Find α+β+αβ: 7 "Solving for α+β+αβ in Arithmetic Progression

[donaldparida]

Homework Statement

P(x) =ax2+bx+c where a, b and c are in arithmetic progression and are positive. α and β are the roots of the equation and are integers. Find the value of α+β+αβ. (Answer is 7)

Homework Equations

x = {−b ± √(b2 − 4ac)} /2a
3. The Attempt at a Solution [/B]
Since a, b and c are in arithmetic progression, b-a=c-b.
α+β+αβ = -b/a+c/a = c-b/a = b-a/a

 

[scottdave]

You know that α and β are roots, so k*(x - α)(x - β) is equal to the polynomial. Expand this out with FOIL k*(x² - (α+β)x + αβ), set this equal to ax² + bx + c. Since they are roots, you do not know what the coefficient of x squared is, yet. So you can represent that k*(x² - (α+β)x + αβ) = ax² + bx + c. Now set the coefficients of the x² terms equal, the x terms, and the constant terms. This should give you a system of equations (along with your knowledge of the arithmetic progression). See if that gets you the answer.

 

[donaldparida]

@scottdave, What is k?

 

[scottdave]

k is a constant which makes the two polynomials equal. You can have several distinct polynomials with the same roots. The difference is a proportionality constant. You will find that k is equal to a.

 

[donaldparida]

@scottdave Do you mean that equations which are related by a multiplicative factor have equal roots?

 

[donaldparida]

Sorry, but i am not getting the answer. I am getting some weird equation.

 

[mfb]

α+β+αβ = -b/a+c/a = c-b/a = b-a/a

Add brackets around the numerator and simplify.

You can also use the condition that the roots are integers.

 

[donaldparida]

@mfb, i cannot understand how to simplify it further.

 

[mfb]

$$\frac{b-a}{a} = \frac b a - \frac a a$$
And the second term can be simplified.

Afterwards you can have a look at the condition that α and β are integers.

 

[donaldparida]

@mfb, I thought you meant some other simplification. I could do that simplification by myself but i cannot simplify it further. I am getting -(α+β )-1=α+β+αβ. I cannot recall any related property of integer which would help me simplify it further.

 

[mfb]

α+β+αβ = b/a - 1. Keep that result for now.

α and β are integers. What does that imply for α+β+αβ?
What does it imply if you look at the formula for α (or β)?

 

[donaldparida]

I found out that if α or β are integers, then [{−b ± √(b² − 4ac)} /2a] must also be integers. ('s' in 'integers' is due to the ± sign).

For them to be integers {−b ± √(b² − 4ac)} and 2a must also be integers and {−b ± √(b2 − 4ac)} must be a multiple of 2a.

For {−b ± √(b² − 4ac)} to be integer and an multiple of 2a, -b and √(b² − 4ac) should be integers and should leave the same remainder on dividing by 2a.
For -b to be an integer b should be an integer (positive, given in the question) and for √(b² − 4ac) to be an integer (b² − 4ac) has to be a positive integer as well as a perfect square.

The most important conclusion is (b² − 4ac) is a perfect square.

 

[mfb]

I found out that if α or β are integers, then [{−b ± √(b² − 4ac)} /2a] must also be integers. ('s' in 'integers' is due to the ± sign).

That directly follows from the definition, yes.

For them to be integers {−b ± √(b² − 4ac)} and 2a must also be integers

Why? You can show that, but it is not trivial.

and {−b ± √(b2 − 4ac)} must be a multiple of 2a.

Sure.

The most important conclusion is (b² − 4ac) is a perfect square.

Right so far.

 

[donaldparida]

α and β are integers. What does that imply for α+β+αβ?
What does it imply if you look at the formula for α (or β)?

(i)If α and β are integers then α+β+αβ is also an integer.
(ii) b²>=4ac
=>(b/a)(b/c)>=4
=>-(α+β)(b/c)>=4
=>-(α+β){(b*a)/(c*a)}>=4
=>-(α+β)*[{-(α+β)}/αβ]>=4
=>(α+β)²/αβ>=4
=>(α+β)²/>=4αβ
=>α²+β²+2αβ-4αβ>=0
=>(α-β)²>=0
=>α>=β

 

[mfb]

=>(α-β)²>=0

A squared real number is always >=0, there is no need to derive this over 9 lines.

=>α>=β

That does not follow from the previous line.

(i)If α and β are integers then α+β+αβ is also an integer.

Correct. And you know that α+β+αβ = b/a - 1. That tells you if b/a has to be an integer.

If b/a is an integer, you can rewrite your arithmetic progression and express c in terms of a and b. That should lead to some interesting results in the square root - which needs a perfect square in it.

 

[donaldparida]

Did you mean this
b-a=c-b
=>b-a+b=c
=>2b-a=c
=>2ak-a=c
=>a(2k-1)=c
or this
b-a=ak-b
=>2b-a-ak=0
=>2b-a(1+k)=0

 

[Chestermiller]

If a, b, and c are the coefficients of a quadratic equation, what is the sum of the roots and what is the product of the roots?

 

[donaldparida]

@Chestermiller, sum of roots=-b/a and product of roots=c/a.

 

[Chestermiller]

@Chestermiller, sum of roots=-b/a and product of roots=c/a.

I'm not sure you used this, so $$\alpha + \beta +\alpha\beta=\frac{c-b}{a}$$Certainly, this isn't generally equal to 7, even for an arithmatic progression.

 

[mfb]

Certainly, this isn't generally equal to 7, even for an arithmatic progression.

We also know $\alpha$ and $\beta$ are integers.
The problem is solvable, but it needs a few steps to get there.

Did you mean this
b-a=c-b
=>b-a+b=c
=>2b-a=c
=>2ak-a=c
=>a(2k-1)=c
or this
b-a=ak-b
=>2b-a-ak=0
=>2b-a(1+k)=0

Both will work, but the first definition of k is much more intuitive.

 

[donaldparida]

I am unable to understand how to proceed. Should i solve for k and then use it's value for determining a,b and c and thus α+β+αβ?

 

[donaldparida]

Ok. I think i got it.
Since α, β are integers, α+β+αβ is also an integer.
Since α+β+αβ is an integer, (b/a-1) is also an integer.
Since (b/a-1) is an integer, b/a is also an integer.
Since b/a is an integer, b is a multiple of a.
Therefore b=ak
Since a, b and c are in A.P. in the same order, b-a=c-b
=>2b-a=c
=>2ak-a=c
=>a(2k-1)=c
Therefore, P(x)=ax²+bx+c=ax²+akx+a(2k-1)
=>ax²+akx+a(2k-1)=0
Dividing both sides by a, we get,
x²+kx+(2k-1)=0
Since this equation is the same as the original one, it also has integer roots and since it has integer roots, the discriminant is a perfect square.
Therefore, k²-4.1.(2k-1)=q² (say)
=>k²-8k+4=q²
=>(k-4)²-12=q²
=>(k-4)²-q²=12
=>(k-4+q)(k-4-q)=12
=>k-4+q=6 or k-4+q=4 ---(1)
and
k-4-q=2 or k-4-q=3 ---(2)
Adding equations(1) and (2) we get,
2k-8=8 or 2k-8=7
=>k=8 or k=15/2
For k=8, b=8a, c=15a and (c-b)/a=(15a-8a)/a=7a/a =7.
Therefore α+β+αβ=7.

Why did we get k=15/2?(The value of k for which an invalid value for α+β+αβ is obtained)

 

[mfb]

Looks great!
Just some details:

Since b/a is an integer, a and b are both integers

4.2/2.1 = 2 is an integer, but 4.2 and 2.1 are not. You don't need it to write b=ak with an integer k.

Why did we get k=15/2?

It leads to 12 (with a non-integer q), but it violates the earlier condition that k has to be an integer.

 

[donaldparida]

4.2 and 2.1 are 8 and 2 respectively which are integers. Aren't they?

 

[mfb]

That is a decimal dot, not a multiplication sign ("*").