If coefficients of the Quadratic Equation are in AP find $\alpha+\beta +\alpha\beta$.

Question

$ax^2+bx+c=0$ is a quadratic equation with integer roots $\alpha,\beta$ and the positive coefficients are in AP.Find $\alpha+\beta +\alpha\beta$.

My attempt:

$\alpha+\beta +\alpha\beta=\frac {-b}{a}+\frac {c}{a}$ and we know $\frac{a+c}{2}=b$.

But I cannot eliminate the $a,b,c$ terms to get the answer $7$ given.

How to do it?Any quick,short,simple solutions?

Answer 1

First of all we need to that if the roots are integer then $S=\alpha+\beta+\alpha\cdot \beta$ is also integer then

$$S=\frac{c-b}{a}=\frac{b-a}{a}=\frac{b}{a}-1\to a|b$$

and once $2b=a+c$ we get that $a|c$ so we can write the terms $(a,b,c)$ like $(a,a(1+r),a(1+2r))$ and then

$$S=r$$

We also have the discriminant:

$$\Delta=b^2-4ac=a^2(1+r)^2-4a\cdot a(1+2r)=a^2[(1+r)^2-4(1+2r)]$$

and once the roots are integer then $\Delta$ should be a square and then

$$(1+r)^2-4(1+2r)=k^2\to r^2-6r-(3+k^2)=0\quad (1)$$

and looking this new quadratic equation (in $r$) we have the discriminant

$$\Delta '=36-4[-(3+k^2)]=4(12+k^2)$$

and $\Delta'$ has also to be an integer, then:

$$12+k^2=p^2\to (p-k)(p+k)=12$$

spliting $12$ as a product of two numbers we get the solutions $(k,p)\in\{(2,4),(-2,-4),(2,-4),(-2,4)\}$. So, $k=\pm 2\to k^2=4$ and then backing to $(1)$ we get $r=7$ or $r=-1$.

For $r=-1$ we get the sequence $(a,0,a)$ and the equation $ax^2+a=a(x^2+1)$ what not give us real solution.

For $r=7$ we get the equation $ax^2+8ax+15a=0$ what give us the roots $-3$ and $-5$, then

$$S=r=7$$

Answer 2

If the given quadratic has integer roots $\alpha,\beta$, then $$\begin{align} ax^2+bx+c &=a\left(x^2+\frac bax+\frac ca\right)\\ &=a(x-\alpha)(x-\beta)\\ &=a[x^2-(\alpha+\beta)x+\alpha\beta]=0\\ \begin{cases} \frac ba=-(\alpha+\beta)\\\frac ca=\alpha\beta\end{cases}\end{align}$$

Given that $a,b,c$ are in AP, then $1, \frac ba, \frac ca$ are also in AP.

The problem reduces to finding $\alpha+\beta+\alpha\beta$ where $1, -(\alpha+\beta), \alpha\beta$ are positive integers in AP.

For the three numbers to be positive, $\alpha, \beta<0 \; (\alpha, \beta\in\mathbb Z)$.

For them to be in AP,

$$\begin{align} -2(\alpha+\beta)&=1+\alpha\beta\\ \beta&=-2+\frac 3{\alpha+2} \end{align}$$

Since $\beta$ is an integer, $\frac 3{\alpha+2}$must be an integer.

As $\alpha,\beta <0$ and $\alpha, \beta\in\mathbb Z$, possible values for $\alpha, \beta$ are $(-3,-5), (-5,-3)$.

Hence $$\alpha+\beta+\alpha\beta=\color{red}7$$