawing a graph.

Oops, hadn't realised. edited mine too (again).

for the x^2 > a^2, I still think you would solve as firstly:
|x| > |a|
Then write out the solutions with the case where it is true:
x > a, where x≥0 and a≥0
x > -a, where x≥0 and a≤0
x < a, where x≤0 and a≤0
x < -a, where x≤0 and a≥0


I haven't learnt to solve like that though, so I'm not sure if that's how the answer would be expressed; that's just how I would think to express it.

 

[Finx]

I don't get it. What do you mean by algebraic form? If you use the testing method, is that doing it in algebraic form?

It probably is, but I don't even know how to test the method in the question.

Some people at school were trying one way, but I didn't see how it worked;

x^2 > a^2
x^2 - a^2 > 0
(x+a)(x-a) > 0

x+a > 0
x > -a

and

x-a > 0
x > a

Except the answer is x > a, x < -a

=/

 

[lolokay]

it doesn't follow that (x+a)(x-a) > 0, therefore x+a > 0 or x-a > 0, which is why the answer is wrong

+However it would follow that either x+a and x-a > 0,
or x+a and x-a <0
Therefore,
x>a and x>-a
or x<a and x<-a
that sounds much better than what I had before