- Thread starter
- Admin
- #1
- Feb 14, 2012
- 3,816
The real numbers \(\displaystyle x\) and \(\displaystyle y\) satisfy \(\displaystyle x^3-3x^2+5x-17=0\) and \(\displaystyle y^3-3y^2+5y+11=0\). Determine the value of \(\displaystyle x+y\).
(Shakes his head.) Wow!Here's my solution
Let $x = u +1$ and $y = v + 1$ giving
$u^3+2u-14=0$ and $v^3+2v+14=0$.
Adding these two gives
$u^3+v^3+2u+2v = 0$ or $ (u+v)(u^2-uv+v^2+2) = 0$.
Since the second term is strictly positive then $u+v = 0$ so that $x + y = u+v+2 = 2$.
Thanks for participating, Jester and WOW!!!Here's my solution
Let $x = u +1$ and $y = v + 1$ giving
$u^3+2u-14=0$ and $v^3+2v+14=0$.
Adding these two gives
$u^3+v^3+2u+2v = 0$ or $ (u+v)(u^2-uv+v^2+2) = 0$.
Since the second term is strictly positive then $u+v = 0$ so that $x + y = u+v+2 = 2$.