1. ## ( IrMO- 93)

The real numbers α,β satisfy the equations :

α³ - 3α² +5α -17 = 0
β³ - 3β² +5β +11 = 0

Find α + β

Spoiler:
α + β = 2

2. ## Re: ( IrMO- 93)

Please show the effort made by you and where did you get stuck?

3. ## Re: ( IrMO- 93)

a³ + b³ -3a² -3b² +5(a+b) -6 = 0 .I can't go any further than this.

4. ## Re: ( IrMO- 93)

Originally Posted by chronoss
a³ + b³ -3a² -3b² +5(a+b) -6 = 0 .I can't go any further than this.
the fact that α³ - 3α² +5α -17 = 0 and β³ - 3β² +5β +11 = 0 doesn't mean α³ - 3α² +5α -17 = β³ - 3β² +5β +11 ;

you have to solve for a in α³ - 3α² +5α -17 = 0, and for b in β³ - 3β² +5β +11 = 0 first, aren't you?

dokrbb

5. ## Re: ( IrMO- 93)

I think you're wrong 0=0 ,but that isn't what i said , i said that 0 + 0= 0 , a³ + b³ -3a² -3b² + 5a +5b -6 = 0.

6. ## Re: ( IrMO- 93)

Originally Posted by chronoss
I think you're wrong 0=0 ,but that isn't what i said , i said that 0 + 0= 0 , a³ + b³ -3a² -3b² + 5a +5b -6 = 0.
I will give you a more simple example, in order to show you that I'm not wrong -

3x - 4 = 0 & 4y - 4 = 0, in the first one, the root is x = 3/4, while in the second one y = 4/4 = 1, does in this case 1 = 3/4, because from your assumption it does

the equations have each of them a root, with a different value - a, and b, good luck,

dokrbb

7. ## Re: ( IrMO- 93)

3x - 4 + 4y -4 = still remains 0 ; but I find the solution in another forum , here it is :

Let's see what happens if we add the two equations:

a³ + b³ - 3a² - 3b² + 5a + 5b² - 6 = 0

Moreover, since the equation (a + b)³ = a³ + 3a²b + 3ab² + b³ , we obtain : a³ + b³ = ( a + b)³ -3a²b - 3ab².

So the equation a³ + b³ - 3a² - 3b² + 5a + 5b - 6 = 0 , is equivalent to : (a + b)³ - 3a²b -3ab² -3a² -3b² +5a +5b -6 = 0 , or : (a + b)³ -3(a +b)² -3ab(a + b) + 5(a + b) - 6 + 6ab = 0.

Thus, the value of a + b corresponds to the root of the polynomial :

p(x) = x³ -3x² -3abx +5x - 6 + 6ab. => p(x) = x³ -3x² -3ab(x-2) + 5x -6 =
x³ - 2x² - x² -3ab(x -2) +5x -6 = x²(x - 2) + - 3ab(x -2) - x² + 5x -6 =
x²(x - 2) + - 3ab(x -2) - ( x² -2x -3x + 6) = x²(x - 2) + - 3ab(x -2) - (x[x - 2] -3[x-2].

Letting x -2 in evidence => p(x) = (x -2)(x² -3ab - [x-3]). p(2) = 0 concluding that a + b = 2.

I would appreciate if you had another solution.

8. ## Re: ( IrMO- 93)

You have done it well. I could not think of another method. Well Done

9. ## Re: ( IrMO- 93)

the credits are not mine, who solved this problem was the user santhiago of the forum ajudamatematica.com .I solved only the first part, I solved only the first part, but I decided to post the resolution to prove that the argument posted by the User dokrbb was incorrect.

10. ## Re: ( IrMO- 93)

Originally Posted by dokrbb
I will give you a more simple example, in order to show you that I'm not wrong -

3x - 4 = 0 & 4y - 4 = 0, in the first one, the root is x = 3/4, while in the second one y = 4/4 = 1, does in this case 1 = 3/4, because from your assumption it does

the equations have each of them a root, with a different value - a, and b, good luck,

dokrbb
I'm not sure why you concluded that his assumption was implying 1=3/4. It was not stated anywhere that $\displaystyle \alpha = \beta$. In fact, x =3/4 and y=1 agrees with 3x-4 = 4y-4, and 3x-4 - (4y-4) = 0. The caveat is that this is not true for every x, y, obviously. However, 3x-4 = 4y-4 has an infinite number of solutions.

11. ## Re: ( IrMO- 93)

Originally Posted by chronoss

I would appreciate if you had another solution.
I agree with @idbutt, I don't think there is an easier way around it. If either equation had rational roots, it would be an easier problem, but even the alpha equation has two complex roots to start with, so the solution you presented is the approachable, "student" friendly one.

12. ## Re: ( IrMO- 93)

thanks everyone