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christinaxxx

  • 3 years ago

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

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

  2. karinewoods17
    • 3 years ago
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    try changing all the p's into x's if you see it better that way.

  3. christinaxxx
    • 3 years ago
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    oh no! i meant for the p to be squared!

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

  5. TuringTest
    • 3 years ago
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    find the x-intercepts you mean then, so solve the equation

  6. christinaxxx
    • 3 years ago
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    I'm not allowed to use the quadratic formula. i can only factor it

  7. TuringTest
    • 3 years ago
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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

  8. christinaxxx
    • 3 years ago
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    okayyy and how do i find those solutions mathmatically?

  9. TuringTest
    • 3 years ago
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    what are the only two integers that can multiply to be 1 ?

  10. christinaxxx
    • 3 years ago
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    1 aaand 1 of course

  11. christinaxxx
    • 3 years ago
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    or -1 and -1

  12. TuringTest
    • 3 years ago
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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 ?

  13. christinaxxx
    • 3 years ago
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    1 & 6 2&3 aanndd thats it

  14. TuringTest
    • 3 years ago
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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.

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

  16. TuringTest
    • 3 years ago
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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.

  17. christinaxxx
    • 3 years ago
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    dude, how I wish you were my math teacher

  18. TuringTest
    • 3 years ago
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    hehe, I do too sometimes welcome!

  19. TuringTest
    • 3 years ago
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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.

  20. christinaxxx
    • 3 years ago
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    haha, yes i understand! thank you so much!

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