schrodinger
  • schrodinger
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purplemexican
  • purplemexican
Nnesha
  • Nnesha
how did you get -6 ?
purplemexican
  • purplemexican
im not entirely sure im still half asleep

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Nnesha
  • Nnesha
show ur work. same like the last one 1st) take square both sides
Nnesha
  • Nnesha
\[\huge\rm ( \sqrt{-3x-2})^2=(x+2)^2\] when you take square on left side square root would cancel out \[\huge\rm -3x-2=(x+2)^2\] now solve for x (x+2)^2 is same as (x+2)(x+2) apply the foil method
purplemexican
  • purplemexican
\[-3x -2=(x+2)^2 \]
purplemexican
  • purplemexican
x=-6, -1
Nnesha
  • Nnesha
hmm
purplemexican
  • purplemexican
-1 works as a solution and so does -6 how am i still getting these
Nnesha
  • Nnesha
yes so answer is what ?
purplemexican
  • purplemexican
i don't know I'm so confused right now
Nnesha
  • Nnesha
\(\color{blue}{\text{Originally Posted by}}\) @purplemexican -1 works as a solution and so does -6 how am i still getting these \(\color{blue}{\text{End of Quote}}\) okay substitute x for -6
Nnesha
  • Nnesha
\[\sqrt{-3(-6)-2}=-6x+2\] both sides are e qual ?
purplemexican
  • purplemexican
thats where i messed up
purplemexican
  • purplemexican
i did -3(-6)-2=(-6+2)^2
Nnesha
  • Nnesha
no don't take square root on right side
Nnesha
  • Nnesha
we need to plug in -6 into the original equation :=)
Nnesha
  • Nnesha
\(\color{blue}{\text{Originally Posted by}}\) @Nnesha \[\sqrt{-3(-6)-2}=-6x+2\] both sides are e qual ? \(\color{blue}{\text{End of Quote}}\) correction \[\sqrt{-3(-6)-2}=-6+2\] both sides are e qual ?
Nnesha
  • Nnesha
1sT) multiply -3 times -6 2nd) subtract -2 take square root of that number
purplemexican
  • purplemexican
4
Nnesha
  • Nnesha
okay so 4=-6+2 both sides are equal ?
purplemexican
  • purplemexican
no
Nnesha
  • Nnesha
so that's an extraneous solution
purplemexican
  • purplemexican
thank you again
Nnesha
  • Nnesha
my pleasure. good job!

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