- anonymous

can anyone help??

- schrodinger

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- anonymous

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- anonymous

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- anonymous

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

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## More answers

- anonymous

can anyone help or understand how to do that?

- xapproachesinfinity

what you think u should be?

- xapproachesinfinity

any ideas?

- xapproachesinfinity

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

- anonymous

its all one problem

- xapproachesinfinity

okay! first pages looks like an example to me

- anonymous

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

- xapproachesinfinity

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

- xapproachesinfinity

any idea?

- anonymous

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

- xapproachesinfinity

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

- xapproachesinfinity

they want you to suggest a u for that equation

- anonymous

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

- xapproachesinfinity

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

- xapproachesinfinity

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

- xapproachesinfinity

because that first page explains it all!

- anonymous

i did read it.

- xapproachesinfinity

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

- xapproachesinfinity

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

- anonymous

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

- xapproachesinfinity

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

- xapproachesinfinity

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

- xapproachesinfinity

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

- xapproachesinfinity

following?

- anonymous

yea

- xapproachesinfinity

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

- xapproachesinfinity

agree with this step or no?

- xapproachesinfinity

i took square of both sides

- anonymous

to get rid of the sqrt, yea

- xapproachesinfinity

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

- xapproachesinfinity

we got a quadratic!

- xapproachesinfinity

can we factor that!

- anonymous

wait

- xapproachesinfinity

yes! any questions

- anonymous

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

- xapproachesinfinity

yes!

- anonymous

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

- xapproachesinfinity

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

- xapproachesinfinity

yes excellent

- anonymous

and use zpp?

- xapproachesinfinity

solve for u
what is zpp lol never heard this haha

- anonymous

zero product property lol

- xapproachesinfinity

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

- anonymous

yep

- xapproachesinfinity

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

- xapproachesinfinity

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

- xapproachesinfinity

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

- anonymous

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

- xapproachesinfinity

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

- xapproachesinfinity

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

- xapproachesinfinity

we are actually doing all those 4 question in one go

- anonymous

ohh okay

- xapproachesinfinity

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

- xapproachesinfinity

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

- anonymous

okay

- xapproachesinfinity

2 we did we wrote the equATION WITH ONLY U

- anonymous

gotcha

- xapproachesinfinity

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

- xapproachesinfinity

we are in 4 now :)

- xapproachesinfinity

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

- xapproachesinfinity

two quadratic equations

- xapproachesinfinity

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

- xapproachesinfinity

looks good to you?

- anonymous

no lol

- xapproachesinfinity

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

- anonymous

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

- xapproachesinfinity

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

- xapproachesinfinity

clear now i guess haha

- xapproachesinfinity

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

- xapproachesinfinity

good or nor good?

- anonymous

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

- anonymous

i got it

- xapproachesinfinity

hmm you are good learner as i see it :)

- anonymous

aw thanks aha :))

- xapproachesinfinity

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

- xapproachesinfinity

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

- xapproachesinfinity

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

- xapproachesinfinity

zpp lol

- anonymous

x=1 and 2

- anonymous

-1,+2

- xapproachesinfinity

yes great job

- xapproachesinfinity

that for step 4 and we are good!

- anonymous

cool thanks so much

- xapproachesinfinity

no problem

- anonymous

you are a great teacher aha

- xapproachesinfinity

not really :) i suck at some stuff haha

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