Find the value of $\alpha + \beta + \alpha\beta$ from the given data

Question

Suppose the quadratic polynomial $P(x) = ax^2 + bx +c $ has positive coefficients $ a, b, c $ in an arithmetic progression in that order. If $P(x) = 0$ has integer roots $\alpha $ and $\beta $, then $\alpha + \beta + \alpha\beta $ equals?

1) $3$

2) $5$

3) $7$

4) $14$

It's a single choice correct problem. All genuine answers are welcome :)

The question is from a famous Indian Scholarship test 'KVPY' for high school students. The official website doesn't provide any solutions to it that is why I am asking for help.

My go on the question

we know that in a quadratic equation $ax^2 + bx +c$ with roots $\alpha, \beta$ the sum of roots is $\alpha +\beta = \frac{(-b)}{a} $ and product is $\alpha\beta = \frac{c}{a} $

So $\alpha + \beta + \alpha\beta = \frac{(-b+c)}{a}$ and since $a,b,c$ are in arithmetic progression $\frac{(a+c)}{2} = b$

Something that I tried was also this -

$ a, b, c $ are in AP (arithmetic progression)

$a-b , 0, c-b$ are in AP (subtracting by $b$)

$1-\frac{b}{a}, 0, \frac{(c-b)}{a}$ are in AP (dividing by $a$)

Now I can't proceed further. After a bit of plugging and chugging I still cant get an integer as an answer...

Answer 1

Any quadratic (or for that matter polynomial) can be written in terms of its roots:

$$P(x)=a(x-\alpha)(x-\beta)=ax^2-a(\alpha+\beta)x+a\alpha\beta$$

So $\alpha+\beta+\alpha\beta=-\frac{b}{a}+\frac{c}{a}=\frac{-b+c}{a}$

We also know $a,b,c$ form an AP so lets rewritten them as $a,a+d,a+2d$ where $d$ is the common difference.

So $\alpha+\beta+\alpha\beta=\frac{-(a+d)+a+2d}{a}=\frac{d}{a}$

As the roots are integers then $d$ is a multiple of $a$ and hence so is $b$ and $c$. Without loss of generality we can let $a=1$.

Also we need $b^2-4ac=n^2,n\in\mathbb{Z}$

Subbing in $a=1$ and $d$ into this gives:

$$(1+d)^2-4(1+2d)=n^2$$

$$1+2d+d^2-4-8d=n^2$$

$$d^2-6d-3=n^2$$

$$d^2-6d-9+12=n^2$$

$$(d-3)^2+12=n^2$$

$$(d-3)^2-n^2=12$$

$$(d-3+n)(d-3-n)=12$$

Let the factors of $12$ be $\left(f,\frac{12}{f}\right)$.

So $d-3+n=f$ and $d-3-n=\frac{12}{f}$ hence $2(d-3)=f+\frac{12}{f}$

So $d=\frac{f}{2}+\frac{6}{f}+3$

Possible values of $f$ can only be $2$,$6$ as $d$ is a positive integer.

Both leads to $d=1+3+3=7$ and $d=3+1+3=7$.

Hence $\alpha+\beta+\alpha\beta=\frac{-(a+d)+a+2d}{a}=\frac{d}{a}=7$

Answer 2

Let $d$ denote the common increment of an Arithmetic Progression .

Now $\alpha + \beta + \alpha\beta = \frac{d}{a}$ , now put the values for $\frac{d}{a}$ from the options given , the first two options will contradict since the roots will be complex but as per the question they have to be integer . so the integer roots are satisfied by the option (c) .

hope this helps.

Answer 3

A roadmap: Without loss of generality we can assume that $a=1$. Then

• $c=2b-1$ (why?)
• $b$ is an integer (why?)
• The discriminant is $D=(b-4)^2-12$ (why?)
• Both $(b-4)^2$ and $(b-4)^2-12$ must be squares (why?). Can you name two squares of integers that differ from each other by twelve?
• Deduce the possible values of $b$, solve for $\alpha$ and $\beta$. Check your list of alternatives and/or the remaining unused assumption.