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ronocthede

  • 2 years ago

Now sure how to get to answer: A quadratic function, f, has zeros P and Q such that P+Q=5 and (1/P)+(1/Q)=8. What function would best describes f?

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  1. hartnn
    • 2 years ago
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    so you have sum of roots find product of roots PQ from (1/P)+(1/Q)=8. can u ?

  2. ronocthede
    • 2 years ago
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    Okay so I assume I use substitution here?

  3. hartnn
    • 2 years ago
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    you can, yes. but see this : 1/P + 1/Q =8 (Q+P)/PQ = 8 we know P+Q = 5 so, 5/(PQ) = 8 got this ?

  4. hartnn
    • 2 years ago
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    find PQ now

  5. ronocthede
    • 2 years ago
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    Hmm I think so. So PQ = 5/8 ?

  6. hartnn
    • 2 years ago
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    yes.

  7. hartnn
    • 2 years ago
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    now , equation will be \(x^2 -(P+Q)x + PQ = 0\)

  8. hartnn
    • 2 years ago
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    then simplify

  9. ronocthede
    • 2 years ago
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    okay so 8x^2-40x+5, which I know is the right answer. I'm still a bit confused where "x^2−(P+Q)x+PQ=0" Is that a formula I forgot?

  10. hartnn
    • 2 years ago
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    thats a standard relation. like if you have x^2 +ax+b = 0 sum of roots will be -a and product of roots = b thats where it came from

  11. ronocthede
    • 2 years ago
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    Oh yeah, I vaguely remember that. One more thing sorry, how difficult would the substitution be?

  12. hartnn
    • 2 years ago
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    to find PQ ? not much, you can try

  13. hartnn
    • 2 years ago
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    a longer way will be to actually find the values of P and Q. then equation will be (x-P)(x-Q) = 0 then simplify

  14. whpalmer4
    • 2 years ago
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    \[f(x)=(x-P)(x-Q)\]is just the polynomial in factored form \[f(x)=(x-P)(x-Q) = x^2-Qx-Px+PQ = x^2-(P+Q)x + PQ\] after expansion and collecting like terms

  15. ronocthede
    • 2 years ago
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    Okay that makes a lot of sense. Was just blanking out there. Thanks for the help guys!

  16. whpalmer4
    • 2 years ago
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    and substitution isn't too bad: \[P+Q=5\]\[P=5-Q\]\[\frac{1}P+\frac{1}Q =8\]\[\frac{1}{5-Q}+\frac{1}Q =8\]Multiply everything by \(Q(5-Q)\) to get \[Q+5-Q = 8Q(5-Q)\]\[-8Q^2+40Q-5=0\]Etc.

  17. whpalmer4
    • 2 years ago
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    I still like @hartnn's approach best, though

  18. ronocthede
    • 2 years ago
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    Thanks again, really helped clear it up.

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