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boomerang285 Group Title

sqrt x+10 = x-2

  • one year ago
  • one year ago

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  1. boomerang285 Group Title
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    \[\sqrt{x+10}=x-2\]

    • one year ago
  2. Nameless Group Title
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    you first need to get rid of the radical.

    • one year ago
  3. Nameless Group Title
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    so square both sides.

    • one year ago
  4. boomerang285 Group Title
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    ok

    • one year ago
  5. Nameless Group Title
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    so you should have x+10 = x^2-4x-4 ?

    • one year ago
  6. Nameless Group Title
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    now move your terms all to one side and set it equal to 0 and factor or use the quadratic to solve for x

    • one year ago
  7. boomerang285 Group Title
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    ok

    • one year ago
  8. Nameless Group Title
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    actualy... i typoed.. it should be \[x+10=x^{2}-4x+4\]

    • one year ago
  9. boomerang285 Group Title
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    ok i was confused trying to work that out

    • one year ago
  10. Nameless Group Title
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    and the only reason i move to the right is cause I dont want to multiply by a negative 1.

    • one year ago
  11. Nameless Group Title
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    \[x ^{2}-5x-6=0\]

    • one year ago
  12. Nameless Group Title
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    you know how to factor or do the quadratic?

    • one year ago
  13. boomerang285 Group Title
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    i have my text book here, these confuse me so much

    • one year ago
  14. Nameless Group Title
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    ok. i would factor this. so you set up parens.. ( )( ) right? fill them with (x )(x ). now we are looking for 2 numbers that multiply to 6 and add to -5

    • one year ago
  15. Nameless Group Title
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    so 2 and 3 seem to work nice.. but they are not gonna get us to where we want because of the negative 5. so we try 1 and 6. (x 1)(x 6) right?

    • one year ago
  16. boomerang285 Group Title
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    so 2 and 3

    • one year ago
  17. boomerang285 Group Title
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    ok

    • one year ago
  18. Nameless Group Title
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    now since we know we a negative 5.. we know that 6 has to be negative. so (x+1)(x-6)

    • one year ago
  19. boomerang285 Group Title
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    so (x+1)(x-6)

    • one year ago
  20. Nameless Group Title
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    solve those as separate equations .. x-6=0 and x+1=0

    • one year ago
  21. boomerang285 Group Title
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    x=6 x=-1

    • one year ago
  22. Nameless Group Title
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    yep. now.. the most important.. you need to plus these numbers back into the equation.. to check if both answers work.

    • one year ago
  23. boomerang285 Group Title
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    ok

    • one year ago
  24. Nameless Group Title
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    \[\sqrt{-1+10} =-1 - 2\]

    • one year ago
  25. Nameless Group Title
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    \[\sqrt{6+10} = 6-2\]

    • one year ago
  26. boomerang285 Group Title
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    sqrt9=-3 sqrt16 = 4

    • one year ago
  27. Nameless Group Title
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    so x = 6 is your only answer? does that make sense?

    • one year ago
  28. Nameless Group Title
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    \[\sqrt{9} =3 \neq -1-2 = -3\] \[\sqrt{16}=4 = 6-2=4\]

    • one year ago
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