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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.2012-04-15
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    Did the book really use lower case letters for the coefficients in the question, and upper case letters for the answer?2012-04-15
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    It's probably just me, but this question seems strikingly unnatural. What book did you take this from, if we may know?2012-04-15
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    @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.2012-04-15
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    *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.2012-04-15
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    @M.Na'el Discrement? What is it? Is it _Discriminant_ we are talking about?2012-04-15
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    @M.Na'el It is considered impolite to grab the attention of a particular user by using the @ system in a question. Also, note that @ system does not work in the questions or in the answers. I, in particular, don't see how the comments directed at Raymond are helpful to the users of this site in general. So, I'll remove that in this edit of mine.2012-04-15
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    He asked for my trial on the second one and so I did... I never intended it to be un-respectful and I knew it doesn't work in questions yet I only wanted it visible as the answer he requested.2012-04-15
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    @KannappanS, Discriminant! I wrote it the other way as I remembered its pronunciation in class... So many faults in my educational system... Anyway, it's ($\Delta = b^2-4ac$).2012-04-15
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    @M.Na'el Now, I have given some hints for $(2)$. Are you done solving $(1)$? Do you need me to check your answers or some such thing?2012-04-15
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    let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/3108/discussion-between-m-nael-and-kannappan-sampath)2012-04-15

3 Answers 3

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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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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.2012-04-15
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    @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!2012-04-15
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    I tried it with the second and can't get the $x$ and $n+3$ out... Where can I post a photo?2012-04-15
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    @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).2012-04-15
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    Now what? {Extra Characters}2012-04-15
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    @M.Na'el: I hope that my edit will help you more!2012-04-15
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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...2012-05-12