The real numbers α,β satisfy the equations :

α³ - 3α² +5α -17 = 0

β³ - 3β² +5β +11 = 0

Find α + β

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- Apr 18th 2013, 10:01 AMchronoss( IrMO- 93)
The real numbers α,β satisfy the equations :

α³ - 3α² +5α -17 = 0

β³ - 3β² +5β +11 = 0

Find α + β

__Spoiler__: - Apr 18th 2013, 08:15 PMibduttRe: ( IrMO- 93)
Please show the effort made by you and where did you get stuck?

- Apr 19th 2013, 06:39 AMchronossRe: ( IrMO- 93)
a³ + b³ -3a² -3b² +5(a+b) -6 = 0 .I can't go any further than this.

- Apr 19th 2013, 06:43 AMdokrbbRe: ( IrMO- 93)
- Apr 19th 2013, 07:05 AMchronossRe: ( 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.

- Apr 19th 2013, 07:21 AMdokrbbRe: ( IrMO- 93)
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 - Apr 19th 2013, 08:43 AMchronossRe: ( 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. - Apr 19th 2013, 08:09 PMibduttRe: ( IrMO- 93)
You have done it well. I could not think of another method. Well Done

- Apr 20th 2013, 05:32 AMchronossRe: ( 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.

- Apr 20th 2013, 06:50 AMmajaminRe: ( IrMO- 93)
I'm not sure why you concluded that his assumption was implying 1=3/4. It was not stated anywhere that . 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.

- Apr 20th 2013, 06:54 AMmajaminRe: ( IrMO- 93)
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.

- Apr 20th 2013, 08:08 AMchronossRe: ( IrMO- 93)
thanks everyone