Question

can anyone help??

Answer 1

i think i just neeed to do the last 4 steps its asking at the bottom

Answer 2

can anyone help or understand how to do that?

Answer 3

what you think u should be?

Answer 4

any ideas?

Answer 5

oh there two pages last 4 you are reffering to page 2 right?

Answer 6

its all one problem

Answer 7

okay! first pages looks like an example to me

Answer 8

i think i need to answer the 4 questions at the bottom of pg 2

Answer 9

anyways we are doing this \(\sqrt{(x^2-x)+7}=2(x^2-x)-1\)
1) is about the u

Answer 10

any idea?

Answer 11

there is no u there, but in pg 11 it says let u= t^2-3t is that right?

Answer 12

it say what would u represent i may reading this incorrectly lol

Answer 13

they want you to suggest a u for that equation

Answer 14

idk im very confused i dont see a U?? :S

Answer 15

no there is no u in the problem!
what they are telling you is do substitution instead of solving for x we solve for a new variable u
but we need first to make some kind of substitution
if you looked at the first page you have they give you an example

Answer 16

i think you didn;t read what that page is saying otherwise you would be confused what u is lol

Answer 17

because that first page explains it all!

Answer 18

i did read it.

Answer 19

lol! okay no problem
so we need to do substitution to some variable u
I'm speaking of this equation now \(\sqrt{(x^2-x)+7}=2(x^2-x)-1\)

Answer 20

this equation looks a mess for now that's why we do u to have only one good looking variable

Answer 21

so we need to replace (x^2-x) with u right

Answer 22

so we can solve easily
ah yes you are catching good going so far

Answer 23

\(\sqrt{u+7}=2u-1\) letting u=x^2-x yes

Answer 24

now we can solve for u before we go back to x

Answer 25

following?

Answer 26

yea

Answer 27

\(\sqrt{u+7}=2u-1 \Longrightarrow u+7=(2u-1)^2\)

Answer 28

agree with this step or no?

Answer 29

i took square of both sides

Answer 30

to get rid of the sqrt, yea

Answer 31

good!
so now \(u+7=4u^2-4u+1 \Longrightarrow 4u^2-5u-6=0\)

Answer 32

we got a quadratic!

Answer 33

can we factor that!

Answer 34

wait

Answer 35

yes! any questions

Answer 36

to get 4u^2 -5u-6, you just did -u-7 from both sides right?

Answer 37

yes!

Answer 38

and then we need to factr that to (4u+3)(u-2)?

Answer 39

now we factor to \((u-2)(4u+3)=0\)

Answer 40

yes excellent

Answer 41

and use zpp?

Answer 42

solve for u
what is zpp lol never heard this haha

Answer 43

zero product property lol

Answer 44

i guess you mean u-2=0 or 4u+3=0
u=2 or u=-3/4

Answer 45

yep

Answer 46

eh i see i never named anyhing it just makes sense to me with no name haha

Answer 47

now we solved for u but we still not finished for x

Answer 48

remember that we let u=x^2-x right

Answer 49

so for the first question its u= *3/4 and u=2 right

Answer 50

so we have \(x^2-x=2 ~~or~~ x^2-x=-3/4\)

Answer 51

oh no
first question to just answer u=x^2-x that is it

Answer 52

we are actually doing all those 4 question in one go

Answer 53

ohh okay

Answer 54

once we finish you put it in order 1 put what we did
2 put what we did

Answer 55

here i went to answer those as one question lol
they supposed to be one it is just that they want you to understand it part by part

Answer 56

okay

Answer 57

2 we did we wrote the equATION WITH ONLY U

Answer 58

gotcha

Answer 59

3 we did we solved for u
u=2 u=-3/4

Answer 60

we are in 4 now :)

Answer 61

\(x^2-x=2 ~~or x^2-x=-3/4 \) we solve this now

Answer 62

two quadratic equations

Answer 63

\(x^2-x-2=0 ~~~or ~~~ 4x^2-4x+3=0\)

Answer 64

looks good to you?

Answer 65

no lol

Answer 66

may be i wen too fast haha
x^2-x=2
x^2-x-2=0 after subtracting 2 from both ssides

Answer 67

how did you get 4x^2-4x+3

Answer 68

the other one x^2-x=-3/4 ===> 4(x^2-x)=-3 (multiplied 4 both sides)
4x^2-4x=-3
4x^2-4x+3=0 added 3 to both sides

Answer 69

clear now i guess haha

Answer 70

i just wanted to have a nice format i dislike fractions haha

Answer 71

good or nor good?

Answer 72

i was just trying to understand it sorry, im a slow learner lol i got it tho

Answer 73

i got it

Answer 74

hmm you are good learner as i see it :)

Answer 75

aw thanks aha :))

Answer 76

we solve \(x^2-x-2=0 \Longrightarrow (x+1)(x-2)=0\)

Answer 77

this one 4x^2-4x+3=0 does not have real solutions

Answer 78

so we just solve (x+1)(x-2)=0

Answer 79

zpp lol

Answer 80

x=1 and 2

Answer 81

-1,+2

Answer 82

yes great job

Answer 83

that for step 4 and we are good!

Answer 84

cool thanks so much

Answer 85

no problem

Answer 86

you are a great teacher aha

Answer 87

not really :) i suck at some stuff haha