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Ben103

t^2+6t-7. Is their a 2 numbers that equael product -7 and sum 6. Is it just me or thats not possible?

  • one year ago
  • one year ago

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  1. KingGeorge
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    That's just you :) There are indeed two numbers that multiply to be -7, and add to be 6. Think about it a bit and I'm sure you'll get it.

    • one year ago
  2. Ben103
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    Well the equation is set up as (a+1) (a-7) thats what I got. but 1-7=-6 so that doesnt work.

    • one year ago
  3. miszery
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    remember that - and + in front of the numbers

    • one year ago
  4. Azteck
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    Wrong way around mate

    • one year ago
  5. KingGeorge
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    That's very close.

    • one year ago
  6. Ben103
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    ok got it switched them. Now it works.

    • one year ago
  7. KingGeorge
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    Excellent!

    • one year ago
  8. Ben103
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    Thanks lol I cant believe i cudnt figure tthat out

    • one year ago
  9. KingGeorge
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    No problem. We've all had something obvious pass us by before :P

    • one year ago
  10. Ben103
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    would the solution set be written as (a+7) (a-1) or just {7,-1}

    • one year ago
  11. miszery
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    are you factorizing? normally it wuold be written as full.

    • one year ago
  12. UnkleRhaukus
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    you could always use the quadratic formula \[t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]

    • one year ago
  13. miszery
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    (a+7)(a-1) that is.... you will need to use the full factors to use in equations.

    • one year ago
  14. KingGeorge
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    The solution set would be {-7,1}. Note that you have -7 and +1 because if \(a+7=0\), then \(a=-7\). Similarly, \(a-1=0\implies a=1\).

    • one year ago
  15. miszery
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    @Ben103 sorry the questions look like what I'm doing at school at the moment which is factorizing :( I could be wrong

    • one year ago
  16. miszery
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    @KingGeorge knows more than me i'm sure! lol I'm only doing high school maths

    • one year ago
  17. KingGeorge
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    As another tip, if you're asked for a solution SET, your final answer should look like {_,_,_,...,_}. I.e., you have a list of things in the curly brackets.

    • one year ago
  18. Ben103
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    Ok thanks I understand it now. :)

    • one year ago
  19. KingGeorge
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    You're welcome.

    • one year ago
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