maths help (2 Viewers)

[tommykins]

Re: maths again

You can't square to get rid of absolute value.

|x| = surd (x^2)

|x| =/= (x^2)

|x| = sqrt(x^2)
|x|^2 = x^2 = x^2 (lol just wrote it twice )

 

[untouchablecuz]

Re: maths again

|x| = sqrt(x^2)
|x|^2 = x^2 = x^2 (lol just wrote it twice )

O! Yer I see, it's just squaring both sides. LOL.

 

[-tal-]

Re: ASAP help!!!!!!!!!!!!!!!!!!!!!!!!!!11

Your working out is not very logical. You don't just multiply it by 100.

I'm pretty sure you do.

Since multiplying anything with base 10 adds another digit to the number, multiplying 3 by 100 would work properly to get the amount of digits.

 

[bubblesss]

inequation

hey guys im a bit stuck on this. any help?????

2/2-x >= 3
please show relevant working.

 

[tommykins]

Re: inequation

2/(2-x) ≥ 3
2(2-x) ≥ 3(2-x)²
0 ≥ 3(2-x)² -2(2-x)
0 ≥ (2-x)[3(2-x) - 2]
0 ≥ (2-x)[6-3x - 2]
0 ≥ (2-x)[4-3x]

Draw the graph, when the graph is 0>, that is your domain.
Thus x ≥2 and x ≤ 4/3
but x =/= 2

Solution x >2 and x ≤ 4/3

 

[bubblesss]

Re: inequation

2/(2-x) ≥ 3
2(2-x) ≥ 3(2-x)²
0 ≥ 3(2-x)² -2(2-x)
0 ≥ (2-x)[3(2-x) - 2]
0 ≥ (2-x)[6-3x - 2]
0 ≥ (2-x)[4-3x]

Draw the graph, when the graph is 0>, that is your domain.
Thus x ≥2 and x ≤ 4/3
but x =/= 2

Solution x >2 and x ≤ 4/3

hey but if x>2 then say we sub x=3. it doesnt work out rite?
2/2 - 3 >= 3
and -2 is not greater than 3???????

 

[tommykins]

Re: inequation

I'll do the question again, sorry. Hard to keep track of it on the computer.

How silly of me, I was looking at the positive domains.

4/3 ≤ x < 2

 

[bubblesss]

Re: inequation

I'll do the question again, sorry. Hard to keep track of it on the computer.

thanks so much!!!!!!11

 

[tommykins]

Re: inequation

Look at the editted post.

 

[bubblesss]

Re: inequation

Look at the editted post.

so when u substitute to check the value which equation do we sub it in???? the actual question or the most recent one we formed???
sry again for bugging u.

 

[bored of sc]

Re: inequation

hey guys im a bit stuck on this. any help?????

2/2-x >= 3
please show relevant working.

2/(2-x) > 3

2(2-x) = 3(2-x)²

4-2x = 3(4-4x+x²)
3x² - 10x + 8 = 0
x = 4/3, 2

Sub in x = 0

1 > 3 False

But 2-x cannot equal 0.

Therefore x cannot be 2.

Therefore,

4/3 < x < 2

 

[bored of sc]

Re: inequation

so when u substitute to check the value which equation do we sub it in???? the actual question or the most recent one we formed???
sry again for bugging u.

The first equation i.e. the question.

 

[tommykins]

Re: inequation

so when u substitute to check the value which equation do we sub it in???? the actual question or the most recent one we formed???
sry again for bugging u.

We sub in the maximum/minimum domains into the original question.

ie, for a ≤ x ≤ b, you sub in a and b to make sure the inequality works.
HOWEVER, if it's a < x < b, you need to sub a value abit higher than a, and also a value abit lower than be

For example, some random question - we have as the final answer
2≤x ≤3

You'd sub in x = 2 and x = 3 into the original equation.

but if we have -
2 < x < 3, you'd sub in x = 2.00001 and x = 2.99999 into the original equation.

If we have something like, x < 3/2 and x > 5, same protocol applies. Look out for the equal signs.

I rarely ever check as normally I'm confident with all my working out, and if it works out nicely (mainly does) then you should simply just check if your domains do not make the denominator = 0

Like in your question, we had 2/(2-x) ≥ 3

We know that x =/= 2, so our solution of 4/3 ≤ x ≤ 2 is false as x =/= 2.

So the solution becomes 4/3 ≤ x < 2

 

[bubblesss]

Re: inequation

The first equation i.e. the question.

thanks a lot to everyone.

 

[bored of sc]

Re: inequation

The denominator with the variable (x) in it is ALWAYS a factor when you are solving (I think).

 

[tommykins]

Re: inequation

The denominator with the variable (x) in it is ALWAYS a factor when you are solving (I think).

They normally make it so to catch out the students who don't check solutions.

But change a few numbers at the numerator and you'll realise that it isn't always a factor.

 

[youngminii]

Re: inequation

Can I just ask,
Why do you times both sides by (2-x)^2 at the start?
Can't you just times both sides by (2-x)?
I know it comes out wrong when you do it by the way I said to, but shouldn't you be able to?

 

[bubblesss]

Re: inequation

Can I just ask,
Why do you times both sides by (2-x)^2 at the start?
Can't you just times both sides by (2-x)?
I know it comes out wrong when you do it by the way I said to, but shouldn't you be able to?[/quotes

we times it by (2-x)^2 so that we can get rid of any negative terms. ur ans will be wrong if u just times it by (2-x)

 

[tommykins]

Re: inequation

Can I just ask,
Why do you times both sides by (2-x)^2 at the start?
Can't you just times both sides by (2-x)?
I know it comes out wrong when you do it by the way I said to, but shouldn't you be able to?

The reason we multiply both side by (2-x)² is to eliminate the possibility that x can be negative. Multiplying both sides by the denominator squared makes it positive, and positives are easier to work with.

 

[lolokay]

Re: inequation

you don't know if (2-x) is positive or negative. If it was negative you would have to change the sign around. However, (2-x)² is always positive so this is not a problem.

hey guys im a bit stuck on this. any help?????

2/2-x >= 3
please show relevant working.

another way to solve this is to use the critical points method - finding where the expression changes from true to false
these points will be when the denominator = 0; x=2
and when 2/(2-x)=3
2=6-3x
x = 4/3

and you can see that the inequation is false when x is very small or very large, or the inequation is undefined (when x=2), so
4/3 =< x < 2

sorry if this was a bit confusing