# Quadratic Equations

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• Jan 11th 2010, 08:50 PM
jarman007
Quadratic Equations
Standard form of quadratic equations is ax^2+bx+c=0

if a=1 , b and c are integers and the roots are rational numbers

then prove that roots must be integers?
• Jan 12th 2010, 12:46 AM
tonio
Quote:

Originally Posted by jarman007
Standard form of quadratic equations is ax^2+bx+c=0

if a=1 , b and c are integers and the roots are rational numbers

then prove that roots must be integers?

Suppose $\frac{r}{s}$ is a root of $x^2+bx+c=0$ , and that $(r,s)=1$ (this means: the fraction $\frac{r}{s}$ is reduced) , then

$\frac{r^2}{s^2}+b\frac{r}{s}+c=0\Longrightarrow r^2+brs+cs^2=0$ (multiplication by common denominator). As $s|(brs+cs^2)$ then also $s|r^2\Longleftrightarrow s|r$ , which is possible (by the reduced fraction assumption) iff $s=\pm 1$ ... end now the argument

Tonio
• Jan 12th 2010, 01:34 AM
crossbone
Tonio,
what does "|" symbol mean?
• Jan 12th 2010, 01:46 AM
tonio
Quote:

Originally Posted by crossbone
Tonio,
what does "|" symbol mean?

"|" = divide : $a|b\Longrightarrow \exists c\in\mathbb{Z}\,\,\,s.t.\,\,\,b=ac$

Tonio
• Jan 12th 2010, 02:12 AM
crossbone
how you get http://www.mathhelpforum.com/math-he...85bf90ab-1.gif from the equation? i'm just trying to follow your thought process and argument.
• Jan 12th 2010, 03:45 AM
tonio
Quote:

Originally Posted by crossbone
how you get http://www.mathhelpforum.com/math-he...85bf90ab-1.gif from the equation? i'm just trying to follow your thought process and argument.

$brs+cs^2=s(br+cs)$ ,and everything's integer here, of course.

Tonio
• Jan 12th 2010, 03:55 AM
HallsofIvy
A slightly different proof. Let $\frac{m}{n}$ and $\frac{p}{q}$ be the rational roots (m, n, p, and q integer, of course), reduced to lowest terms. Then we must have $\left(x- \frac{m}{n}\right)\left(x- \frac{p}{q}\right)= 0$. Multiplying that out, $x^2- \left(\frac{m}{n}- \frac{p}{q}\right)x+ \frac{mp}{nq}$ $= x^2- \frac{mq- np}{nq}x+ \frac{mp}{nq}= x^2+ bx+ c= 0$.

Since b and c are integers, we must have nq= 1. Since n and q are integers, we must have $n= \pm 1$ and $q= \pm 1$. Then the roots $\frac{m}{n}$ and $\frac{p}{q}$ are integer.
• Jan 12th 2010, 04:03 AM
Moo
Hi,
Quote:

Originally Posted by HallsofIvy
A slightly different proof. Let $\frac{m}{n}$ and $\frac{p}{q}$ be the rational roots (m, n, p, and q integer, of course), reduced to lowest terms. Then we must have $\left(x- \frac{m}{n}\right)\left(x- \frac{p}{q}\right)= 0$. Multiplying that out, $x^2- \left(\frac{m}{n}- \frac{p}{q}\right)x+ \frac{mp}{nq}$ $= x^2- \frac{mq- np}{nq}x+ \frac{mp}{nq}= x^2+ bx+ c= 0$.

Since b and c are integers, we must have nq= 1. Since n and q are integers, we must have $n= \pm 1$ and $q= \pm 1$. Then the roots $\frac{m}{n}$ and $\frac{p}{q}$ are integer.

Why the sentence in red ?
We could have nq|(mq-np) and nq|(mp) without nq being 1 (or -1)
• Jan 12th 2010, 06:46 AM
tonio
Quote:

Originally Posted by Moo
Hi,

Why the sentence in red ?
We could have nq|(mq-np) and nq|(mp) without nq being 1 (or -1)

Not if the fractions are reduced:

$nq\mid(mq-np)\Longrightarrow mq-np=xnq$

$nq\mid mp\Longrightarrow mp =ynq$

multiply the first eq. above by m and the second one by n:

$m^2q-mnp=xmnq$

$mnp=yn^2q$ , and now add these two eq's:

$m^2q=(xm+yn)nq\Longrightarrow m^2=(xm+yn)n$ .Since n divides the right hand n divides the left hand as well, contradicting the fact that $\frac{m}{n}$ is reduced...unless $n=\pm 1$ ,and the same for q, of course.

Tonio
• Jan 12th 2010, 07:01 AM
Jhevon
Quote:

Originally Posted by jarman007
Standard form of quadratic equations is ax^2+bx+c=0

if a=1 , b and c are integers and the roots are rational numbers

then prove that roots must be integers?

Another alternative: We wish to show that the roots of $x^2 + bx + c = 0$ are integers. By the quadratic formula, it suffices to show that the expression $\frac {-b \pm \sqrt{b^2 - 4c}}2$ is always an integer provided it is rational (which means $b^2 - 4c$ is a perfect square, call it $k^2$) and $b,c \in \mathbb{Z}$.

It suffices to show that $-b \pm \sqrt{b^2 - 4c}$ is even. We consider two cases: (1) $b$ is even and (2) $b$ is odd. (Notice that the parity of $c$ is immaterial, since $4c$ would be even whether $c$ is even or odd).

case 1: $b$ is even.

Then, $b^2$ is even, so that $k^2 = b^2 - 4c$ is even (since it is the difference of two even integers), which implies $k$ is even. But then, $-b \pm \sqrt{b^2 - 4c} = -b \pm k$ is even, since it is the sum of two even integers.

case 2 is no worse and will be left to the OP.
• Jan 12th 2010, 01:21 PM
crossbone
Quote:

Originally Posted by tonio
$brs+cs^2=s(br+cs)$ ,and everything's integer here, of course.

Tonio

the original equation is $r^2+brs+cs^2=0$
why did you ignore $r^2$?
• Jan 12th 2010, 01:31 PM
Jhevon
Quote:

Originally Posted by crossbone
the original equation is $r^2+brs+cs^2=0$
why did you ignore $r^2$?

he didn't. you asked about one part though, so he answered that part. he did state that $s|r^2$...
• Jan 12th 2010, 02:16 PM
crossbone
that's the problem. i'm trying to understand his logic from going from $r^2+brs+cs^2=0$ to $s/(brs+cs^2)$. it's like some leap of logic unless he's using some mathematical properties of polynomial coefficients that I'm missing. btw, I'm new here so quite curious why is everybody using "|" instead of "/" to signify division? there's no using the shift key to choose the latter.
• Jan 12th 2010, 02:28 PM
Jhevon
Quote:

Originally Posted by crossbone
that's the problem. i'm trying to understand his logic from going from $r^2+brs+cs^2=0$ to $s/(brs+cs^2)$. it's like some leap of logic unless he's using some mathematical properties of polynomial coefficients that I'm missing.

actually, he was applying basic properties of integers that you can learn about in a number theory or abstract algebra class.

showing that $s|r$ was important to tonio's argument. he could show that $s|r$ by showing that $s|r^2$

we have
$r^2 + brs + cs^2 = 0$

$\Rightarrow r^2 = -(brs + cs^2)$

by dividing through by $s$, we have

$\frac {r^2}s = -(br + cs)$

since the right side is an integer (because $s|(brs + cs^2)$), the left side must be also, because we have equality here. but $\frac {r^2}s$ is an integer if and only if $s|r^2$

Quote:

btw, I'm new here so quite curious why is everybody using "|" instead of "/" to signify division? there's no using the shift key to choose the latter.
they are actually quite different symbols. "|" means "divides" while "/" means "divided by".

the difference?

2|4 means that 4 = 2k for some integer k. it means that 2 divides into 4 without leaving a remainder. while,

2/4 = 1/2 = 0.5

see the difference? one is making a statment about the relationship, namely a divisibility relationship, between two integers, while the other signifies performing and arithmetical operation, namely division, on two integers.
• Jan 12th 2010, 04:10 PM
crossbone
Quote:

Originally Posted by Jhevon
actually, he was applying basic properties of integers that you can learn about in a number theory or abstract algebra class.

showing that $s|r$ was important to tonio's argument. he could show that $s|r$ by showing that $s|r^2$

we have
$r^2 + brs + cs^2 = 0$

$\Rightarrow r^2 = -(brs + cs^2)$

by dividing through by $s$, we have

$\frac {r^2}s = -(br + cs)$

since the right side is an integer (because $s|(brs + cs^2)$), the left side must be also, because we have equality here. but $\frac {r^2}s$ is an integer if and only if $s|r^2$

they are actually quite different symbols. "|" means "divides" while "/" means "divided by".

the difference?

2|4 means that 4 = 2k for some integer k. it means that 2 divides into 4 without leaving a remainder. while,

2/4 = 1/2 = 0.5

see the difference? one is making a statment about the relationship, namely a divisibility relationship, between two integers, while the other signifies performing and arithmetical operation, namely division, on two integers.

i've never encountered the use of | in my high school math. i know the stuff here are all pre-calc but isn't the use of these esoteric explanation beyond the average high-school student?
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