Stupid year 11 math problem... (1 Viewer)

[Brodie28]

In the extension 1 text book there was this problem...

Solve: X² > A²

The only solution I can see from that is

|X| > |A|

But the text book answer is

X < -A, X > A

Is the text book wrong or am I missing something?

 

[beccaxx]

im really sorry. i thought yr 11 would b easy maths 2 explain.
but u do extension im sorry. im a general girl.

 

[Dreamerish*~]

X² > A²

+X > +A

if X and A are both positive then X > A
if X is negative and A is positive then -X > A and so X < -A

i'm not really sure. i'm confused now '-_-

 

[Brodie28]

but musn't they both be positive because they are squared?

 

[Dreamerish*~]

no because -x times -x is still x²
the problem is i don't know exactly how x can be less than -a, when -a is negative but x is positive.... it's confusing me a lot
i'm ashamed... yr12 3-unit person can't do yr11 question.

 

[Dumsum]

no because -x times -x is still x²
the problem is i don't know exactly how x can be less than -a, when -a is negative but x is positive.... it's confusing me a lot
i'm ashamed... yr12 3-unit person can't do yr11 question.

X is a variable. Meaning it can be positive or negative. "-A" doesn't strictly imply that that is less than 0. So it works.

Like, ok. Say a is 3. For X to be less than -a, then x must be less than -3, say -5.

(-5)^2 = 25
3^2 = 9

so the statement holds.

 

[thunderdax]

The easiest way to do it is:
X^2-A^2>0
Try to think of this as a parabola.
The parabola cuts the x axis at A and -A and is concave up
Therefore, X>A or X<-A

 

[Will Hunting]

Guys, don't make this harder than it needs to be. Common sense is just about all you need in your arsenal to get this done.

Think of an equality first:

x^2 = 4
x = 2, -2

Obviously, making x larger than 2 will make the answer larger than 4. Remembering that any number squared is positive, making x LESS than -2 will make the answer LARGER than 4.

This summarises to |x| > 2, so you were close, Brodie.

All your question is is my example in general terms.

Therefore, applying this logic, x > a, x < -a, OR, |x| > a

 

[Dumsum]

Guys, don't make this harder than it needs to be. Common sense is just about all you need in your arsenal to get this done.

Think of an equality first:

x^2 = 4
x = 2, -2

Obviously, making x larger than 2 will make the answer larger than 4. Remembering that any number squared is positive, making x LESS than -2 will make the answer LARGER than 4.

This summarises to |x| > 2, so you were close, Brodie.

All your question is is my example in general terms.

Therefore, applying this logic, x > a, x < -a, OR, |x| > a

Does this work though? Isn't a a variable too?
e.g., x = -1/2, a = -1. Your statement there is true, but subbing the same values for the original question is not.

 

[Will Hunting]

Yeah, man, but it's a dependent variable that "responds" to changes in x. The pronumeral, x, is usually always taken as the independent variable in mathematics (Like a function of x, where y, the ordinate, responds to changes in x, the abscissa). In other words, the question is to be solved with respect to, of for, x.

 

[dawma88]

X^2 > A^2

X^2 - A^2 > 0

(X-A)(X+A) > 0

by drawing simple concave up parabola, and letting the intercepts be -A (LHS of origin) & +A (RHS of origin), it can be clearly seen that X < - A and X > A if the solution is (X-A)(X+A) > 0

hope it helps, my son

 

[Spadge]

X^2 > A^2

X^2 - A^2 > 0

(X-A)(X+A) > 0

by drawing simple concave up parabola, and letting the intercepts be -A (LHS of origin) & +A (RHS of origin), it can be clearly seen that X < - A and X > A if the solution is (X-A)(X+A) > 0

hope it helps, my son

Expanding on this, if you've got the parabola crossing the xaxis at -A and A, it will cross the yaxis (x=0) at:
(0-A)(0+A)
=-A * A
=-(A^2)

So therefore, it will cross the yaxis =<0, as A^2 has to be positive, and there's the negative out front, and therefore the parabola has it's head down.

 

[shafqat]

nah mate its concave up
think abt it some more: as x goes to +/- infinity, y goes to?

 

[dawma88]

nah mate its concave up
think abt it some more: as x goes to +/- infinity, y goes to?

yeah man .....