solving inequalities (1 Viewer)

[ratherbesleepin]

okay - was taught to do these by timesing through by the bottom squared
Eg

1/(x+2)>3x
would then becme 1(X+2) > 3x(X+2)^2

and then draw the graph and solve the inequality. This worked all the time.
Now i do the same sorta qtns, the same way and i cant get any right.

Can someone tell me how i'm supposed to do it and why my way doesn't work anymore??

 

[...]

give us some of the questions and see...

the example method u got there is right...or u can do it graphically...

 

[redslert]

here is the way i do it

1/(x+2)>3x
move everything to one side
1/(x+2) - 3x > 0
put everything over the one denominator
[1 - 3x^2 - 6x] / (x + 2) > 0
SIMPLIFY IF POSSIBLE!
but did u just make up this question, because it doesnt exactly work out.......

SBS - Square Both Sides with denominator
[1 - 3x^2 - 6x] * (x + 2) > 0


and you should be able to do the rest

 

[iambored]

i do it a different way


u see what the bottom can't equal (in that case i can't equal -2)
then u let the 2 sides equal each other, and find x
1/(x+2) = 3x
1 = 3x(x+2)
0 = 3x^2 + 6x - 1
factorise and find x

then u draw it on a number line, and test the points between each

anyway, it might be stressful to learn a new way now, and your way is just an alternative method

just do what "..." said and post some questions up

 

[...]

hmm..haven't seen either iambored or redslert way b4...

but similiar to iambored..

1/(x+2) >3x

1/(x+2) * (x+2)^2 > 3x * (x+2)^2

x+2 cancel

1 * (x+2) > 3x(x+2)(x+2)

solve for x...

 

[ND]

Originally posted by ...
hmm..haven't seen either iambored or redslert way b4...

but similiar to iambored..

1/(x+2) >3x

1/(x+2) * (x+2)^2 > 3x * (x+2)^2

x+2 cancel

1 * (x+2) > 3x(x+2)(x+2)

solve for x...

I wouldn't use this method or the square method for this question, cubics are not nice. Here's what i'd do:

Firstly you have to remember that when to * or / by a negative number, you have to reverse the sign. So because (x+2) could be either +ve or -ve, we have to take both cases:

1/(x+2)>3x

when x+2>0 i.e. x>-2:

1>3x(x+2)
3x^2+6x-1<0
3x^2+6x-1 has no real roots and is concave up so is never < 0.

when x+2<0 i.e. x<-2:
1<3x(x+2)
3x^2+6x-1>0
3x^2+6x-1 has no real roots and is concave up so is >0.
.'. the condition for 1/(x+2)>3x is x<-2

 

[iambored]

Originally posted by redslert
but did u just make up this question, because it doesnt exactly work out.......

i was wondering as well, because my way wouldnt't have worked with whole numbers (that's why i stopped the question where i did, unless i didn'y try enough)

edit - how many ways????!!!

 

[redslert]

ok do this question

1/x >= 1/(x+1)

my approach

1. move everything to one side
1/x - 1/(x+1) >= 0

2. put everything over one denominator
[(x+1) - x] / x(x+1) >= 0

3. Simplify
1 / x(x+1) >= 0

4. Sqaure both sides with denom
x(x+1) >= 0

5. Solve inequality
<<<<_____>>>>
.......(-1).....(0)......

.'. x <= -1 & x>= 0
but x cannot equal -1 or 0

.'. x < -1 & x > 0

 

[iambored]

Originally posted by iambored
u see what the bottom can't equal (in that case i can't equal -2)
then u let the 2 sides equal each other, and find x
1/(x+2) = 3x
1 = 3x(x+2)
0 = 3x^2 + 6x - 1
factorise and find x

i don't know y mine words out differently to yours


continuing:
facotrise to find x
cann't be done so have to use formula
x = -b+-(squareroot b2-2ac))/(2a)

you get:
-1-sqrt8

or -1+sqrt8

that's approx = to 3.8 and -3.8

so ur 3 points are 3.8, -2 and -3.8
and then u draw these on a no line

-3.8_____-2_____3.8
and then test points between
e.g. test 1

ahh, crap, i just realised there is an x on the other side, i don't know if that would make a difference, i'm gonna try a different way

edit: makes no difference, i don't understand why it's not working

 

[Rahul]

Originally posted by redslert
ok do this question

1/x >= 1/(x+1)

my approach....

1/x >= 1/(x+1)

multiply each side by x^2 and (x+1)^2

x.(x+1)^2 >= x^2.(x+1)
...
x(x+1) >= 0

then solve by graphing.
you can treat these as two different finction to make it easier to understand.
y=x(x+1) and g=0

now graph where 'y is greater than or equal to g'.

answer: x<=-1, x>=0

 

[Rahul]

Originally posted by iambored
i do it a different way


u see what the bottom can't equal (in that case i can't equal -2)
then u let the 2 sides equal each other, and find x
1/(x+2) = 3x
1 = 3x(x+2)
0 = 3x^2 + 6x - 1
factorise and find x

then u draw it on a number line, and test the points between each

anyway, it might be stressful to learn a new way now, and your way is just an alternative method

just do what "..." said and post some questions up

iambored, dont do it! i doubt you will get marks for the working. marks are often assiciated with some trick with inequalities(multiply by the square of the denominator).

 

[ratherbesleepin]

Yeah i made up the qtn just to demonstrate my point

Real qtn - that i couldn't get to work my way
2/x>x-1

i ended up with -1 < x < 2 but x cant be 0

the answer in the back says x<-1 and 0< x<2

All yoru versions except ND's are pretty much the same way as i do them, give or take.

I thought there must have been some other ways to do it.

Originally posted by iambored

anyway, it might be stressful to learn a new way now, and your way is just an alternative method

just do what "..." said and post some questions up

I've done the 3u course over 3 years so i kinda am relearning all the basic stuff anyway - so learning something completely different isn't gonna be any more stressful

 

[SoCal]

I think this is a similar question to the ones you are doing. How would you answer this question? I have never been taught to answer questions with an x on top and bottom. Stupid teachers.

x/1-x^2> or equal to 0

How would you answer this question? I have never been taught to answer inequalities with an x on bottom and top. Stupid teachers.

 

[...]

Originally posted by Merethrond

x/1-x^20

wtf is that...

edit:
oh opps..lol

 

[SoCal]

Originally posted by ...
wtf is that...

edit:
oh opps..lol

Ha ha, I went into Character Map to get the greater than or equal to sign and then copied and pasted it onto this message but it didn't work, oh well.

 

[...]

getting back to ur question...

x/1 = x

and then u would factorise it and get somethng like 1>=x>=0

or am i wrong?

 

[Rahul]

Originally posted by Merethrond
x/1-x^2> or equal to 0

hmmmm i would do it by graphing. this way of graphing should be covered in 3u.

 

[SoCal]

"...", the answer is x<-1 or 0 = < x < 1


Rahul, I have never seen a question anything like this, so no we have not ever learnt to graph something like this.

I am thinking maybe times through by (1-x^2)^2

so you get x(1-x^2)>=0

Factorise

x(1-x)(1+x)>=0

So you get

x=0 or x=1 or x=-1

but where would you go after that to get the answer above?

 

[redslert]

you would draw the graph!!!!!1

hold on while i open paint

 

[redslert]

you draw the graph here:

and then u would shade in everything up top because it is 'greater than' (>)

and then the answer will be


-1 <= x <= 0 & x >= 1


and the condition is that x cannot equal 1 or -1

.'. -1 < x <= 0 & x > 1