anonymous
  • anonymous
can anyone help??
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
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anonymous
  • anonymous
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anonymous
  • anonymous
i think i just neeed to do the last 4 steps its asking at the bottom

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anonymous
  • anonymous
can anyone help or understand how to do that?
xapproachesinfinity
  • xapproachesinfinity
what you think u should be?
xapproachesinfinity
  • xapproachesinfinity
any ideas?
xapproachesinfinity
  • xapproachesinfinity
oh there two pages last 4 you are reffering to page 2 right?
anonymous
  • anonymous
its all one problem
xapproachesinfinity
  • xapproachesinfinity
okay! first pages looks like an example to me
anonymous
  • anonymous
i think i need to answer the 4 questions at the bottom of pg 2
xapproachesinfinity
  • xapproachesinfinity
anyways we are doing this \(\sqrt{(x^2-x)+7}=2(x^2-x)-1\) 1) is about the u
xapproachesinfinity
  • xapproachesinfinity
any idea?
anonymous
  • anonymous
there is no u there, but in pg 11 it says let u= t^2-3t is that right?
xapproachesinfinity
  • xapproachesinfinity
it say what would u represent i may reading this incorrectly lol
xapproachesinfinity
  • xapproachesinfinity
they want you to suggest a u for that equation
anonymous
  • anonymous
idk im very confused i dont see a U?? :S
xapproachesinfinity
  • 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
  • xapproachesinfinity
i think you didn;t read what that page is saying otherwise you would be confused what u is lol
xapproachesinfinity
  • xapproachesinfinity
because that first page explains it all!
anonymous
  • anonymous
i did read it.
xapproachesinfinity
  • 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
  • xapproachesinfinity
this equation looks a mess for now that's why we do u to have only one good looking variable
anonymous
  • anonymous
so we need to replace (x^2-x) with u right
xapproachesinfinity
  • xapproachesinfinity
so we can solve easily ah yes you are catching good going so far
xapproachesinfinity
  • xapproachesinfinity
\(\sqrt{u+7}=2u-1\) letting u=x^2-x yes
xapproachesinfinity
  • xapproachesinfinity
now we can solve for u before we go back to x
xapproachesinfinity
  • xapproachesinfinity
following?
anonymous
  • anonymous
yea
xapproachesinfinity
  • xapproachesinfinity
\(\sqrt{u+7}=2u-1 \Longrightarrow u+7=(2u-1)^2\)
xapproachesinfinity
  • xapproachesinfinity
agree with this step or no?
xapproachesinfinity
  • xapproachesinfinity
i took square of both sides
anonymous
  • anonymous
to get rid of the sqrt, yea
xapproachesinfinity
  • xapproachesinfinity
good! so now \(u+7=4u^2-4u+1 \Longrightarrow 4u^2-5u-6=0\)
xapproachesinfinity
  • xapproachesinfinity
we got a quadratic!
xapproachesinfinity
  • xapproachesinfinity
can we factor that!
anonymous
  • anonymous
wait
xapproachesinfinity
  • xapproachesinfinity
yes! any questions
anonymous
  • anonymous
to get 4u^2 -5u-6, you just did -u-7 from both sides right?
xapproachesinfinity
  • xapproachesinfinity
yes!
anonymous
  • anonymous
and then we need to factr that to (4u+3)(u-2)?
xapproachesinfinity
  • xapproachesinfinity
now we factor to \((u-2)(4u+3)=0\)
xapproachesinfinity
  • xapproachesinfinity
yes excellent
anonymous
  • anonymous
and use zpp?
xapproachesinfinity
  • xapproachesinfinity
solve for u what is zpp lol never heard this haha
anonymous
  • anonymous
zero product property lol
xapproachesinfinity
  • xapproachesinfinity
i guess you mean u-2=0 or 4u+3=0 u=2 or u=-3/4
anonymous
  • anonymous
yep
xapproachesinfinity
  • xapproachesinfinity
eh i see i never named anyhing it just makes sense to me with no name haha
xapproachesinfinity
  • xapproachesinfinity
now we solved for u but we still not finished for x
xapproachesinfinity
  • xapproachesinfinity
remember that we let u=x^2-x right
anonymous
  • anonymous
so for the first question its u= *3/4 and u=2 right
xapproachesinfinity
  • xapproachesinfinity
so we have \(x^2-x=2 ~~or~~ x^2-x=-3/4\)
xapproachesinfinity
  • xapproachesinfinity
oh no first question to just answer u=x^2-x that is it
xapproachesinfinity
  • xapproachesinfinity
we are actually doing all those 4 question in one go
anonymous
  • anonymous
ohh okay
xapproachesinfinity
  • xapproachesinfinity
once we finish you put it in order 1 put what we did 2 put what we did
xapproachesinfinity
  • 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
  • anonymous
okay
xapproachesinfinity
  • xapproachesinfinity
2 we did we wrote the equATION WITH ONLY U
anonymous
  • anonymous
gotcha
xapproachesinfinity
  • xapproachesinfinity
3 we did we solved for u u=2 u=-3/4
xapproachesinfinity
  • xapproachesinfinity
we are in 4 now :)
xapproachesinfinity
  • xapproachesinfinity
\(x^2-x=2 ~~or x^2-x=-3/4 \) we solve this now
xapproachesinfinity
  • xapproachesinfinity
two quadratic equations
xapproachesinfinity
  • xapproachesinfinity
\(x^2-x-2=0 ~~~or ~~~ 4x^2-4x+3=0\)
xapproachesinfinity
  • xapproachesinfinity
looks good to you?
anonymous
  • anonymous
no lol
xapproachesinfinity
  • xapproachesinfinity
may be i wen too fast haha x^2-x=2 x^2-x-2=0 after subtracting 2 from both ssides
anonymous
  • anonymous
how did you get 4x^2-4x+3
xapproachesinfinity
  • 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
  • xapproachesinfinity
clear now i guess haha
xapproachesinfinity
  • xapproachesinfinity
i just wanted to have a nice format i dislike fractions haha
xapproachesinfinity
  • xapproachesinfinity
good or nor good?
anonymous
  • anonymous
i was just trying to understand it sorry, im a slow learner lol i got it tho
anonymous
  • anonymous
i got it
xapproachesinfinity
  • xapproachesinfinity
hmm you are good learner as i see it :)
anonymous
  • anonymous
aw thanks aha :))
xapproachesinfinity
  • xapproachesinfinity
we solve \(x^2-x-2=0 \Longrightarrow (x+1)(x-2)=0\)
xapproachesinfinity
  • xapproachesinfinity
this one 4x^2-4x+3=0 does not have real solutions
xapproachesinfinity
  • xapproachesinfinity
so we just solve (x+1)(x-2)=0
xapproachesinfinity
  • xapproachesinfinity
zpp lol
anonymous
  • anonymous
x=1 and 2
anonymous
  • anonymous
-1,+2
xapproachesinfinity
  • xapproachesinfinity
yes great job
xapproachesinfinity
  • xapproachesinfinity
that for step 4 and we are good!
anonymous
  • anonymous
cool thanks so much
xapproachesinfinity
  • xapproachesinfinity
no problem
anonymous
  • anonymous
you are a great teacher aha
xapproachesinfinity
  • xapproachesinfinity
not really :) i suck at some stuff haha

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