inequalities with x in the denominator (1 Viewer)

[kr73114]

eg...
1/x>4
or
4/2x-1<2/3
or
6/5x-2<2

 

[x jiim]

Because they are inequalities, you have to multiply both sides by the denominator squared [as multiplying by a negative means you have to change the signs around]. Eg:
1/x>4
x > 4x^2
x(4x-1)<0
0<x<1/4

4/(2x-1)<2/3
24x-12<8x^2-8x+2
4x^2-16x+7>0
etc.

Or you could graph it, depending on how yucky the inequality is =\

 

[Aquawhite]

These are Mathematics Extension 1 problems. It would be much better if you could repost these questions in the Mathematics Extension 1 forums.

+1

I also prefer to approach these questions by using a squaring method in which we would multiply each side by x^2 (if x is the pro numeral) and then you will end up with a nice quadratic to solve most times...

 

[cutemouse]

Could also use the case method... although it appears that alot of people don't get it.

I remember in Year 11 the teacher of another class taught this method to his students. Only one of them (out of 30) got it in the test, whereas in our class everyone got the correct answer using the above method.

 

[jet]

Could also use the case method... although it appears that alot of people don't get it.

I remember in Year 11 the teacher of another class taught this method to his students. Only one of them (out of 30) got it in the test.

From its title I can guess how the method works. Though it does seem more difficult.

The squaring method is nice and simple, as long as students understand that they may be introducing solutions.

 

[Aquawhite]

From its title I can guess how the method works. Though it does seem more difficult.

The squaring method is nice and simple, as long as students understand that they may be introducing solutions.

That only generally occurs when dealing with trigonometric functions though - to my knowledge.

 

[jet]

That only generally occurs when dealing with trigonometric functions though - to my knowledge.

Nope. If there is a less/greater than or equal to sign, then the boundaries mightn't be included even if the solutions say they are.

The reason being that quadratics have 2 solutions whilst linear equations only have one. When you square a linear equation, you introduce another solution, even if it might just be a double root.

That's why you always always always substitute the boundaries back into the original inequality. If a solution is introduced, it generally means that it will cause an undefined quantity in the equation. (This is analogous to what Aquawhite said, as you might find (with ASTC in trig) that a solution in the wrong quadrant is introduced. t-results are also bad because they skip the solution 180 degrees which is undefined because t = tan(x/2)).

 

[The Nomad]

Could also use the case method... although it appears that alot of people don't get it.

I remember in Year 11 the teacher of another class taught this method to his students. Only one of them (out of 30) got it in the test, whereas in our class everyone got the correct answer using the above method.

The case method is quite useful...but I doubt most teachers would understand it

 

[Aquawhite]

Nope. If there is a less/greater than or equal to sign, then the boundaries mightn't be included even if the solutions say they are.

The reason being that quadratics have 2 solutions whilst linear equations only have one. When you square a linear equation, you introduce another solution, even if it might just be a double root.

That's why you always always always substitute the boundaries back into the original inequality. If a solution is introduced, it generally means that it will cause an undefined quantity in the equation. (This is analogous to what Aquawhite said, as t-solutions generally introduce the solution 180 degrees which is undefined when t = tan(x/2)).

Ah, I'd always subbed the answers back into the equation at the end (unless it was super easy and simple) to ensure I was right and didn't make a silly error for that very reason... Generally you'll find in HSC questions (because these appear in the first or second question) that they are simple and don't require much work.

 

[cutemouse]

From its title I can guess how the method works. Though it does seem more difficult.

The squaring method is nice and simple, as long as students understand that they may be introducing solutions.

But for some other types you need to use the case method.

Eg. |5x-2|/x > 0 or |x+1| + |2x-1| < 5

 

[jet]

Yeah, I use the case method there. I've always enjoyed that with absolute values No idea why though.

 

[Aquawhite]

What's the case method? Either I know it by a different name or have never learnt it before.

 

[jet]

What's the case method? Either I know it by a different name or have never learnt it before.

"When denominator ≥ 0...

When denominator < 0 ..."

 

[Aquawhite]

"When denominator ≥ 0...

When denominator < 0 ..."

Oh, okay I just didn't know it by that name haha

We learnt this method first before I heard about the squaring method (which I prefer) and I didn't know it had a name

 

[Tofuu]

since your not given any info on weather x is >0 or <0
this is a problem because if you move negatives the sign will change (<,>,=<,=>)
so to take the problem away, multiply both sides by the denominator squared

 

[The Nomad]

But for some other types you need to use the case method.

Eg. |5x-2|/x > 0 or |x+1| + |2x-1| < 5

I'm not sure what you mean by this?

|5x-2|/x > 0, automatically x > 0.

And for |x+1| + |2x-1| < 5, I just do ++, +-, -+, -- and then substitute back into the equation to see if it works.

Case method does not have to be used.

 

[shady145]

Re: solving inequalities with x in the denominator

you have to know the rule that... if you time or divide both sides by a negative number the inequality sign will flip...
so you make 2 cases... then work through it...
eg first one
1/x>4
case 1, when x>0 ie a positive (this will not make the sign change)
1>4x
1/4>x
x<1/4
so x>0, and x<1/4
so the solution is in that region... 0<x<1/4
case 2, when x<0 ie a negative number (this will make the sign flip)
1<4x
1/4<x
... you can finish this + the rest

 

[pman]

Re: solving inequalities with x in the denominator

you just have to multiply both sides by the denominator squared, can''t remember why squared

 

[Aquawhite]

Re: solving inequalities with x in the denominator

you just have to multiply both sides by the denominator as shown above

For that method you have to either use a set of rules or multiply by the square of the denominator (realizing that this can in fact create more answers), and then substitute the answers back in or simply evaluate the answers in your head.

 

[shady145]

Re: solving inequalities with x in the denominator

Oo my post is retarded... when i press edit there is mroe stuff in there than what you can see normally