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There is only one real value of $'a'$ for which the quadratic equation $$ax^2 + (a+3)x + a-3 = 0$$ has two positive integral solutions.The product of these two solutions is :

Since the solutions are positive, therefore the product of roots and sum of root will be positive. This will give us two inequalities in $a$. Substitute the values of $a$ in the quadratic but I'm not getting my answer correct.

The two inequalities will be $$\frac{a-3}{a} > 0$$ and the other one will be $$\frac {a+3}{a} < 0$$

From the first one we get $a>3$ and from the second one we get $a< -3$.

I don't know how to proceed after this.

Kindly help.

Did
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Heisenberg
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  • Where did you use that both solutions must be integers? – Did Aug 25 '16 at 09:05
  • Additionally, you solved $\frac{a-3}{a} > 0$ incorrectly. – Did Aug 25 '16 at 09:06
  • @Did, he solved both incorrectly. OP, if you have $\frac{a}{b}>0$ there are two options - $a>0\text{ and }b>0$ or $a<0\text{ and }b<0$. Can you solve now? Also, don't forget that you want two real solutions, thus you have $\Delta>0$. – Galc127 Aug 25 '16 at 09:15
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    As @Did hinted you must verify the solution of your inequalities (for ie $a=-6$ they are both true), then you must have $a-3$ and $a+3$ both divisibile by $a$. – N74 Aug 25 '16 at 09:18
  • @Galc127 Indeed. – Did Aug 25 '16 at 09:18
  • @N74, for $a=-6$ the second one doesn't hold. Actually, the solution for inequalities is $-3<a<0$. Also, why must $a|a-3\wedge a|a+3$? – Galc127 Aug 25 '16 at 09:25
  • @Galc Right for the $a$ value... To reply to you question, if the solutions are integer, so must be its sum and product. – N74 Aug 25 '16 at 09:30
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    @N74. Unfortunately, your conditions on divisibility may hold not only for integer $a$, so we may not consider a small number of $a$ manually. Also do this conditions guarantee the solutions (not their product and sum) being integer? – rtmd Aug 25 '16 at 09:31
  • @N74, as suggested by rtmd, $a$ doesn't have to an integer (also, it is easy to check that $a=-1$ and $a=-2$ give no integer solutions). Using the fact that $\Delta>0$ we can write that $a\in (1-2\sqrt{2},0)$, but not more than that. I don't know how one should find the exact value of a. – Galc127 Aug 25 '16 at 09:35
  • @Galc127 I also obtained $\Delta=(a+3)^2-4(a-3)a=-3a^2+18a+9$. The condition $\Delta>0$ gave me a diffirent inteval: $a\in(3-2\sqrt{3};0)$. Though I do not know yet how to move further. – rtmd Aug 25 '16 at 09:46
  • One inequality is incorrect. – jimpix Aug 25 '16 at 09:46
  • @rtmd, you are correct, I had a mistake. – Galc127 Aug 25 '16 at 09:54
  • Actually, you solved the first inequality incorrectly.

    Sum of roots = $\frac{-b}{a} = \frac{-(a + 3)}{a} = \frac{-a-3}{a}$. According to the condition given in the problem, $\frac{-a-3}{a} > 0$ $\therefore a < -3$.

    Product of roots = $\frac{c}{a} = \frac{a + 3}{a}$. According to the condition given in the problem $\frac{a + 3}{a} > 0 :\therefore a > -3$ I don't get it.

     – jimpix Aug 25 '16 at 09:54
  • @jimpix, no condition guarantees that $a>0$. Also, please notice that $\frac{-a-3}{a}>0$ is the same as $\frac{a+3}{a}<0$. Inequalities are written correctly. – Galc127 Aug 25 '16 at 10:05
  • @Galc127 The only possible solution is at $a=-\frac{3}{7}$. I'll try to write the solution now. – rtmd Aug 25 '16 at 10:24
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    @Galc127 My solution is not necessary, see solution of iamvegan below. – rtmd Aug 25 '16 at 10:29

2 Answers2

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We need $$\frac{a+3}{a}\in\Bbb{Z}\ , \frac{a-3}{a}\in\Bbb{Z}$$ or $$1+\frac{3}{a}\in\Bbb{Z}\ , \ 1-\frac{3}{a}\in\Bbb{Z}$$ thus $\displaystyle \frac{3}{a}\in\Bbb{Z}$, means that $\displaystyle a=\frac{3}{m}$ where $m\in\Bbb{Z}$.

Now we can write the equation as $$\frac{3}{m}x^2+\left(\frac{3}{m}+3\right)x+\frac{3}{m}-3=0\implies x^2+(m+1)x+1-m=0$$ Using quadratic formula we have $$x_{1,2}=\frac{-m-1\pm\sqrt{m^2+6m-3}}{2}=\frac{-m-1\pm\sqrt{(m+3)^2-12}}{2}$$We need the expression under the square root to be an integer. There are only two squares with difference $12$ (those are $4,16$), hence we want $(m+3)^2=16\implies m=1\text{ or }m=-7$, thus $$a=3 \text{ or } a=-\frac{3}{7}$$ and as $a<0$ we have

$$a=-\frac{3}{7}$$

Did
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Galc127
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1

Thanks to @rtmd, I fixed the solution. This only holds when $a=-\frac{3}{7}$. Here is the proof.

$\frac{a-3}{a}, \frac{a+3}{a} \in \mathbb{Z}$ implies $\frac{6}{a} \in \mathbb{Z}$. We can write $a=\frac{6}{n}$ for some integer $n$. \begin{align*} \frac{a-3}{a}= 1-\frac{n}{2} \in \mathbb{Z}\\ \frac{a+3}{a}= 1+\frac{n}{2} \in \mathbb{Z} \end{align*} Therefore $n$ is even. Assume $p,q$ are the positive integer solutions of the given quadratic equation. \begin{align*} pq= 1-\frac{n}{2} \in \mathbb{Z}\\ p+q= -1-\frac{n}{2} \in \mathbb{Z} \end{align*} This gives us \begin{align*} pq-p-q&=2\\ (p-1)(q-1)&=3 \end{align*} Therefore WLOG $p=4,q=2$. This only holds when $n=-14$ and $a=-\frac{3}{7}$

iamvegan
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