MHB Finding the Sum of Real Numbers Satisfying Cubic Equations

Click For Summary
The discussion focuses on finding the sum of real numbers x and y that satisfy the cubic equations x^3 - 3x^2 + 5x - 17 = 0 and y^3 - 3y^2 + 5y + 11 = 0. Participants share their solutions and approaches, with MarkFL receiving praise for a quick and accurate resolution. Other contributors express admiration for the problem-solving methods presented. The overall tone is supportive, highlighting the collaborative effort in tackling the mathematical challenge. The goal is to determine the value of x + y.
anemone
Gold Member
MHB
POTW Director
Messages
3,851
Reaction score
115
The real numbers $$x$$ and $$y$$ satisfy $$x^3-3x^2+5x-17=0$$ and $$y^3-3y^2+5y+11=0$$. Determine the value of $$x+y$$.
 
Mathematics news on Phys.org
Let $$z=x+y$$ and so the first equation becomes:

$$(z-y)^3-3(z-y)^2+5(z-y)-17=0$$

Which becomes:

$$z^3-3yz^2+3y^2z-y^3-3z^2+6yz-3y^2-5y+5z-17=0$$

$$z^3-3yz^2+3y^2z-3z^2+6yz-3y^2+5z-6-\left(y^3+5y+11 \right)=0$$

Using the second equation, this becomes:

(1) $$z^3-3yz^2+3y^2z-3z^2+6yz-6y^2+5z-6=0$$

Now, the second equation may be written:

$$(z-x)^3-3(z-x)^2+5(z-x)+11=0$$

Which becomes:

$$z^3-3xz^2+3x^2z-x^3-3z^2+6xz-3x^2-5x+5z+11=0$$

$$z^3-3xz^2+3x^2z-3z^2+6xz-3x^2+5z-6-\left(x^3+5x-17 \right)=0$$

Using the first equation, this becomes:

(2) $$z^3-3xz^2+3x^2z-3z^2+6xz-6x^2+5z-6=0$$

Adding (1) and (2), and simplifying, we have:

$$(z-2)\left(3\left(x^2+y^2 \right)-\left(z^2+2z-6 \right) \right)=0$$

Hence:

$$z=x+y=2$$
 
Well done MarkFL for submitting a complete and correct solution in such a short period of time!(Clapping)

I just love your approach so so much!(Inlove)
 
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 $x + y = u+v+2 = 2$.
 
Jester said:
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 $x + y = u+v+2 = 2$.
(Shakes his head.) Wow! (Bow)

-Dan
 
Jester said:
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!(Clapping) This is surely another impressive and great way to tackle this problem!(Nerd)
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

Similar threads

Replies
4
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 15 ·
Replies
15
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K