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Maths Question ^_^ (2 Viewers)

bored of sc

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I've got a question for you guys: is x^2 - 3 a difference of two squares?
 

bored of sc

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aMUSEd1977 said:
(x - (3)0.5)(x + (3)0.5)

You roots would be +/- root(3), so yes.

Guys, use the sup tags.

x{sup}2{/sup}

will give x2...just replace {} with []
you are a wise one

x2 yay, it worked, i think
 

Aplus

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aMUSEd1977 said:
Guys, use the sup tags.

x{sup}2{/sup}

will give x2...just replace {} with []
Thanks for the tip.
 

selablad

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aMUSEd1977 said:
Guys, use the sup tags.

x{sup}2{/sup}

will give x2...just replace {} with []
Wow, that's useful, thanks10

And the other way? Does that work? H2O2

Yay it does!

Sorry, pointless post :shy:
 

Pink Oni

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I got stuck in my studies again :( More Inequalities :(

How do you solve for x in these types (just giving the specific question):

22x - 2(2x) <= -1

BTW, I'm using Fitzpatrick 3U textbook, pg 60 if anyone was interested.

Thanks for the help everyone :D
 

Aplus

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Pink Oni said:
22x - 2(2x) <= -1
22x - 2(2x) ≤ -1
22x - 4x ≤ -1
2x(12 - 2) ≤ -1
2x(1 - 2) ≤ -1
2x(-1) ≤ -1
2x ≤ -1/-1
2x ≤ 1
Since anything to the power of 0 is 1, then the value of x must be less than or equal to 0.
 

Pink Oni

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Aplus said:
Since anything to the power of 0 is 1, then the value of x must be less than or equal to 0.
Thanks so much, I'm REALLY greatful. Is their a way I can rearrange 2x ≤ 1 so that it says x ≤ 0 or in a test would I need to write it out in words, as you just mentioned to me?
 

Pink Oni

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OK, to check my understanding, I did the only other similar question in the exercise:

22x - 5(2x) + 4 < 0

22x - 10x < -4

2x(12 - 5) < -4

2x(-4) < -4

2x < -4/-4

2x < 1

x < 0?
 

lyounamu

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Aplus said:
22x - 2(2x) ≤ -1
22x - 4x ≤ -1
2x(12 - 2) ≤ -1
2x(1 - 2) ≤ -1
2x(-1) ≤ -1
2x ≤ -1/-1
2x ≤ 1
Since anything to the power of 0 is 1, then the value of x must be less than or equal to 0.
Your working out is incorrect. It is impossible to go from 2(2^x) to 4^x. It is against the Index Law.
If you also test your solution, it won't fit the question (e.g. check x=-1, it does not satisfy the equation). Only possible solution is x=0.

2^2x – 2(2^x) <= - 1
Let 2^x = m
M^2 – 2m <= -1
M^2 – 2m + 1 <= 0
(m-1)^2 <=0
Therefore, (2^x – 1)^2 <=0
(Draw the graph here)
Therefore (2^x-1)^2 <= 0 when x=0
 
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lyounamu

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Pink Oni said:
OK, to check my understanding, I did the only other similar question in the exercise:

22x - 5(2x) + 4 < 0

22x - 10x < -4

2x(12 - 5) < -4

2x(-4) < -4

2x < -4/-4

2x < 1

x < 0?
As I mentioned above, it is impossible to move from 5(2^x) to 10^x. It just cannot happen.

Let 2^x = m

Therefore, m^2 - 5m +4 <= 0
(m-4)(m-1)<=0
So (2^x-4)(2^x-1) <=0
(Draw the graph here or test the points)
Therefore, x<= 0 or 0<=x<=2
 

tommykins

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I highly suggest for questinos like these- you use a dummy variable.

That is, let m = a term and it should work out as a quadratic.
 

selablad

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tommykins said:
I highly suggest for questinos like these- you use a dummy variable.

That is, let m = a term and it should work out as a quadratic.
You mean like this?

22x - 5(2x) + 4 < 0

(2x)2 - 5(2x) + 4 < 0

Let m = 2x

m2 - 5m + 4 < 0

and solve?

Because my teacher told me to do something like that, but I didn't really understand what she meant...
 

Pink Oni

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lyounamu said:
As I mentioned above, it is impossible to move from 5(2^x) to 10^x. It just cannot happen.

Let 2^x = m

Therefore, m^2 - 5m +4 <= 0
(m-4)(m-1)<=0
So (2^x-4)(2^x-1) <=0
(Draw the graph here or test the points)
Therefore, x<= 0 or 0<=x<=2
Exphate said:
I should've also mentioned, that after 3 or 4 goes, you should be able to pick up the pattern

y = +-mx2+2k - parabola
y = +-mx3+2k - cubic

When you start plugging in k's (positive terms only) your gradient becomes steeper and the curve increases faster. You'll do a few questions and realise "hay, a cubic is this shape, and a parabola is this shape". Then you just have to adjust it according to your m and k values. (k=2, m=1 ---> x4)
Thank you both for these solutions and methods, I have an Extension Maths Half Yearly tomorrow and I want to be able to knock out any question I encounter :D
 

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Pink Oni said:
Thank you both for these solutions and methods, I have an Extension Maths Half Yearly tomorrow and I want to be able to knock out any question I encounter :D
Good luck :karate:
 

Aplus

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lyounamu said:
Your working out is incorrect. It is impossible to go from 2(2^x) to 4^x. It is against the Index Law.
If you also test your solution, it won't fit the question (e.g. check x=-1, it does not satisfy the equation). Only possible solution is x=0.

2^2x – 2(2^x) <= - 1
Let 2^x = m
M^2 – 2m <= -1
M^2 – 2m + 1 <= 0
(m-1)^2 <=0
Therefore, (2^x – 1)^2 <=0
(Draw the graph here)
Therefore (2^x-1)^2 <= 0 when x=0
.... How shameful :(
 

Aplus

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Aerath said:
LOL, we're all bound to make stupid mistakes. In my Maths half yearly, I wrote this:

k/3 = 1
k = 1

Teacher marking the exam wrote: "Haha" over it.
I admire who that teacher is :eek:
 

Pink Oni

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Aplus said:
Because you always get full marks :p
...And they don't give a half-ark in the real thing ;D

I did my test and it was pretty bad to say the least. Mostly geometry which got me a little cheesed but at least I'm not alone (everyone I spoke to said they failed for sure, I'm hoping that wasn't modesty :().
 

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