Inequalities Questions (1 Viewer)

[Twickel]

Hi
Could you please give me some extentsion level inequalities and could you solve the following for me.

x+1/x is greater then or equal to 2

1/x < 1/x+1

 

[Iruka]

Hi
Could you please give me some extentsion level inequalities and could you solve the following for me.

x+1/x is greater then or equal to 2

1/x < 1/x+1

For the first one, we want to multiply throught by x to get rid of the algebraic fraction, which means that we have to consider two cases:

if x>0, then
x^2 -2x +1 >=0
(x-1)^2>=0, which is always true.

If x<0, then we must reverse the inequality, so we get (x-1)^2<=0, which is never true.

So the solution is x>0

For the second one, it is always true except for x=0, in which case the expression 1/x is undefined.

 

[leoyh]

for the first one, you times throughout by x^2 since x can either be positive or negative, so multipying throughout by x^2 considers both cases. so you can (x+1)x > or equal to 2x^2, then bring everything to one side, factorise and solve and you should get 0 < x < or equal to 1

btw, i interpreted the Q as (x+1)/x > or equal to 2

for the second one, times throughout by x^2 and hten by (x+1)^2 and then simplify and solve as you would do in number 1. theres probably a shorter way to that one though
and hopefully you get -1 < x < 0

btw, i interpreted the Q as 1/x < 1/(x+1)

hopefully my answers are right though lol, if in interpreted the Q's right

 

[Slidey]

Um. 1/x is never less than 1/(x+1).

Give me some real examples and I'll help.

x+1/x >= 2
x+1/x-2 >= 0
(x^2-2x+1)/x >= 0
x(x-1)^2 >= 0 (must multiply by x^2 because x could be negative which ruins inequalities)
Thus when 0<=x<=1 and when x>=1.
Or, really, x<=0.
But since x never equals zero:
x<0

Edit: Note how if you had of said x+1/x > 2, the answer would then be:
0 < x <1 and x>1

 

[Twickel]

See the way I do it is multiply the LHS and RHs by the demoinator not times it byitself sqaured and then test numbers.

I thoguht that for the first question we would factorise then test

 

[Iruka]

How about you people use some brackets so we know what question you actually mean?

 

[Slidey]

See the way I do it is multiply the LHS and RHs by the demoinator not times it byitself sqaured and then test numbers.

I thought that for the first question we would factorise then test

No, you can't do that.

You must ALWAYS multiply by a square.

(x+4)/(-8x+7) >= -1
x+4 >= 8x-7 (right???)
-7x+11 >=0, so
x<=11/7

Let's test that:
x=1
5/-1 = -5
Is -5 greater than -1?
No.

So let's try again this time multiplying by the SQUARE of the denominator:
(x+4)(-8x+7) >= -(-8x+7)^2
(x+4)(-8x+7)+(-8x+7)^2 >= 0
(-8x+7)(x+4-8x+7) >= 0
(8x-7)(7x-11) >= 0
7/8 < x and x >= 11/7
Yay it works. (Note the < sign for 7/8 rather than <= this is because x=7/8 is an asymptote)

Why does this work? Because a square is always positive. Whilst (-8x+7) could be anything. It looks mostly negative to me, though, which means unpredictable behaviour when it comes to inequalities signs.

 

[SpinCobra]

Cambridge. They go from easy to hard to impossibly hard.

 

[Iruka]

You can either multiply by a square, consider cases, or you could try graphing the function if you're good at curve sketching.

The important thing is not to treat it like an equality.

 

[tommykins]

See the way I do it is multiply the LHS and RHs by the demoinator not times it byitself sqaured and then test numbers.

I thoguht that for the first question we would factorise then test

You must multiply it by a square, as the x value can be positive or negative, multiplying it by a square gets rid of the issue.

I can't be botherd doing it at the moment but the steps are normally -

• Multiply both sides by denominator squared.
• Algebraic manipulation, get the whole experssion on one side and make it < or > to 0.
• Collect like terms, and sketch the graph (this is particularly easy with quadratics).
• If the expression is <0, the answer is when x is BELOW the x axis (that is, when y is negative) but if it is >0, the answer is when x is ABOVE the x axis.

 

[Twickel]

I undersand, so all the questions in my textbook have the denominator as positive, in this case the multiplying by the denominator works, if the denominator is negative we must use the squairng method.

Thanks for the help.

 

[Slidey]

I undersand, so all the questions in my textbook have the denominator as positive, in this case the multiplying by the denominator works, if the denominator is negative we must use the squairng method.

Thanks for the help.

No offence, but it doesn't seem as though you understand the method (anybody who says they'll only use the square/cases sometimes is going to get lots of questions wrong).

Try these:

1) 1/(x-1) <= 3
2) x/(x^2-1) >= 0
3) x-1/x < 0
4) (x^2-9)/x >= 0
5) 2/x <= 3/(x+2)

 

[tommykins]

I undersand, so all the questions in my textbook have the denominator as positive, in this case the multiplying by the denominator works, if the denominator is negative we must use the squairng method.

Thanks for the help.

Know WHY you 're doing it, it is because x is an unknown, for all we know, x could be negative, so the sure thing we can only do is square each side by the denominator that has an x.

If the denominator was simply 6 or a Real number, you can just simply multiply each side by that value.

 

[Twickel]

x>=4/3, x< 1 ???? I think I did it wrong, thats the answer for the first one.


But about the sqauring method questions such as 1/n-3>=2
5m+4/2m < 1/4

The multiplying by the denominator method works and in all those questions the denominator was positive, had it been a negavtive it would not have worked only the saquring method would have.

The question in Maths in Focus all have +denominators e.g 2m 2p-9 3t-8.

Im not saying what Im doing is right, its just the moethod I have been using has worked, if the denominaotr was -29-p it would not work right, thats when i would have to use the sqauiring method.

Sorry for spelling.

 

[Slidey]

Fair enough mate. I'd obviously be a fool to tell you not to do something a certain way if it gets you the right answer.

But personally I do recommend multiplying by the square of the denominator, regardless of the question. That way you can guarantee your answer is correct, instead of worrying about negatives or cases.

Because if Maths In Focus truly does only have questions that have positive denominators, that's a big flaw on its part; you can be certain the exams won't be the same.

 

[Twickel]

Yeah I guess. Thanks alot for your help. Do you think you could give a cuple questions with a negative demoniator with solotutions.

 

[Slidey]

Yeah I guess. Thanks alot for your help. Do you think you could give a cuple questions with a negative demoniator with solotutions.

Try the 5 I posted using your method, then I'll post solutions.

 

[Twickel]

x^2-1/x>0 is the answer x> then 1 or x is < -1

 

[Slidey]

It's greater than or equal to. Answer: x<=0 and x>1

 

[leoyh]

multiply it by a square. thats exactly what i did from the beginning T__T "