URGENT half yearly tomorrow :( (1 Viewer)

[velox]

if i have an inequality like this (a+4/3-a) < 1
how do i go about solving it? i know you should multiply both sides by the square of the denominator, but i dont know what to do next. thanks

 

[zahid]

i luv these questions (inequalities with the unknown in the demoninator). All you have to do is consider two possible cases for the denominator either:
3-a > 0 or
3-a < 0

then see which one holds simultaneously.....Simple

 

[grimreaper]

I prefer the method wrx was thinking about. Once youve multiplied by the square of the denominator, you bring everything to one side and then factorise (its pretty easy since you already have a common factor of something, 3-a in the case you have). Once thats done, sub in a value to see where the inequality holds

 

[Estel]

Multiplying by the square of the denominator, considering both cases of x being positive or negative, are both methods which pale against the graph...

Go the graph!

Actually I use all three methods.

For your q i'd multiply by square
(a+4)/(3-a) < 1
(a+4)(3-a) < (3-a)^2
(3-a)(a+4-3+a) <0
(3-a)(2a-1) <0
Graph.

 

[velox]

would u just use quadratic formula instead of subbing in a test value?

Estel, dont you have to sqaure it first blah blah and then do the graph?

 

[Estel]

You square the denominator.
The reason is if you multiply by a negative number the sign changes.
If you multiply by a squared number this is not a concern.

 

[velox]

i wasnt debating that, anyway thanks guys. its cleared things up for tomorrow.

 

[Estel]

Oh you meant the pure graphical method?
That involves breaking the graph up and using your curve sketching menu...
1 quick easy step... but less accurate. Test for critical values if needed.

For some inequalities this is by far and away the easiest and least painful method. 5 mark inequalities that would take a page shrivel away in insignificance.
That's why graphs are good.

 

[velox]

yes, of course i know that.........lol dont worry
cos i am crap at that method

 

[Estel]

Well why didn't you do it with the orginal q?

 

[capsicum]

this is what i'd do for this q:
1. square the denominator and multiply both sides by the square
2. bring everything to one side so the other equals zero
3. solve the quadratic
4. graph to see where a satisfies inequality

 

[CrashOveride]

Originally posted by capsicum
this is what i'd do for this q:
1. square the denominator and multiply both sides by the square
2. bring everything to one side so the other equals zero
3. solve the quadratic
4. graph to see where a satisfies inequality

I think you've nicely summed up what everyone was saying

In any case, you should get x < -0.5 or x > 3