MrBrightside
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2^2x -9.2^x +8=0
2^2x -9.2^x +8=0
holyshit thank you , i put 1 instead of 8 in the last line thnx....btw in the book it has no brackets, its like that.use brackets please
im pretty sure you mean
2 ^ (2x) - 9 . 2^x +8
= ( 2^x)^2 - 9 . 2^x +8 { when you raise a power to anther power, multiply powers}
let u = 2^x
u^2 -9u +8=0
(u-1)(u-8)=0
u=1, u= 8
2^x= 1 ---> x=0
2^x= 8--> x=3
x=0, x=3
holyshit thank you , i put 1 instead of 8 in the last line thnx....btw in the book it has no brackets, its like that.
so basically the 9.2 is round down...correct?
okay thx, wow wtf how come the book doesn't mention that....i bet the book was designed to make people fail.no, what i meant was 2^2x, even though it was fairly obvious you meant "2 to the power of 2x"
someone could read it as 2^2 times x, ie 4x, so put brackets around powers and stuff when there is more than just one number/term in the index
and the 9.2 means 9 times 2^x, you dont ROUND, its another notation for multiplication, just make sure you are aware of it
okay thx, wow wtf how come the book doesn't mention that....i bet the book was designed to make people fail.
wouldnt do anythingi would've logged the whole thing
teacher never told us tht...yea maths in focus (shit book at explaining things)the teacher should have mentioned that, common notation ( especially considering there are already x's in the eqn, if they put an x for "9 times 2^x" that would make it seem even more confusing)
its not the books fault lol, which book is it from by the way , maths in focus?
teacher never told us tht...yea maths in focus (shit book at explaining things)
yeah he does, but like when u look at the book the questions r different.i used it , its alright, well i had a good teacher.
doesnt your teacher go over some examples before you do the textbook exercises.
well same sort of thingyeah he does, but like when u look at the book the questions r different.
anyways i'm stuck on this one ...
Solve:
(x^2 - x)^2 + (x^2 - x) - 2 = 0 giving exact values. thx
ummm yea thx so the correct ans is 1 -+ root 5 / 2 ....only 2 ans for this question...as the next one has no solutions root - 7 (can't do).well same sort of thing
everything is in brackets yes, makes it much easier to spot the substitution we make, we have a squared term , a linear term and a constant term =0
u= x^2 -x
u^2 +u -2=0
(u+2) ( u-1) =0 etc
u=-2, u-1
so x^2 -x =-2 and x^2 -x = 1
two eqns, solve using quadratic formula, etc, should get 4 answers
Let u = (x^2-x)yeah he does, but like when u look at the book the questions r different.
anyways i'm stuck on this one ...
Solve:
(x^2 - x)^2 + (x^2 - x) - 2 = 0 giving exact values. thx
ummm yea thx so the correct ans is 1 -+ root 5 / 2 ....only 2 ans for this question...as the next one has no solutions root - 7 (can't do).
PS: why is the standard letter u? my teacher used m. is there a meaning behind it?
hmm thats different...what i did wasdoesnt matter what letter, if you do 3 unit maths you will do integration by substitution and generally the letter used is u, just easy to write a u compared to any other letter i think, doesnt matter.
just noticing x^2 + x +c =0
e.g. if you had 3( sinx)^2 + 5sinx +3=0, you would substitute u= sinx,
heres one for you solve e^(x) + e^(-x) +3 =0, its a little different and theres a little trick
hmm thats different...what i did was
let m = x^e
m-m+3=0
3=0 (wrong i guess)
I remember seeing that question in a 2 unit paper but it was logx instead of e^xmm yes, dont worry, its a common trick hopefully you will learn it now and will remember it forever lol
ok, now e^-x = 1/ e^x
what if we changed it to e^x + 1 / e^x +3 = 0
how would you go about solving that