Can someone just help me solve this please?? i need an extraneous solution and it solved. i chose this but im not sure its correct.
2√x+5+7=4

- anonymous

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- stefrheart

Is that the original equation?

- anonymous

no it said to choose your own using a√x+b+c=d

- stefrheart

Ah I understand hold for a minute please

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## More answers

- anonymous

okay.

- stefrheart

I dont think it has an extraneous solution because I got x=16

- anonymous

well could you help me find one with one please? using the a√x+b+c=d formula thingy

- stefrheart

Sure!

- stefrheart

So you do understand what an extraneous solution is correct?

- anonymous

i think it is how it solves the problem but if you plug it back in it doesnt work out

- anonymous

like if you solve the radical but if you plug it in to check the answer will not work out

- stefrheart

Yes that is correct! :)

- anonymous

im not sure how to like create and solve a radical like that though... ;(

- anonymous

Create two radical equations: one that has an extraneous solution, and one that does not have an extraneous solution. Use the equation below as a model.
a√x+b+c=d
Use a constant in place of each variable a, b, c, and d. You can use positive and negative constants in your equation.

- stefrheart

is it \[a \sqrt{x+b+c}=d\] or \[a \sqrt{x} + b + c=d\]

- phi

try something like
\[ \sqrt{x+2} + 1 = -1 \]

- anonymous

the first one @stefrheart

- anonymous

@phi the only problem is is that i don't know how to solve that equation to check if it has an extraneous solution

- stefrheart

To solve that all you need to do is subtract the one on both sides to create
\[\sqrt{x+2}=-2\]
and then square both sides
\[(\sqrt{x+2})^{2}=(-2)^{}\]
and then solve

- stefrheart

just for phi's equation

- dinamix

x=-8 @stefrheart

- dinamix

@kenzi041

- anonymous

i dont even know. i never learned radicals

- stefrheart

oh you dont know what a radical is or how to work with it?

- anonymous

we had a large learning gap because they changed curriculums

- anonymous

not really.

- stefrheart

Ah okay so you should learn what a radical is first then, before proceeding with the problem

- anonymous

oh.

- stefrheart

It would make alot more sense before solving this. Do you know what a \[x ^{2}\] is? It would be easier to teach you if you knew what it is

- anonymous

x squared?

- stefrheart

yes you know what to do with it if someone gave you the equation \[4^{2}\] ?

- anonymous

16, i know like how to do that i just dont understand how to use like larger equations like a√x+b+c=d

- stefrheart

Ah okay so you know how to do that. Well the larger equations are just the same ting as a regular square root.

- anonymous

okay, so how do i find one with an extraneous solution

- anonymous

because i have one for without but i just need one with one

- stefrheart

I think I may have one hold on

- anonymous

ok

- dinamix

x=-8 u have check in the equation ans see it

- stefrheart

Whats the extraneous solution then? @dinamix

- anonymous

like how do you plug it in to check if it has one @dinamix

- stefrheart

Try this for help
https://www.khanacademy.org/math/algebra2/radical-equations-and-functions/radical-equations/e/extraneous-solutions-to-radical-equations

- anonymous

i already tried but it was in a different format than what the question asked

- stefrheart

Yeah I noticed

- anonymous

and i got more confused..

- stefrheart

Hmmm im trying to figure out equations because in order for it to have both an actual and extraneous there needs to be 2 solutions which is a \[x ^{2}+bx+c\]
unless we can do that you cant get both an actual and extraneous

- anonymous

no it needs to just be a extraneous solution.. you write two problems and one should be regular and one should have an extraneous solution

- stefrheart

Ohhh then I already have one then

- stefrheart

\[-5\sqrt{-3x-5}=-25\]

- anonymous

but doesnt it have to be x+b+c=d

- stefrheart

if you wwant it you can just put -3x+-5+0 for under the radical

- anonymous

okay, one sec im checking a problem i just did.

- stefrheart

I will be back in a few minutes

- anonymous

i think i found one, the problem i chose originally

- anonymous

and okay

- stefrheart

AH okay

- stefrheart

Message me if need you still need help

- anonymous

thanks for helping

- anonymous

how do i give u a medal?

- stefrheart

@kenzi041 click the best response

- anonymous

@stefrheart i failed, but thanks anyway

- stefrheart

Aw im sorry :( @kenzi041

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