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christinaxxx Group Title

okay, I struggle hard core with these and I don't know why. so I need to put 0=6^2p+5p+1 into intercept form

  • one year ago
  • one year ago

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  1. TuringTest Group Title
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    slope intercept of\[0=6^2p+5p+1\]are you sure you typed it right?

    • one year ago
  2. karinewoods17 Group Title
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    try changing all the p's into x's if you see it better that way.

    • one year ago
  3. christinaxxx Group Title
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    oh no! i meant for the p to be squared!

    • one year ago
  4. christinaxxx Group Title
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    it should be 6p^2+5p+1=0 and i need to change that into x-intercept form

    • one year ago
  5. TuringTest Group Title
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    find the x-intercepts you mean then, so solve the equation

    • one year ago
  6. christinaxxx Group Title
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    I'm not allowed to use the quadratic formula. i can only factor it

    • one year ago
  7. TuringTest Group Title
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    6p^2+5p+1=0 ( _p + _ )( _p + _ )=0 we need to fill in the blanks such that the last terms multiply to 1 and the first terms multiply to 6

    • one year ago
  8. christinaxxx Group Title
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    okayyy and how do i find those solutions mathmatically?

    • one year ago
  9. TuringTest Group Title
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    what are the only two integers that can multiply to be 1 ?

    • one year ago
  10. christinaxxx Group Title
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    1 aaand 1 of course

    • one year ago
  11. christinaxxx Group Title
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    or -1 and -1

    • one year ago
  12. TuringTest Group Title
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    good observation that it could be negative, but since everything is postive here we will just assume it is positive for now (_p+1)(_p+1) what pairs of integers multiply to 6 ?

    • one year ago
  13. christinaxxx Group Title
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    1 & 6 2&3 aanndd thats it

    • one year ago
  14. TuringTest Group Title
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    right, and since we know that (3p+1)(2p+1)=0 that implies that either 3p+1=0 or 2p+1=0, or both so solve those two equations and you get your two values of x for which y is zero, i.e. where the graph intercepts the x-axis.

    • one year ago
  15. christinaxxx Group Title
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    so (3p+1)(2p+1)=0 is tha answer? and i can use this process for every problem?

    • one year ago
  16. TuringTest Group Title
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    you should always double-check that you get the correct cross term to be sure you have done it right (3p+1)(2p+1)=0 6p^2+3p+2p+1=0 6p^2+5p+1=0 ^^ cross term when factoring something that looks like ax^2+bx+c=0 sometimes you can find numbers that multiply to a and c, but do not produce the correct b, so you should FOIL out what you have to be sure also, you are not done with the statement (3p+1)(2p+1)=0 you still need to follow what I said above and solve for p: since we know that (3p+1)(2p+1)=0 that implies that either 3p+1=0, or 2p+1=0, or both equal zero so solve those two equations and you get your two values of x for which y is zero, i.e. where the graph intercepts the x-axis.

    • one year ago
  17. christinaxxx Group Title
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    dude, how I wish you were my math teacher

    • one year ago
  18. TuringTest Group Title
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    hehe, I do too sometimes welcome!

    • one year ago
  19. TuringTest Group Title
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    oh, typo... should say: so solve those two equations and you get your two values of p for which the function is zero, i.e. where the graph intercepts the p-axis.

    • one year ago
  20. christinaxxx Group Title
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    haha, yes i understand! thank you so much!

    • one year ago
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