$a+b+c = 13$; if $b/a=c/b$, find the maximum and minimum values of $a$ and the corresponding $b$ and $c$

Question

Question :

The sum of $3$ integers $a,b$ and $c$ is $13$. If $\dfrac{b}{a}=\dfrac{c}{b}$, find the maximum and minimum values of $a$ and the corresponding $b$ and $c$.

To tackle this problem I let $x=\dfrac{b}{a}=\dfrac{c}{b}$ because I wanted to create a quadratic equation in order to use the discriminant theorem. From the equation above I can deduce that $b=ax$ and $c=ax^2$. Because $a+b+c=13$.

Therefore; $$a+ax+ax^2=13$$ $$\implies 1+x+x^2-\frac{13}{a} = 0 $$ (where $a \ne 0$, $b \ne 0$, $c \ne 0$)

I can only work up to here. I do not know how to use the discriminant theorem to work out the maximum and minimum of $a$, $b$ and $c$.

Answer 1

∵ $x=b/a$ is a solution of (1)

∴ (1) has at least one solution.

$△=1-4(1-(13/9))≥0$

$52/a≥3...(2)$

From (2) $a>0$ and $52/3≥a$

$∴0<a≤52/3$

$1<a≤17$

Minimum of $a=1$ and maximum of $a =17$.

Sub these values of $a$ into (1) to find the corresponding maximum and minimum of b and c.

Answer 2

So if a, b and c are integers (as per latest edit to question) then x is real and rational. And if the solution to $$ax^2 + ax + (a-13)=0$$ is real and rational then its discriminant must be a perfect square. So you are looking for values of a that make $$a^2 - 4a(a-13)$$ a perfect square. As shown elsewhere, a must be between 0 and 17, so there are a finite number of possibilities to check. And you can remove any solution that has a=0 or b=0. Don't forget that for each candidate value for a that you find there are 2 possible values for x, and so 2 possible values for b and c.