Inequality difficulties D= (1 Viewer)

[Finx]

Hey again >.>

This time I'm having trouble with inequalities in extension 1 maths;

Q1:

Solve x^2 > a^2

~ I have no idea how to write this down in equation form.


Q2:

Solve |t+2| + |3t-1| < 5

~ My working;

t+2 + 3t-1 < 5
4t + 1 < 5
4t < 4
t < 1

The answer is -1<t<1 :: how?

Ty =]

 

[Aplus]

Q1.
x^2 > a^2
x^2 - a^2 > 0
(x+a)(x-a) > 0
x+a = 0 or x-a = 0
x= -a or x = a
a= -x a = x

Q2.
This is how you solve absolute inequalities with a less than sign:
|t+2| + |3t-1| < 5
-5 < t+2+3t-1 < 5
-5 < 4t+1 < 5
-6 < 4t < 4
-6/4 < t < 1
-1 1/2 < t < 1

Unfortunately, for Q2. I did not get the same answer as -1<t<1

 

[Finx]

For Q1, the answer was " x < -a, x > a "

How do I achieve this answer?

I'm still scratching my head at Q2 - I'm pretty sure the answer sheet is 100% right.

 

[iCpurplehippos]

Q1.
x^2 > a^2
x^2 - a^2 > 0
(x+a)(x-a) > 0
x+a = 0 or x-a = 0
x= -a or x = a
a= -x a = x

Q2.
This is how you solve absolute inequalities with a less than sign:
|t+2| + |3t-1| < 5
-5 < t+2+3t-1 < 5
-5 < 4t+1 < 5
-6 < 4t < 4
-6/4 < t < 1
-1 1/2 < t < 1

Unfortunately, for Q2. I did not get the same answer as -1<t<1

The answer to the first one is actually x<-a, x>a
=P

In the second question, don't you have to consider 4 cases seeing as there are 2 absolute values in it? We did a similiar question to that at school and the working out took a whole A4 page. :S What we did first was work out the boundary points, plotted it on a number line and then used the testing method to see what region it's in.

 

[Aplus]

You have to draw a parabola for Q1.

 

[Finx]

The answer to the first one is actually x<-a, x>a
=P

In the second question, don't you have to consider 4 cases seeing as there are 2 absolute values in it. We did a similiar question to that at school and the working out took a whole A4 page. :S What we did first was work out the boundary points, plotted it on a number line and then used the testing method to see what region it's in.

Could you please demonstrate how you got the answer to Q1? Thanks =)

 

[iCpurplehippos]

What Aplus said...sketch a parabola and then find the regions where the parabola is greater than 0.

 

[cwag]

Q1 x² - a² > 0

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

if we draw a parabola with concave positive we find that x > 0 when it is less that -a or greater than a. therefore solutions are x > a, x < -a

Q2. yea i always thought u have to consider all posibilties, then delete the wrong ones..

for both positive
x+2+3x-1 <5
4x<4
x<1

first one negative
-x-2+3x-1 <5
2x<8
x<4 (doesn't work)

both negative
-x-2-3x+1<5
-4x-1<5
-4x<6
x>3/2 (doesn't work)

second one negative
x+2-3x+1<5
-2x+3<5
-2x<2
x>-1

there fore -1<x<1

 

[Aplus]

Yeah 4 answers, I was lazy.

 

[Continuum]

for both positive
x+2+3x-1 <5
4x<4
x<1

first one negative
-x-2+3x-1 <5
2x<8
x<4 (doesn't work)

both negative
-x-2-3x+1<5
-4x-1<5
-4x<6
x>3/2 (doesn't work)

second one negative
x+2-3x+1<5
-2x+3<5
-2x<2
x>-1

there fore -1<x<1

Ahh, the algebraic method.

I reckon that the graphical method of solving the inequality would be easier for the OP, since he/she can actually see the equation and see when |t+2|+|3t-1| is below 5. Here it is:

 

[st1m]

easy Finx

I will explain it to u in the easiest way

i will be back

bye

 

[Finx]

Actually you do need to do all four cases because there is a number on the right.

I still can't express Q1 as an equation (do I really have to draw the graph to prove my answer?)

 

[lolokay]

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.

 

[bored of sc]

absolute value and squaring something is very similar - the negatives go in both

 

[iCpurplehippos]

You can do the testing method.
I prefer the graphing method as it is far more easy.

 

[Sarah182]

OK I have question thats a bit annoying
it was something like this
lx+2l + lx+5l<4
anyways I can do if they are both negative and both positive but when one is pos and one is neg then the x's die.. so im only left with numbers
for example:
(first one is neg, second is pos)
-(x+2)+x+5<4
= -x-2+x+5<4
=3<4
so does that make the statement true, even though I didnt have to test with x?
If any of that really made any sense haha.

 

[Finx]

Yes actually, it means the statement is true.

 

[lolokay]

-(x+2)+x+5<4 works for any -5 < [or =] x < [or =] -2.

for the other neg/pos case..

lx+2l + lx+5l<4
x + 2 -(x + 5)
x + 2 -x -5
-3 < 4

There are no solutions for this one.
(As [x>-2 and x<-5] makes no sense/has no values)

hopefully got that right

 

[Finx]

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.

 

[bored of sc]

OK I have question thats a bit annoying
it was something like this
lx+2l + lx+5l<4
anyways I can do if they are both negative and both positive but when one is pos and one is neg then the x's die.. so im only left with numbers
for example:
(first one is neg, second is pos)
-(x+2)+x+5<4
= -x-2+x+5<4
=3<4
so does that make the statement true, even though I didnt have to test with x?
If any of that really made any sense haha.

when the x's die it means that answer doesn't work

i think