Maths Question ^_^ (1 Viewer)

[bored of sc]

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

 

[bored of sc]

(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]

Guys, use the sup tags.

x{sup}2{/sup}

will give x²...just replace {} with []

Thanks for the tip.

 

[selablad]

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

 

[Pink Oni]

I got stuck in my studies again More Inequalities

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

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

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

Thanks for the help everyone

 

[Aplus]

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

2^2x - 2(2^x) ≤ -1
2^2x - 4^x ≤ -1
2^x(1² - 2) ≤ -1
2^x(1 - 2) ≤ -1
2^x(-1) ≤ -1
2^x ≤ -1/-1
2^x ≤ 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]

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 2^x ≤ 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]

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

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

2^2x - 10^x < -4

2^x(1² - 5) < -4

2^x(-4) < -4

2^x < -4/-4

2^x < 1

x < 0?

 

[lyounamu]

2^2x - 2(2^x) ≤ -1
2^2x - 4^x ≤ -1
2^x(1² - 2) ≤ -1
2^x(1 - 2) ≤ -1
2^x(-1) ≤ -1
2^x ≤ -1/-1
2^x ≤ 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

 

[lyounamu]

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

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

2^2x - 10^x < -4

2^x(1² - 5) < -4

2^x(-4) < -4

2^x < -4/-4

2^x < 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]

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]

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?

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

(2^x)² - 5(2^x) + 4 < 0

Let m = 2^x

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]

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

I should've also mentioned, that after 3 or 4 goes, you should be able to pick up the pattern

y = ⁺_-mx^2+2k - parabola
y = ⁺_-mx^3+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

 

[selablad]

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

Good luck

 

[Aplus]

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

 

[Pink Oni]

Good luck

Thank you very much

.... How shameful

Don't feel down, you got me further than nowhere didn't you? Chill out

 

[Aplus]

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

 

[lyounamu]

He still gave me 1/2 a mark. So I got 2.5/3.

That's really generous. I never get half mark. :angry:

 

[Aplus]

That's really generous. I never get half mark. :angry:

Because you always get full marks

 

[Pink Oni]

Because you always get full marks

...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 ).