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icalibear

  • one year ago

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

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  1. icalibear
    • one year ago
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    \[\sqrt{3a+1}= a - 3\]

  2. egbeach
    • one year ago
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    how do you get rid of a square root?

  3. icalibear
    • one year ago
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    square root both sides?

  4. icalibear
    • one year ago
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    i mean square both sides

  5. egbeach
    • one year ago
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    correct. what do you get when you do that?

  6. icalibear
    • one year ago
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    \[3a + 1 = a^2 + 9\]

  7. egbeach
    • one year ago
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    you should actually get \[3a+1=\left( a-3 \right)^{2}\] you want to square it all together

  8. icalibear
    • one year ago
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    oh okay, and then FOIL that?

  9. egbeach
    • one year ago
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    yep what do you get after you FOIL?

  10. icalibear
    • one year ago
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    \[(a - 3)(a-3)\] \[a^2 - 3a - 3a + 9\] \[a^2 - 6 + 9\]

  11. icalibear
    • one year ago
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    6a not 6

  12. egbeach
    • one year ago
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    okay good. so now the whole equation is \[3a+1= a ^{2}-6a+9\] Now set the eqation to 0

  13. icalibear
    • one year ago
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    \[a^2 - 6a + 9 - 3a - 1 = 0\]

  14. egbeach
    • one year ago
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    simplfy

  15. icalibear
    • one year ago
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    \[a^2 - 9a + 10 = 0\]

  16. egbeach
    • one year ago
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    9-1+8 not 10

  17. icalibear
    • one year ago
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    oh i thought it wa 9 - (-1)

  18. egbeach
    • one year ago
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    = 8*

  19. icalibear
    • one year ago
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    \[a^2 - 9a + 8 = 0\]

  20. egbeach
    • one year ago
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    yep. can you factor it out by yourself or do you need help?

  21. icalibear
    • one year ago
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    do i use an 8 and a 1

  22. egbeach
    • one year ago
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    yep!

  23. icalibear
    • one year ago
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    \[(a-8)(a+1)\] \[(a - 8) = 0\] \[a = 8\] \[(a+1) = 0\] \[a =-1\]

  24. egbeach
    • one year ago
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    it should be (a-1)=0 so both answers will be positive

  25. icalibear
    • one year ago
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    oh okay. thanks

  26. egbeach
    • one year ago
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    welcome

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