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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.

$$ m=2n \\ ax^2(x-m)(x-n) \\ a(x-[n+3] \hspace{3pt})(x-n) \\ ax^2-2anx-3ax+an^2+3an \\ \hspace{-.9cm} = \hspace{.3cm} ax^2-ax(2n-3)+an(n+3) \\ *n=\frac {c} {a(n+3)} \\ *-ax \left[ \frac{2c}{2an+6a} \right] = b \\ x \left( \frac {c}{n+3}+3a \right)=b$$

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Do you know Vieta's Formula? This will help you solve the problem. – Johannes Kloos Apr 15 '12 at 11:56
Did the book really use lower case letters for the coefficients in the question, and upper case letters for the answer? – Gerry Myerson Apr 15 '12 at 12:54
It's probably just me, but this question seems strikingly unnatural. What book did you take this from, if we may know? – 000 Apr 15 '12 at 13:07
@GerryM, I'd said it was weird from the get go... It did actually use different case letters! - It's from my curriculum book, 10th grade equiv., Egypt. – Noein Apr 15 '12 at 13:45
Can't delete prev. comments. - @JohannesK, Isn't Vieta's formula where we use a discrement to extract the roots? I tried but that didn't work. – Noein Apr 15 '12 at 13:46
up vote 3 down vote accepted

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

  1. $m=2n$
  2. $m=n+3$

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.
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Tomorrow's the exam. I'm keeping all this in mind and hoping for he best... – Noein May 12 '12 at 20:06

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!

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But that would make $9a^2c = 2b^2$ which isn't the answer provided. It does seem logical and correct in steps but here they look for how good you followed the book even if it's incorrect. So I just need a confirmation that there's no other way to re-produce the answer provided. - I'll follow your steps and post the second one. – Noein Apr 15 '12 at 13:44
@M.Na'el: Oops I forgot the 'a' factor in $2n^2$ (corrected). There are many ways to find these solutions. I have the feeling that you let the words used block your progression, we all guess the meaning first and sometimes we are wrong but that shouldn't stop us! – Raymond Manzoni Apr 15 '12 at 13:51
I tried it with the second and can't get the $x$ and $n+3$ out... Where can I post a photo? – Noein Apr 15 '12 at 14:09
@M.Na'el: you may edit your question or provide an answer (I don't know if you have enough points to add a picture but you may provide any link even if $\LaTeX$ is much more appreciated here). – Raymond Manzoni Apr 15 '12 at 14:13
Now what? {Extra Characters} – Noein Apr 15 '12 at 15:04

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.

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