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anonymous

  • 5 years ago

solve for x: x^2 + 5x is > or = 6

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  1. myininaya
    • 5 years ago
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    x^2+5x-6>=0 (x+6)(x-1)>=0 when is this LHS zero?

  2. myininaya
    • 5 years ago
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    when x=___ and x=___?

  3. anonymous
    • 5 years ago
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    What is LHS..?

  4. anonymous
    • 5 years ago
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    You would use the (x + 6)(x - 1) >= 0 When x <=-6 or x >= 1

  5. anonymous
    • 5 years ago
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    from the (X+6) you would then solve for x for both giving you the positive 1 and negative 6 correct?

  6. anonymous
    • 5 years ago
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    That would give you when it = 0, but when would it be > 0... see my earlier response

  7. anonymous
    • 5 years ago
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    Then where are you getting the positive and negative...... your doing x+6=0 or x-1+0 ... right

  8. myininaya
    • 5 years ago
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    (-inf,-6]u[1,inf)

  9. anonymous
    • 5 years ago
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    What.... does that say

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