Find the solution of 3 times the square root of the quantity of x plus 6 equals negative 12, and determine if it is an extraneous solution.

- iwanttogotostanford

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- iwanttogotostanford

@ospreytriple @BloomLocke367 @freckles

- BloomLocke367

First off, can you write that in an equation?

- iwanttogotostanford

yes

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## More answers

- BloomLocke367

Do that first. Tell us what you get.

- anonymous

You have\[3\sqrt{x+6} = -12\]What do you think we should do first?

- iwanttogotostanford

\[3\sqrt{x+6}=-12\]

- iwanttogotostanford

not sure

- BloomLocke367

There are a few ways to do this, but the easiest would be to isolate the radical. Do you know how you would do that?

- anonymous

Well we have to get the 'x' stuff all by itself, so how would you get rid of that 3?

- iwanttogotostanford

square it?

- BloomLocke367

you have to get rid of the 3 first. Right now, the 3 is being multiplied with \(\sqrt{x+6}\). How do you "get rid of" something that is being multiplied?

- iwanttogotostanford

divide

- BloomLocke367

Yes! now divide both sides by 3 and tell me what you're left with

- iwanttogotostanford

i don't have a calulator with me right now

- BloomLocke367

You don't know \(-12\div 3\)?

- iwanttogotostanford

yes its -4

- BloomLocke367

There you go. so right now, you have \(\sqrt{x+6}=-4\) correct?

- iwanttogotostanford

yes

- BloomLocke367

Now, how would you get rid of that radical?

- iwanttogotostanford

hmm im not sure

- BloomLocke367

Well, what's the opposite of taking the square root of something?

- iwanttogotostanford

squaring it?

- iwanttogotostanford

im clueless on this sorry

- BloomLocke367

yes! so you have to square both sides :D

- BloomLocke367

Don't apologize. You're doing great :D

- iwanttogotostanford

okay, thanks!

- iwanttogotostanford

so it would be x^2+36=16?

- BloomLocke367

No, because when you squared it, it just cancelled out the radical. For instance, when you take \(\sqrt1^2\), you don't get \(1^2\) do you? No, you just get 1. Try again. The right side is correct.

- iwanttogotostanford

so, x+6=16?

- BloomLocke367

Yep!

- iwanttogotostanford

ok, thanks! so then it would be x=10 right?

- iwanttogotostanford

and would it be extraneous or not extraneous though?

- BloomLocke367

mhm, I'm pretty certain. Let me ask @Nnesha if I did this correctly. Now we just have to check to make sure it's not extraneous. Do you know what an extraneous solution is?

- Nnesha

did a great job!

- BloomLocke367

Okay, thanks. lol
@iwanttogotostanford do you know what an extraneous solution is?

- iwanttogotostanford

@BloomLocke367 yes

- BloomLocke367

Then graph it and find out if there's a point where x=4

- iwanttogotostanford

do you know if it is or not? I'm sorry I'm just in a very hard place right now to be graphing things (I'm on a plane)

- Nnesha

you can substitute x for 10 into the original equation to find out if it's extraneous or not. :=)

- iwanttogotostanford

thanks! @Nnesha

- BloomLocke367

Sorry, I was afk. Thanks @Nnesha

- Nnesha

np

- Nnesha

if you get equal sides then it's not extraneous
if both sides are not equal then 10 would be an extraneous

- iwanttogotostanford

ok thanks do u know if it is or not? @Nnesha

- BloomLocke367

oops, I don't know why I said 4 earlier, I meant to say 10. lol

- Nnesha

no i don't. you have to figure it out!

- iwanttogotostanford

@Nnesha well i don't have paper with me so its really hard

- Nnesha

welll, you can use `paint`

- Nnesha

that would be fun! trust me

- iwanttogotostanford

ok, thank you @Nnesha

- iwanttogotostanford

i got extraneous:-)

- Nnesha

both sides are equal ?

- iwanttogotostanford

no they aren't equal

- BloomLocke367

There you go!

- iwanttogotostanford

thanks!

- Nnesha

o^_^o

- iwanttogotostanford

they aren't equal= not extraneous?

- BloomLocke367

You were correct the first time, it is extraneous because they are not equal.

- BloomLocke367

wait....

- Nnesha

both sides are not equal = extraneous
both sides are equal = non-extraneous

- BloomLocke367

when you take the square root of something, it could be positive or negative

- BloomLocke367

Right @Nnesha?

- Nnesha

`something`
if you mean
square root of negative number then you will get `imaginary` solution
if you take square root of positive then you would always get positive answer

- iwanttogotostanford

i think its non-extraneous

- BloomLocke367

how so? because -4*-4=16 AND 4*4=16

- iwanttogotostanford

oh whoops i did -4 times 4

- Nnesha

\[\huge\rm 3\sqrt{10+16}=-12\]
whet did you get at left side ?

- iwanttogotostanford

i get it now

- BloomLocke367

I gotta go, sorry!

- Nnesha

i said `number`
if -4 times -4 = positive number so square root would be positive

- Nnesha

so it's extraneous or non-extraneous ?

- iwanttogotostanford

ok it is extraeous then

- Nnesha

yes right \[12\cancel{=}-12\]

- iwanttogotostanford

thank you @Nnesha

- Nnesha

np

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