I only have two left that I'm a little unclear on.

- BloomLocke367

I only have two left that I'm a little unclear on.

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- BloomLocke367

\(x+1-2\sqrt{x+4}=0\)

- BloomLocke367

I have to solve that too

- BloomLocke367

I know the 4 can come out of the radical

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## More answers

- Nnesha

no

- Nnesha

it's (x+4)

- BloomLocke367

ohhhhhhhhhhhh

- Nnesha

you can't separate it :(

- BloomLocke367

yeah, when you wrote it that way I noticed.

- Nnesha

so move all the terms to the right side instead sqrt{(x+4}

- BloomLocke367

Why am I moving them?

- 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

OH THAT MAKES SENSE

- BloomLocke367

i have \(\frac{1}{2}x+\frac{1}{2}=\sqrt{x+4}\)

- BloomLocke367

right?

- Nnesha

looks right

- BloomLocke367

so \(x+4=\frac{1}{4}x+\frac{1}{4}\), right?

- Nnesha

mhmm

- BloomLocke367

so 0.75x=-3.75?

- Nnesha

there is an easier way to do it

- Nnesha

i hate fractions :P \[\huge\rm x+1-2\sqrt{x+4}=0\]
move the x+1 to the right side

- BloomLocke367

\(-2\sqrt{x+4}=-x-1\)

- Nnesha

looks good now move the -2 to the right side
remember we need just radical sign on one side

- BloomLocke367

I already did that.

- Nnesha

\(\color{Red}{-2}\sqrt{x+4}=-x-1\)
we just need radical at left side

- BloomLocke367

oops

- BloomLocke367

\(\sqrt{x+4}=\frac{1}{2}x+\frac{1}{2}\)

- 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

yes \[\sqrt{x+4}=\frac{ (x+1) }{ 2 }\]
which is same as 1/2x+1/2
but keep that to one fraction

- Nnesha

when we finish with this question
please pnch on my forehead okay

- BloomLocke367

hahaha okay XD

- Nnesha

that's right but i don't how u got the negative sign

- Nnesha

i'm sorry i'm tired actually
so i apologies

- Nnesha

okay so how would you cancel out the square root ?

- BloomLocke367

square both sides

- Nnesha

yes right

- BloomLocke367

so \(x+4=\frac{1}{4}x+\frac{1}{4}\)

- BloomLocke367

which I also said already

- Nnesha

\[\sqrt{x+4}= (\frac{x+1 }{ 2 })^2\]
that's how you should take square root at left side

- BloomLocke367

so \(x+4=\frac{x^2+1}{4}\)

- Nnesha

that's why i said it's better to combine them together
and no

- BloomLocke367

*facepalm*

- Nnesha

\[\huge\rm (\frac{ x+1 }{ 2 }) = \frac{ (x+1)^2 }{ 2^2 }\] both should be squared

- BloomLocke367

sorry, everyone is talking around me and I just found out my boyfriend is in the ER... I'm a little distracted

- Nnesha

mah don't `facepalm` :P

- Nnesha

it's okay! :=)

- BloomLocke367

oh right, that makes sense

- BloomLocke367

x+2x+1 should be the numerator, right?

- Nnesha

right now (x+1)^2 is same as (x+1)(x+1) you're an expert when it comes to foil method

- Nnesha

typo

- Nnesha

you sure it's x+2x +1 ?

- BloomLocke367

X^2

- BloomLocke367

UGH sorry

- Nnesha

yes right :=) \[x+4=\frac{ x^2+2x+1 }{ 4 }\] now oslve for x

- Nnesha

solve*

- BloomLocke367

4x+16=x^2+2x+1?

- BloomLocke367

to start with, that is

- Nnesha

:=)

- Nnesha

yes that's right

- BloomLocke367

okay. so x^2-2x-15?

- Nnesha

mhmm error!

- BloomLocke367

what?

- BloomLocke367

what error?

- Nnesha

\[\color{ReD}{4x}+16=x^2\color{ReD}{+}2x+1\]

- Nnesha

you would subtract 2x both sides right so
4x-2x = 2x not -2x :=)

- 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

no... 2x-4x=-2x

- Nnesha

facedesk**

- Nnesha

i moved all the terms to the left side
why i'm still looking on it
you're right god:(

- BloomLocke367

hahaha, we both are having a tough time

- BloomLocke367

meanwhile, I'm about to have a meltdown because I'm so worried...

- BloomLocke367

anyhoo, I have (x-5)(x+3)=0

- Nnesha

ah my fault

- BloomLocke367

so my zeros are 5 and -3, right?

- Nnesha

yes right solve for x
and what is the statement ? you have to find solutions right ?

- BloomLocke367

yep

- BloomLocke367

which I found

- Nnesha

make sure you check your work
plugin 5 and -3 into the equation

- Nnesha

there can be an extraneous solution!

- BloomLocke367

yea, I know. I just have one more.. then I can go and quietly breakdown in my room. ;-;

- Nnesha

i'm pretty you don't want to help anymore ahahhah

- BloomLocke367

5 works. just checked

- Nnesha

me*

- Nnesha

yes right -3 is an extraneous

- BloomLocke367

yep

- Nnesha

alright good luck
btw what's ur next question ?

- BloomLocke367

\(\sqrt x+x=1\)

- Nnesha

is it okay if we do it on this thread ??

- BloomLocke367

yea, I don't mind. I just wanna get this done

- 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

\(\sqrt x=-x+1\)

- Nnesha

yes right take square both sides \[(\sqrt{x})^2= (-x+1)^2\]

- Nnesha

(-x+1)(-x+1) =foil

- BloomLocke367

x=x^2-2x+1

- Nnesha

right solve for x :=)

- 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

well should move the x to the right side
cuz there are like terms that we can combine

- BloomLocke367

ok

- Nnesha

if you factor right side that will not help you you still have to distribute factors with x

- BloomLocke367

x^2-3x+1

- Nnesha

yes right you need to apply quadratic formula to find x

- BloomLocke367

ok

- BloomLocke367

I can do that

- BloomLocke367

gimme a moment

- 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

take your time! :=)

- BloomLocke367

yea I know gimme a sec

- BloomLocke367

so far I have \(x=\frac{3\pm\sqrt5}{2}\)

- BloomLocke367

that's as far as I'm going

- BloomLocke367

is that right?

- BloomLocke367

- Nnesha

looks right!

- BloomLocke367

ok. I'm leaving it as is

- BloomLocke367

I'm gonna go now, thanks for your help. Goodbye!

- Nnesha

alright good luck!

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