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anonymous
 3 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?
anonymous
 3 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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hartnn
 3 years ago
Best ResponseYou've already chosen the best response.2so you have sum of roots find product of roots PQ from (1/P)+(1/Q)=8. can u ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Okay so I assume I use substitution here?

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.2you 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 ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Hmm I think so. So PQ = 5/8 ?

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.2now , equation will be \(x^2 (P+Q)x + PQ = 0\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0okay so 8x^240x+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?

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.2thats 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Oh yeah, I vaguely remember that. One more thing sorry, how difficult would the substitution be?

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.2to find PQ ? not much, you can try

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.2a longer way will be to actually find the values of P and Q. then equation will be (xP)(xQ) = 0 then simplify

whpalmer4
 3 years ago
Best ResponseYou've already chosen the best response.1\[f(x)=(xP)(xQ)\]is just the polynomial in factored form \[f(x)=(xP)(xQ) = x^2QxPx+PQ = x^2(P+Q)x + PQ\] after expansion and collecting like terms

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Okay that makes a lot of sense. Was just blanking out there. Thanks for the help guys!

whpalmer4
 3 years ago
Best ResponseYou've already chosen the best response.1and substitution isn't too bad: \[P+Q=5\]\[P=5Q\]\[\frac{1}P+\frac{1}Q =8\]\[\frac{1}{5Q}+\frac{1}Q =8\]Multiply everything by \(Q(5Q)\) to get \[Q+5Q = 8Q(5Q)\]\[8Q^2+40Q5=0\]Etc.

whpalmer4
 3 years ago
Best ResponseYou've already chosen the best response.1I still like @hartnn's approach best, though

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Thanks again, really helped clear it up.
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