# If x and y are the real roots of x³ - 3x² + 5x -17 = 0, and y³ - 3y² + 5y + 11 = 0., find x + y.?

### 3 Answers

- atsuoLv 63 years agoFavourite answer
Let x = u + 1 .

Given equation x^3 - 3x^2 + 5x - 17 = 0 becomes

(u + 1)^3 - 3(u + 1)^2 + 5(u + 1) - 17 = 0

u^3 + 2u - 14 = 0 ---(#1)

This equation has only one real root .

Let y = v + 1 .

Another given equation y^3 - 3y^2 + 5y + 11 = 0 becomes

(v + 1)^3 - 3(v + 1)^2 + 5(v + 1) + 11 = 0

v^3 + 2v + 14 = 0 ---(#2)

This equation has only one real root , too .

Multiply both sides of (#2) by -1 ,

-v^3 - 2v - 14 = 0

(-v)^3 + 2(-v) - 14 = 0 ---(#3)

This is the same equation as (#1) , so if w is a real root of (#1) then -w is a real root of (#2) .

That is , x = w + 1 and y = -w + 1 are real roots of two given equations .

Therefore , x + y = (w + 1) + (-w + 1) = 2 .

- PinkgreenLv 73 years ago
x^3-3x^2+5x-17=0 has one real root & 2 non real roots which are

x=3.134885

x=-0.06744242+/-2.327724 i

y^3-3y^2+5y+11=0 has one real root & 2 non real roots too which are

y=-1.134885

y=2.067442+/-2.327724 i

approximately.

=>

x+y=3.134885-1.134885=2

- icemanLv 73 years ago
x³ - 3x² + 5x -17 + y³ - 3y² + 5y + 11 = 0

(x + y - 2) (x² - xy - x + y² - y + 3) = 0

x + y = 2