The real numbers α,β satisfy the equations :
α³ - 3α² +5α -17 = 0
β³ - 3β² +5β +11 = 0
Find α + β
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,
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.
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.