2
$\begingroup$

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...

  • 3
    We would appreciate it if you could give some context. Where did you see this question? Care to share your own thoughts? Users generally frown upon multiple choice questions because they smell of homework. It is important that you work a bit to dispel such thoughts. – 2017-01-14
  • 1
    See, already somebody (not me) felt the need to downvote this. Fix that! Quickly! It is an important part of the site culture :-) I hope you spent a few hours roaming, familiarizing yourself with the local do's and do-not's. – 2017-01-14
  • 3
    @YourAverageEuler Excellent edit. – 2017-01-14
  • 1
    Good job editing! Dividing the equation by $a$ won't change the AP property, so we are allowed to assume that $a=1$ when $c=2b-1$. Will quadratic formula help at all? You need both $b$ and the square root of the discriminant to be integers... – 2017-01-14
  • 0
    Exact duplicate: http://math.stackexchange.com/questions/2068329/a-p-terms-in-a-quadratic-equation – 2017-01-15
  • 3
    Possible duplicate of [A.P. terms in a Quadratic equation.](https://math.stackexchange.com/questions/2068329/a-p-terms-in-a-quadratic-equation) – 2018-08-01

3 Answers 3

1

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.

  • 1
    Is it possible to deduce the answer without having the list of answers to choose between? – 2017-01-14
  • 0
    @IanMiller exactly what I wanted to say – 2017-01-14
  • 0
    @Ian Miller $\alpha + \beta + \alpha\beta$ is an integer , and hence so is $\frac{d}{a}$ , now $\frac{d}{a}$ can be written as $\frac{b}{a} - 1$ ,implying $b$ must be a multiple of $a$ . then can we proceed? – 2017-01-14
  • 0
    There are (up to scalar multiple) two arithmetic sequences leading to integer roots. But one of the assumptions, unused by BAYMAX, eliminates the other :-) – 2017-01-14
4

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.
  • 0
    Can you answer your 'whys'? Not all are obvious. – 2017-01-14
  • 0
    If a is an integer, is it necessary that $b$ is an integer? – 2017-01-14
  • 1
    @YourAverageEuler You already listed the equation $\alpha+\beta=-b/a$, so... – 2017-01-14
  • 0
    Oh, thanks for pointing out @JyrkiLahtonen – 2017-01-14
  • 0
    Good job! Keep working! – 2017-01-14
  • 0
    Square of 2 and square of 4 differ by 12 – 2017-01-14
  • 0
    Correct again! So $(b-4)^2$ is... – 2017-01-14
  • 0
    Is it 4 squared? – 2017-01-14
  • 0
    Correct, (and also $(-4)^2$). So the alternatives for $b$ are... – 2017-01-14
  • 0
    -4 is not feasible as it yields $b=0$ and if $(b-4)^2 = 4^2$, $b=8$ then if $a,b,c$ are in AP, $c = 15$ , then the required result is 7 ! – 2017-01-14
  • 0
    Good job! You reached the finish line. – 2017-01-14
  • 0
    Thank you @JyrkiLahtonen can you just give me a hint on how you came up with the fourth bulleted point in your answer – 2017-01-14
  • 0
    Experience. You can simply try and find a value of $b$ such that $(b-4)^2-12$ is a square. Testing $b=1,b=2,\ldots$ will get you there also soon enough. That was a bit of shortcut that my experience with problems like this suggested. – 2017-01-14
  • 0
    That ends it , thank you @JyrkiLahtonen – 2017-01-14
  • 0
    You are welcome @YourAverageEuler Good luck in preparing for the exam! – 2017-01-14
  • 0
    @JyrkiLahtonen already gave it and screwed it :( now just fortifying my algebra and avoiding mistakes – 2017-01-14
3

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

  • 0
    This is extremely vivid! Thank you @IanMiller – 2017-01-14
  • 0
    For it's great clarity I will have to make this the best answer – 2017-01-14