Integer solution.

[man111]

for what integer value of [tex]a[/tex] for which the equation [tex]ax^2 - 4x +9 = 0[/tex] has integer solution.

[inveidar]

Отговор: [tex]a=-5,a=-13[/tex].

[man111]

Sorry for wrong typing.....

The Original Question is .....

for what value of [tex]a[/tex] for which the equation [tex]ax^2-4x+9 = 0[/tex] has Integer solution.

[allier]

It's still not clear. Does the equation need to have at least one integer solution, or do both solutions need to be integers?

[man111]

both solution are Integer..

[mkmarinov]

The roots of the equation are
[tex]x = \frac{2 \pm \sqrt{4-9a}}{a}[/tex]. Let [tex]4-9a=k^2, k \in Z[/tex].
[tex]a=\frac{4-k^2}{9}[/tex], so the roots become:
[tex]x_1=\frac{9(2+k)}{4-k^2}=\frac{9}{2-k}[/tex]
[tex]x_2=\frac{9(2-k)}{4-k^2}=\frac{9}{2+k}[/tex]
which must both be integer.
If k is a solution, the -k is also a solution, so we shall consider k>0. (and we only need the square of k to determine a)
Both [tex]2-k[/tex] and [tex]2+k[/tex] should divide 9, in other words they can be [tex]\pm 1, \pm 3, \pm 9[/tex].
2+k is either 3 or 9.
Let [tex]2+k=3=> 2-k=1 => k=1[/tex] is a solution.
Let [tex]2+k=9 => 2-k=-5[/tex], not a solution.
So, the only positive solution is k=1.
Going back to a:
[tex]a=\frac{4-k^2}{9}=\frac{1}{3}[/tex].

P.S. If at least one root must be integer, this leaves k=1,3,5,7,11. See a pattern?

[allier]

I do not agree mkmarinov! D is not necessarily a square of a rational number, or at least it's not evident from what you've written. The first step in the solution is to show that a is rational, after that - to show that k is an integer. After that, all you've written is almost entirely correct. Note that 2-k might also be a rational number if you have not proved that k is an integer beforehand.

[mkmarinov]

Well, when I think of it, a is obviously rational:
[tex]a=\frac{4x-9}{x^2}[/tex], which is rational for integer (actually, for all rational) x. Therefore k is also rational (and we'll take it positive). (this is why I did not include it)
For k not necessarily being integer, you are right. But another solution, which is actually shorter, came to my mind:
[tex]k=\frac{p}{q}, (p,q)=1[/tex]
[tex]x_1=\frac{9q}{2p+q}[/tex]
[tex]x_2=\frac{9q}{2p-q}[/tex]
[tex]\frac{1}{x_1}+\frac{1}{x_2}=\frac{2p+q+2p-q}{9q}=\frac{4p}{9q}[/tex], but
[tex]\frac{1}{x_1}+\frac{1}{x_2}=\frac{x_1+x_2}{x_1x_2}=\frac{\frac{4}{a}}{\frac{9}{a}}=\frac{4}{9}[/tex].
Therefore [tex]\frac{p}{q}=k=1[/tex].

[man111]

Thanks for nice explanation and answer.

[inveidar]

Отговор: [tex]a=-5,a=-13[/tex].

It is for a-integer and equation has at least one integer solution.

[man111]

Here We have to solve for both Integer Roots.

Let [tex]\alpha[/tex] and [tex]\beta[/tex] be the roots of [tex]ax^2-4x+9=0[/tex], Where [tex]\alpha,\beta \in Z[/tex]

Then [tex]\alpha+\beta =\frac{4}{a}[/tex] and [tex]\alpha.\beta=\frac{9}{a}[/tex]

Here both [tex]\alpha,\beta \in Z[/tex]. So [tex]\alpha+\beta=\frac{4}{a}\in Z[/tex] and [tex]\alpha.\beta=\frac{9}{a}\in Z[/tex]

Let [tex]\frac{1}{a}=n\in Z[/tex] and [tex]\alpha,\beta=\frac{4\pm\sqrt{16-36a}}{2a}=\frac{2\pm\sqrt{4-9a}}{a}\in Z[/tex]

for this [tex]4-9a[/tex]=perfect square Integer

So Let [tex]4-9a=m^2[/tex], Where [tex]m\in Z[/tex], Put [tex]a=\frac{1}{n}[/tex]

[tex]4-\frac{9}{n}=m^2[/tex],Which is posiible only when [tex]n=\pm1,\pm3,\pm 9[/tex]

So for only [tex]n=+3[/tex], [tex]m^2[/tex]=Perfect square Integer.

So [tex]a=\frac{1}{n}=\frac{1}{3}[/tex] is a value for which [tex]ax^2-4x+9=0[/tex] has both Integer Roots.

[ganka simeonova]

The roots of the equation are
[tex]x = \frac{2 \pm \sqrt{4-9a}}{a}[/tex]. Let [tex]4-9a=k^2, k \in Z[/tex].
[tex]a=\frac{4-k^2}{9}[/tex], so the roots become:
[tex]x_1=\frac{9(2+k)}{4-k^2}=\frac{9}{2-k}[/tex]
[tex]x_2=\frac{9(2-k)}{4-k^2}=\frac{9}{2+k}[/tex]
which must both be integer.
If k is a solution, the -k is also a solution, so we shall consider k>0. (and we only need the square of k to determine a)
Both [tex]2-k[/tex] and [tex]2+k[/tex] should divide 9, in other words they can be [tex]\pm 1, \pm 3, \pm 9[/tex].
2+k is either 3 or 9.
Let [tex]2+k=3=> 2-k=1 => k=1[/tex] is a solution.
Let [tex]2+k=9 => 2-k=-5[/tex], not a solution.
So, the only positive solution is k=1.
Going back to a:
[tex]a=\frac{4-k^2}{9}=\frac{1}{3}[/tex].

P.S. If at least one root must be integer, this leaves k=1,3,5,7,11. See a pattern?

Чудесно, но защо при [tex]a=-\frac{1}{ 4}[/tex] също получаваме два цели корена?

[mkmarinov]

Had a technical mistake in my last post.
[tex]x_1=\frac{9}{2-k}, x_2=\frac{9}{2+k}[/tex].
Let [tex]2-k=\frac{p}{q}; (p,q)=1[/tex]
[tex]x_1=\frac{9q}{p}, x_2=\frac{9q}{4q-p}[/tex].
p divides 9, so [tex]p = \pm 1, \pm 3, \pm 9[/tex]
Let p=1: [tex]x_1=9q, x_2=\frac{9q}{4q-1}=2+\frac{q+2}{4q-1}[/tex]
But [tex]|4q-1| \ge |q+2|[/tex], so q=1.

Let p=-1: [tex]x_1=-9q, x_2=2+\frac{q-2}{4q+1}[/tex]
[tex]|4q+1| \ge |q-2| => q=-1[/tex]

Let p=3: [tex]x_1=3q; x_2=2+\frac{q+6}{4q-3}[/tex]
Let [tex]q+6=m(4q-3) => (4-m)q=6+3m => q=\frac{6+3m}{4-m}=-3+\frac{18}{4-m}[/tex], a few cases for m here.

And so on...

[ganka simeonova]

1) Може ли все пак да пишем и на бг?
2) Задачата може да се реши без никакви си к и m. Само на няколко редчета.

[mkmarinov]

1) Неуважително е към ОР.
2) Би ли пказала?

[ganka simeonova]

Оначе е неуважително и към всички останали участници, понеже решението е публично, а голяма част от тях са българи:)
Нямам нищо против решението на чужд език, но нека го даваме и на бг:)
Така, това решение, което ще дам не е мое, ако трябва да съм честна, а на един приятел:)
След прилагане на Виет, елиминиране на а и лека заигравка получаваме:
[tex](4x_1-9)(4x_2-9)=81[/tex]
Да разложим 81 на множители:
[tex]1. 81; (-1).(-81); 3.27; (-3)(-27); 9.9; (-9).(-9)[/tex]
И да приравним двата множителя с корените съответно на множителите в различните произведения.
Ясно е, че само 3. 27 и (-3).(-27) вършат работа

[mkmarinov]

Последното не го разбрах. Защо само 3.27 и (-3)(-27) да вършат работа?
Ако имаш предвид, че [tex]4x_1-9=3[/tex] и [tex]4x_2-9=27[/tex] (както и другия случай), не е така. Вземи например [tex]x_1=2; x_2=-18[/tex]. Тогава се получава (-1)(-81). Което впрочем е при а=-1/4 .

[ganka simeonova]

Последното не го разбрах. Защо само 3.27 и (-3)(-27) да вършат работа?
Ако имаш предвид, че [tex]4x_1-9=3[/tex] и [tex]4x_2-9=27[/tex] (както и другия случай), не е така. Вземи например [tex]x_1=2; x_2=-18[/tex]. Тогава се получава (-1)(-81). Което впрочем е при а=-1/4 .

Сори, да - при х=-1 и х-81 техническа грешка
Така или иначе стойностите на а са две