Conditions on roots of a quadratic The question in my book says:

Find the necessary conditions in which the quadratic equation $ax^2+bx+c$
would have roots (m, n) such that (in individual cases):
(i) m=2n
(ii) m=n+3

It's really annoying because whatever knowledge I had about quad. equ.'s didn't work and the answer the book gave was:

(i) $9AC=2B^2$
and (ii) $B^2=9A^2+4AC$

I gave this 10 minutes of thought; none.
The question form is obviously the problem; I don't have a teacher so I thought that someone might have a better explanation for "conditions" generally and specifically in quadratic roots.

$$ m=2n \\ ax^2(x-m)(x-n) \\ a(x-[n+3] \hspace{3pt})(x-n) \\ ax^2-2anx-3ax+an^2+3an \\ \hspace{-.9cm} =  \hspace{.3cm} ax^2-ax(2n-3)+an(n+3) \\ *n=\frac {c} {a(n+3)} \\ *-ax \left[ \frac{2c}{2an+6a} \right] = b \\ x \left( \frac {c}{n+3}+3a \right)=b$$
 A: Let's help you a bit for the first one :  
(i) $m=2n$ with $m$ and $n$ both roots so that $ax^2+bx+c=a(x-m)(x-n)=a(x-2n)(x-n)=ax^2-(2a+a)nx+2an^2$.
Identifying powers of $x$ we get $b=-3an$ and $c=2an^2$ and the first result (because $n=-\frac{b}{3a}$)
Hoping this will help you too to solve the second question...

Let's edit/correct your update :
$$
m=n+3 \\ a(x-m)(x-n) \\ =a(x-[n+3] \hspace{3pt})(x-n) \\ =ax^2-2anx-3ax+an^2+3an \\ \hspace{-.9cm} =  \hspace{.3cm} ax^2-a(2n+3)x+an(n+3)
$$
At this point you have to identify this with $ax^2+bx+c$ so that :
$$b=-a(2n+3),\ c=an(n+3)$$
From the first equation you may deduce $2n+3=-\dfrac ba$ or $n=-\dfrac b{2a}-\dfrac 32$
Replace $n$ in the expression for $c$ (using $(-u-v) \cdot(-u+v)=u^2-v^2)$ and conclude!
Richard Feynman enjoyed a mathematics book starting with :
"What one fool can do, another can too".
It is the first step that is difficult!
A: That this is a homework and the OP seems to be actively trying to solve the problem, I will hint at some steps for this problem:
You have a quadratic equation $ax^2+bx+c=0$. Now, if $(m,n)$ are the roots of this equation, find a condition on $a,b,c$ such that 


*

*$m=2n$

*$m=n+3$


Hint for $(1)$
Think about sum of the roots and product of the roots. In particular $$\begin{align}\frac{-b}{a}&=m+n=3n\\\frac{c}{a}&=mn=2n^2\end{align}$$ Eliminate $n$ from here...
Hint for $(2)$: (Can be worked like above but lengthy.)
Answer the following questions sequentially and put them together to answer $(2)$:


*

*What are the roots of the equation? (Hint: Quadratic formula)

*Given $m=n+3$, we have that $m-n=3>0$. This means, $m>n$. Of the two roots (a priori two of them! ) you get from the quadratic formula, can you decide which must be $m$ and which one $n$? 

*Put the values of $m$ and $n$ into $m=n+3$. What do you get? 

*Re-arrange your solution if need be to see the answer. 

A: OK, I'm getting somebody is fighting with these quadratic equation problems, now I'm explaining how the answer is coming.
For (1.)
It is given that a quadratic equation is $$ax²+bx+c=0$$ has roots m & n. Where $$m=2n$$
As we know that, sum of roots = $$-b/a$$ and product of roots = $$c/a$$
Now put these,
$$m+n=(-b/a)
2n+n=-b/a
3n=-b/a => n=-b/3a$$
Therefore,
$$m*n=c/a
2n*n=c/a
2n²=c/a$$
Put the value of n from (i) -
$$2(-b/3a)²=c/a
b²/9a²=c/2a
2ab²=9a²c
9ac=2b²$$
Which is the required condition.
