sqrt 3a+ 1 = a-3 Solve for a

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sqrt 3a+ 1 = a-3 Solve for a

Mathematics
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\[\sqrt{3a+1}= a - 3\]
how do you get rid of a square root?
square root both sides?

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i mean square both sides
correct. what do you get when you do that?
\[3a + 1 = a^2 + 9\]
you should actually get \[3a+1=\left( a-3 \right)^{2}\] you want to square it all together
oh okay, and then FOIL that?
yep what do you get after you FOIL?
\[(a - 3)(a-3)\] \[a^2 - 3a - 3a + 9\] \[a^2 - 6 + 9\]
6a not 6
okay good. so now the whole equation is \[3a+1= a ^{2}-6a+9\] Now set the eqation to 0
\[a^2 - 6a + 9 - 3a - 1 = 0\]
simplfy
\[a^2 - 9a + 10 = 0\]
9-1+8 not 10
oh i thought it wa 9 - (-1)
= 8*
\[a^2 - 9a + 8 = 0\]
yep. can you factor it out by yourself or do you need help?
do i use an 8 and a 1
yep!
\[(a-8)(a+1)\] \[(a - 8) = 0\] \[a = 8\] \[(a+1) = 0\] \[a =-1\]
it should be (a-1)=0 so both answers will be positive
oh okay. thanks
welcome

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