iwanttogotostanford
  • iwanttogotostanford
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.
Mathematics
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SOLVED
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jamiebookeater
  • jamiebookeater
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iwanttogotostanford
  • iwanttogotostanford
@ospreytriple @BloomLocke367 @freckles
BloomLocke367
  • BloomLocke367
First off, can you write that in an equation?
iwanttogotostanford
  • iwanttogotostanford
yes

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More answers

BloomLocke367
  • BloomLocke367
Do that first. Tell us what you get.
anonymous
  • anonymous
You have\[3\sqrt{x+6} = -12\]What do you think we should do first?
iwanttogotostanford
  • iwanttogotostanford
\[3\sqrt{x+6}=-12\]
iwanttogotostanford
  • iwanttogotostanford
not sure
BloomLocke367
  • 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
  • anonymous
Well we have to get the 'x' stuff all by itself, so how would you get rid of that 3?
iwanttogotostanford
  • iwanttogotostanford
square it?
BloomLocke367
  • 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
  • iwanttogotostanford
divide
BloomLocke367
  • BloomLocke367
Yes! now divide both sides by 3 and tell me what you're left with
iwanttogotostanford
  • iwanttogotostanford
i don't have a calulator with me right now
BloomLocke367
  • BloomLocke367
You don't know \(-12\div 3\)?
iwanttogotostanford
  • iwanttogotostanford
yes its -4
BloomLocke367
  • BloomLocke367
There you go. so right now, you have \(\sqrt{x+6}=-4\) correct?
iwanttogotostanford
  • iwanttogotostanford
yes
BloomLocke367
  • BloomLocke367
Now, how would you get rid of that radical?
iwanttogotostanford
  • iwanttogotostanford
hmm im not sure
BloomLocke367
  • BloomLocke367
Well, what's the opposite of taking the square root of something?
iwanttogotostanford
  • iwanttogotostanford
squaring it?
iwanttogotostanford
  • iwanttogotostanford
im clueless on this sorry
BloomLocke367
  • BloomLocke367
yes! so you have to square both sides :D
BloomLocke367
  • BloomLocke367
Don't apologize. You're doing great :D
iwanttogotostanford
  • iwanttogotostanford
okay, thanks!
iwanttogotostanford
  • iwanttogotostanford
so it would be x^2+36=16?
BloomLocke367
  • 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
  • iwanttogotostanford
so, x+6=16?
BloomLocke367
  • BloomLocke367
Yep!
iwanttogotostanford
  • iwanttogotostanford
ok, thanks! so then it would be x=10 right?
iwanttogotostanford
  • iwanttogotostanford
and would it be extraneous or not extraneous though?
BloomLocke367
  • 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
  • Nnesha
did a great job!
BloomLocke367
  • BloomLocke367
Okay, thanks. lol @iwanttogotostanford do you know what an extraneous solution is?
iwanttogotostanford
  • iwanttogotostanford
@BloomLocke367 yes
BloomLocke367
  • BloomLocke367
Then graph it and find out if there's a point where x=4
iwanttogotostanford
  • 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
  • Nnesha
you can substitute x for 10 into the original equation to find out if it's extraneous or not. :=)
iwanttogotostanford
  • iwanttogotostanford
thanks! @Nnesha
BloomLocke367
  • BloomLocke367
Sorry, I was afk. Thanks @Nnesha
Nnesha
  • Nnesha
np
Nnesha
  • Nnesha
if you get equal sides then it's not extraneous if both sides are not equal then 10 would be an extraneous
iwanttogotostanford
  • iwanttogotostanford
ok thanks do u know if it is or not? @Nnesha
BloomLocke367
  • BloomLocke367
oops, I don't know why I said 4 earlier, I meant to say 10. lol
Nnesha
  • Nnesha
no i don't. you have to figure it out!
iwanttogotostanford
  • iwanttogotostanford
@Nnesha well i don't have paper with me so its really hard
Nnesha
  • Nnesha
welll, you can use `paint`
Nnesha
  • Nnesha
that would be fun! trust me
iwanttogotostanford
  • iwanttogotostanford
ok, thank you @Nnesha
iwanttogotostanford
  • iwanttogotostanford
i got extraneous:-)
Nnesha
  • Nnesha
both sides are equal ?
iwanttogotostanford
  • iwanttogotostanford
no they aren't equal
BloomLocke367
  • BloomLocke367
There you go!
iwanttogotostanford
  • iwanttogotostanford
thanks!
Nnesha
  • Nnesha
o^_^o
iwanttogotostanford
  • iwanttogotostanford
they aren't equal= not extraneous?
BloomLocke367
  • BloomLocke367
You were correct the first time, it is extraneous because they are not equal.
BloomLocke367
  • BloomLocke367
wait....
Nnesha
  • Nnesha
both sides are not equal = extraneous both sides are equal = non-extraneous
BloomLocke367
  • BloomLocke367
when you take the square root of something, it could be positive or negative
BloomLocke367
  • BloomLocke367
Right @Nnesha?
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
  • iwanttogotostanford
i think its non-extraneous
BloomLocke367
  • BloomLocke367
how so? because -4*-4=16 AND 4*4=16
iwanttogotostanford
  • iwanttogotostanford
oh whoops i did -4 times 4
Nnesha
  • Nnesha
\[\huge\rm 3\sqrt{10+16}=-12\] whet did you get at left side ?
iwanttogotostanford
  • iwanttogotostanford
i get it now
BloomLocke367
  • BloomLocke367
I gotta go, sorry!
Nnesha
  • Nnesha
i said `number` if -4 times -4 = positive number so square root would be positive
Nnesha
  • Nnesha
so it's extraneous or non-extraneous ?
iwanttogotostanford
  • iwanttogotostanford
ok it is extraeous then
Nnesha
  • Nnesha
yes right \[12\cancel{=}-12\]
iwanttogotostanford
  • iwanttogotostanford
thank you @Nnesha
Nnesha
  • Nnesha
np

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