Equations Reducible to Quadratics (2 Viewers)

[MrBrightside]

2^2x -9.2^x +8=0

 

[random-1006]

2^2x -9.2^x +8=0

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

 

[MrBrightside]

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?

 

[random-1006]

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?

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

 

[MrBrightside]

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.

 

[pooshwaltzer]

i would've logged the whole thing

 

[random-1006]

okay thx, wow wtf how come the book doesn't mention that....i bet the book was designed to make people fail.

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?

 

[random-1006]

i would've logged the whole thing

wouldnt do anything

 

[MrBrightside]

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)

 

[random-1006]

teacher never told us tht...yea maths in focus (shit book at explaining things)

i used it , its alright, well i had a good teacher.

doesnt your teacher go over some examples before you do the textbook exercises.

 

[MrBrightside]

i used it , its alright, well i had a good teacher.

doesnt your teacher go over some examples before you do the textbook exercises.

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

 

[random-1006]

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

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

 

[MrBrightside]

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

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?

 

[hscishard]

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

Let u = (x^2-x)

 

[random-1006]

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?

doesnt 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

 

[MrBrightside]

doesnt 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)

 

[random-1006]

hmm thats different...what i did was

let m = x^e

m-m+3=0

3=0 (wrong i guess)

mm 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

 

[Help Galore]

Let u = e^x and 1/u = 1/ e^x and just go about from there.

 

[hscishard]

mm 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

I remember seeing that question in a 2 unit paper but it was logx instead of e^x

 

[MrBrightside]

ummm dude...i'm getting m^2 + 4 - 0 which is incorrect as there is no b. ?

i did (x)^e + 1/ x^e + 3 = 0

X everything by x^e

(x^e)^2 + 1 + 3 = 0

let m = x^e

m^2 + 4 = 0 ...?