BloomLocke367
  • BloomLocke367
I only have two left that I'm a little unclear on.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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BloomLocke367
  • BloomLocke367
\(x+1-2\sqrt{x+4}=0\)
BloomLocke367
  • BloomLocke367
I have to solve that too
BloomLocke367
  • BloomLocke367
I know the 4 can come out of the radical

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More answers

Nnesha
  • Nnesha
no
Nnesha
  • Nnesha
it's (x+4)
BloomLocke367
  • BloomLocke367
ohhhhhhhhhhhh
Nnesha
  • Nnesha
you can't separate it :(
BloomLocke367
  • BloomLocke367
yeah, when you wrote it that way I noticed.
Nnesha
  • Nnesha
so move all the terms to the right side instead sqrt{(x+4}
BloomLocke367
  • BloomLocke367
Why am I moving them?
Nnesha
  • Nnesha
you need to get sqrt{x+4} one one side and all other terms to opposite side so then you can take square both sides to cancel out the square root
BloomLocke367
  • BloomLocke367
OH THAT MAKES SENSE
BloomLocke367
  • BloomLocke367
i have \(\frac{1}{2}x+\frac{1}{2}=\sqrt{x+4}\)
BloomLocke367
  • BloomLocke367
right?
Nnesha
  • Nnesha
looks right
BloomLocke367
  • BloomLocke367
so \(x+4=\frac{1}{4}x+\frac{1}{4}\), right?
Nnesha
  • Nnesha
mhmm
BloomLocke367
  • BloomLocke367
so 0.75x=-3.75?
Nnesha
  • Nnesha
there is an easier way to do it
Nnesha
  • Nnesha
i hate fractions :P \[\huge\rm x+1-2\sqrt{x+4}=0\] move the x+1 to the right side
BloomLocke367
  • BloomLocke367
\(-2\sqrt{x+4}=-x-1\)
Nnesha
  • Nnesha
looks good now move the -2 to the right side remember we need just radical sign on one side
BloomLocke367
  • BloomLocke367
I already did that.
Nnesha
  • Nnesha
\(\color{Red}{-2}\sqrt{x+4}=-x-1\) we just need radical at left side
BloomLocke367
  • BloomLocke367
oops
BloomLocke367
  • BloomLocke367
\(\sqrt{x+4}=\frac{1}{2}x+\frac{1}{2}\)
BloomLocke367
  • BloomLocke367
but, look:\(\color{#0cbb34}{\text{Originally Posted by}}\) @BloomLocke367 i have \(\frac{1}{2}x+\frac{1}{2}=\sqrt{x+4}\) \(\color{#0cbb34}{\text{End of Quote}}\) I already did that
Nnesha
  • Nnesha
yes \[\sqrt{x+4}=\frac{ (x+1) }{ 2 }\] which is same as 1/2x+1/2 but keep that to one fraction
Nnesha
  • Nnesha
when we finish with this question please pnch on my forehead okay
BloomLocke367
  • BloomLocke367
hahaha okay XD
Nnesha
  • Nnesha
that's right but i don't how u got the negative sign
Nnesha
  • Nnesha
i'm sorry i'm tired actually so i apologies
Nnesha
  • Nnesha
okay so how would you cancel out the square root ?
BloomLocke367
  • BloomLocke367
square both sides
Nnesha
  • Nnesha
yes right
BloomLocke367
  • BloomLocke367
so \(x+4=\frac{1}{4}x+\frac{1}{4}\)
BloomLocke367
  • BloomLocke367
which I also said already
Nnesha
  • Nnesha
\[\sqrt{x+4}= (\frac{x+1 }{ 2 })^2\] that's how you should take square root at left side
BloomLocke367
  • BloomLocke367
so \(x+4=\frac{x^2+1}{4}\)
Nnesha
  • Nnesha
that's why i said it's better to combine them together and no
BloomLocke367
  • BloomLocke367
*facepalm*
Nnesha
  • Nnesha
\[\huge\rm (\frac{ x+1 }{ 2 }) = \frac{ (x+1)^2 }{ 2^2 }\] both should be squared
BloomLocke367
  • BloomLocke367
sorry, everyone is talking around me and I just found out my boyfriend is in the ER... I'm a little distracted
Nnesha
  • Nnesha
mah don't `facepalm` :P
Nnesha
  • Nnesha
it's okay! :=)
BloomLocke367
  • BloomLocke367
oh right, that makes sense
BloomLocke367
  • BloomLocke367
x+2x+1 should be the numerator, right?
Nnesha
  • Nnesha
right now (x+1)^2 is same as (x+1)(x+1) you're an expert when it comes to foil method
Nnesha
  • Nnesha
typo
Nnesha
  • Nnesha
you sure it's x+2x +1 ?
BloomLocke367
  • BloomLocke367
X^2
BloomLocke367
  • BloomLocke367
UGH sorry
Nnesha
  • Nnesha
yes right :=) \[x+4=\frac{ x^2+2x+1 }{ 4 }\] now oslve for x
Nnesha
  • Nnesha
solve*
BloomLocke367
  • BloomLocke367
4x+16=x^2+2x+1?
BloomLocke367
  • BloomLocke367
to start with, that is
Nnesha
  • Nnesha
:=)
Nnesha
  • Nnesha
yes that's right
BloomLocke367
  • BloomLocke367
okay. so x^2-2x-15?
Nnesha
  • Nnesha
mhmm error!
BloomLocke367
  • BloomLocke367
what?
BloomLocke367
  • BloomLocke367
what error?
Nnesha
  • Nnesha
\[\color{ReD}{4x}+16=x^2\color{ReD}{+}2x+1\]
Nnesha
  • Nnesha
you would subtract 2x both sides right so 4x-2x = 2x not -2x :=)
Nnesha
  • Nnesha
\(\color{blue}{\text{Originally Posted by}}\) @BloomLocke367 okay. so x^2-2x-15? \(\color{blue}{\text{End of Quote}}\) so it's \[\huge\rm x^2+2x-15=0\]
BloomLocke367
  • BloomLocke367
no... 2x-4x=-2x
Nnesha
  • Nnesha
facedesk**
Nnesha
  • Nnesha
i moved all the terms to the left side why i'm still looking on it you're right god:(
BloomLocke367
  • BloomLocke367
hahaha, we both are having a tough time
BloomLocke367
  • BloomLocke367
meanwhile, I'm about to have a meltdown because I'm so worried...
BloomLocke367
  • BloomLocke367
anyhoo, I have (x-5)(x+3)=0
Nnesha
  • Nnesha
ah my fault
BloomLocke367
  • BloomLocke367
so my zeros are 5 and -3, right?
Nnesha
  • Nnesha
yes right solve for x and what is the statement ? you have to find solutions right ?
BloomLocke367
  • BloomLocke367
yep
BloomLocke367
  • BloomLocke367
which I found
Nnesha
  • Nnesha
make sure you check your work plugin 5 and -3 into the equation
Nnesha
  • Nnesha
there can be an extraneous solution!
BloomLocke367
  • BloomLocke367
yea, I know. I just have one more.. then I can go and quietly breakdown in my room. ;-;
Nnesha
  • Nnesha
i'm pretty you don't want to help anymore ahahhah
BloomLocke367
  • BloomLocke367
5 works. just checked
Nnesha
  • Nnesha
me*
Nnesha
  • Nnesha
yes right -3 is an extraneous
BloomLocke367
  • BloomLocke367
yep
Nnesha
  • Nnesha
alright good luck btw what's ur next question ?
BloomLocke367
  • BloomLocke367
\(\sqrt x+x=1\)
Nnesha
  • Nnesha
is it okay if we do it on this thread ??
BloomLocke367
  • BloomLocke367
yea, I don't mind. I just wanna get this done
Nnesha
  • Nnesha
okay cool! that's the easy one just like we did move the x to the right side (bec we need the radical one side )
BloomLocke367
  • BloomLocke367
\(\sqrt x=-x+1\)
Nnesha
  • Nnesha
yes right take square both sides \[(\sqrt{x})^2= (-x+1)^2\]
Nnesha
  • Nnesha
(-x+1)(-x+1) =foil
BloomLocke367
  • BloomLocke367
x=x^2-2x+1
Nnesha
  • Nnesha
right solve for x :=)
BloomLocke367
  • BloomLocke367
I'm drawing a blank here, you know why. should I factor the right side, or move the x on the left to the right?
Nnesha
  • Nnesha
well should move the x to the right side cuz there are like terms that we can combine
BloomLocke367
  • BloomLocke367
ok
Nnesha
  • Nnesha
if you factor right side that will not help you you still have to distribute factors with x
BloomLocke367
  • BloomLocke367
x^2-3x+1
Nnesha
  • Nnesha
yes right you need to apply quadratic formula to find x
BloomLocke367
  • BloomLocke367
ok
BloomLocke367
  • BloomLocke367
I can do that
BloomLocke367
  • BloomLocke367
gimme a moment
Nnesha
  • Nnesha
cuz one is the only one factor of 1 so 1 time 1 there is not way you can get 3 from 1+1
Nnesha
  • Nnesha
take your time! :=)
BloomLocke367
  • BloomLocke367
yea I know gimme a sec
BloomLocke367
  • BloomLocke367
so far I have \(x=\frac{3\pm\sqrt5}{2}\)
BloomLocke367
  • BloomLocke367
that's as far as I'm going
BloomLocke367
  • BloomLocke367
is that right?
BloomLocke367
  • BloomLocke367
@Nnesha
Nnesha
  • Nnesha
looks right!
BloomLocke367
  • BloomLocke367
ok. I'm leaving it as is
BloomLocke367
  • BloomLocke367
I'm gonna go now, thanks for your help. Goodbye!
Nnesha
  • Nnesha
alright good luck!

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