## ronocthede Group Title 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? 7 months ago 7 months ago

1. hartnn Group Title

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

2. ronocthede Group Title

Okay so I assume I use substitution here?

3. hartnn Group Title

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 Group Title

find PQ now

5. ronocthede Group Title

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

6. hartnn Group Title

yes.

7. hartnn Group Title

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

8. hartnn Group Title

then simplify

9. ronocthede Group Title

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 Group Title

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 Group Title

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

12. hartnn Group Title

to find PQ ? not much, you can try

13. hartnn Group Title

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 Group Title

$f(x)=(x-P)(x-Q)$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 Group Title

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

16. whpalmer4 Group Title

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 Group Title

I still like @hartnn's approach best, though

18. ronocthede Group Title

Thanks again, really helped clear it up.