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Presleyram Group Title

Algebra II Problems (work shown)

  • one year ago
  • one year ago

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  1. Presleyram Group Title
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    |dw:1355298946753:dw|

    • one year ago
  2. nubeer Group Title
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    u can start by squaring both sides.

    • one year ago
  3. nubeer Group Title
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    |dw:1355259732990:dw|

    • one year ago
  4. ZeHanz Group Title
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    Before running off it it smart to take a look at two things: 1. The number under the radical, 6x+2 2. The result of the square root, x-1 If you think about it, both numbers must be >=0! 6x+2 >=0 because there is no (real) square root of a negative number. x+1 >=0 because it's a square root and that is >=0 by definition.

    • one year ago
  5. ZeHanz Group Title
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    This means that you answers are restricted two ways: 1. 6x+2>=0, so 6x>=-2, x>=2/6, x>=1/3 2. x-1>=0, so x>=1 BOTH must hold, so we combine them: x>=1. Whatever "solutions" you may find, they MUST be >= 1. Now square everything in sight: (x-1)² = 6x+2 Work out the brackets Everything to the left side of the =-sign Factorize, or use abc-formula Check the solutions! Only if they are >= 1 they are really solutions. Reason for this nitpicking: by squaring everything, you lose the information about the square root and you could get fake solutions...

    • one year ago
  6. ZeHanz Group Title
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    Oops, typo: 6x>=-2, so x>= -1/3 Makes no difference: x >=1 anyway.

    • one year ago
  7. ZeHanz Group Title
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    @nubeer: I wouldn't *start* with squaring, see above reasoning...

    • one year ago
  8. Presleyram Group Title
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    Thanks @ZeHanz

    • one year ago
  9. ZeHanz Group Title
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    yw! Have you got the answer?

    • one year ago
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