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yociyoci

Can someone guide me through inequality questions? i) 15-2y-y^2<0 ii) 20x-4x^2-25>0

  • one year ago
  • one year ago

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  1. Mertsj
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    \[y^2+2y-15>0\]

    • one year ago
  2. yociyoci
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    so now, (y-3)(y+5)>0? and y>-5 or y<3?

    • one year ago
  3. Mertsj
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    The first part is right. If you multiply two factors and get a positive result, then both factors must be positive or both factors must be negative.

    • one year ago
  4. Mertsj
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    |dw:1361936550944:dw|

    • one year ago
  5. Mertsj
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    So we see that the solution is y <-5 or y>3

    • one year ago
  6. yociyoci
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    oh, alright. What about the second question? I got suck at (2x-3)^2 > 0

    • one year ago
  7. Mertsj
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    \[4x^2-20x+25<0\] \[(2x-5)(2x-5)<0\]

    • one year ago
  8. yociyoci
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    opps, I was looking at a different one. Yes, I got to (2x-5)^2<0

    • one year ago
  9. Mertsj
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    |dw:1361936871505:dw|

    • one year ago
  10. Mertsj
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    Oh, it will never be less than 0 but of course we already knew that a real number raised to the second power will always be positive or 0 and never less than 0

    • one year ago
  11. yociyoci
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    So there will be no solution in this case, but what if it is (2x-5)^2 > 0? Will this work?

    • one year ago
  12. Mertsj
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    Yes. Then it would be all real numbers except 0

    • one year ago
  13. yociyoci
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    Alright! Thank you so much!

    • one year ago
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