Melody

I get      sqrt(405)

Do you still want/need an explanation?

A:B:C=3:2:3 (1)

so    C=A

and 

\(\frac{B}{A}=\frac{2}{3}\\ B=\frac{2}{3}*A\)

Now

15A+6B+9C=20 (2)

15A+6*(2/3)A+9A=20

15A+4A+9A=20

28A=20

\(A=\frac{20}{28}=\frac{5}{7}\\ B=\frac{2}{3}*\frac{5}{7}=\frac{10}{21}\\ C=\frac{5}{7}\)

4ax - 21x = -41x

(4a-21) x = - 41 x

4a - 21 = -41

4a = -20

a = -5 

A+B=17   (1)

B+C=16   (2)

(1)-(2)

A+B-B-C = 17-16

A - C = 1

so

A=1+C      

Now   

A*C = 30    so

(1+C)*C = 30

c^2+C-30=0

(C-5)(C+6)=0

C = 5 or -6

but C is positive so  C=5

so

A=1+C = 6

I think you can take it from there to find  B and D

Note that there is one piece of info in the question that I have not used yet.   

Attn: Helperid18...  and guest asker

Guest has asked the question as I have always requested that people do.

Thanks for that guest.

The only improvement I can see is that the address could be included as a link.

That way people could hyperlink to it more easily.

The asker should also request that people answer on the original and leave just a little note here (as well) stating that they have done so.

Thanks for answering Jugoslav :)

Thanks to gest for putting the link to Jugoslav's answer.

\(f(-4)=\frac{a}{b}*-4=\frac{-4a}{b}\\ \frac{-4a}{b}=\frac{-60}{13}\\ \frac{4a}{b}=\frac{60}{13}\\ \frac{a}{b}=\frac{15}{13}\\ a=\frac{15b}{13}\)

\(f(4)=ab*4^2=16ab\\ 16ab=3120\\ ab=195\\~\\ so\\ \frac{15b}{13}*b=195\\ \frac{b^2}{13}=13\\ b^2=13^2\\ b=\pm13 \)

You can check what I have done and continue from there.

LaTex:

f(-4)=\frac{a}{b}*-4=\frac{-4a}{b}\\
\frac{-4a}{b}=\frac{-60}{13}\\
\frac{4a}{b}=\frac{60}{13}\\
\frac{a}{b}=\frac{15}{13}\\
a=\frac{15b}{13}

This question has been asked numerous times lately. Go look for an old answer.

Very impressive guest :)

Ok I explained a bit more.

Next time please comment on the original.  Answerers do not like to be ignored.

You can still put a new post to say that you have done so.