Question

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?

Answer 1

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

Answer 2

Okay so I assume I use substitution here?

Answer 3

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 ?

Answer 4

find PQ now

Answer 5

Hmm I think so. So PQ = 5/8 ?

Answer 6

yes.

Answer 7

now ,
equation will be
\(x^2 -(P+Q)x + PQ = 0\)

Answer 8

then simplify

Answer 9

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?

Answer 10

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

Answer 11

Oh yeah, I vaguely remember that. One more thing sorry, how difficult would the substitution be?

Answer 12

to find PQ ?
not much, you can try

Answer 13

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

Answer 14

\[f(x)=(x-P)(x-Q)\]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

Answer 15

Okay that makes a lot of sense. Was just blanking out there. Thanks for the help guys!

Answer 16

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.

Answer 17

I still like @hartnn's approach best, though

Answer 18

Thanks again, really helped clear it up.