bored of sc
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I've got a question for you guys: is x^2 - 3 a difference of two squares?
you are a wise oneaMUSEd1977 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 []
Thanks for the tip.aMUSEd1977 said:Guys, use the sup tags.
x{sup}2{/sup}
will give x2...just replace {} with []
Wow, that's useful, thanks10aMUSEd1977 said:Guys, use the sup tags.
x{sup}2{/sup}
will give x2...just replace {} with []
22x - 2(2x) ≤ -1Pink Oni said:22x - 2(2x) <= -1
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?Aplus said: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.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.
As I mentioned above, it is impossible to move from 5(2^x) to 10^x. It just cannot happen.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?
You mean like this?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.
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
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 encounterExphate 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)
Good luckPink 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
.... How shamefullyounamu 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
Thank you very muchselablad said:Good luck
Don't feel down, you got me further than nowhere didn't you? Chill outAplus said:.... How shameful
I admire who that teacher isAerath 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.
That's really generous. I never get half mark. :angry:Aerath said:He still gave me 1/2 a mark. So I got 2.5/3.
Because you always get full markslyounamu said:That's really generous. I never get half mark. :angry:
...And they don't give a half-ark in the real thing ;DAplus said:Because you always get full marks