POTW Solve 3-Variable Equation with Real Solutions | √(z-y^2-6x-26)+x^2+6y+z-8=0

  • Thread starter Thread starter anemone
  • Start date Start date
  • Tags Tags
    Variables
AI Thread Summary
To solve the equation √(z - y² - 6x - 26) + x² + 6y + z - 8 = 0, it is suggested to manipulate the equation to isolate the square root term. Keeping the square root intact may simplify finding real solutions. The discussion emphasizes the importance of rearranging terms effectively to facilitate the solution process. Participants are encouraged to explore various methods to derive the real solutions for the equation. Ultimately, the goal is to find values of x, y, and z that satisfy the equation.
anemone
Gold Member
MHB
POTW Director
Messages
3,851
Reaction score
115
Find the real solutions for ##\sqrt{z-y^2-6x-26}+x^2+6y+z-8=0##.
 
  • Like
Likes bob012345 and topsquark
Physics news on Phys.org
One way to go about it is by moving some terms around to get rid of the square root.
 
It's actually better to keep the square root.
 
Last edited:
anemone said:
Find the real solutions for ##\sqrt{z-y^2-6x-26}+x^2+6y+z-8=0##.
Here is my solution;

Given $$\sqrt{z-y^2-6x-26}+x^2+6y+z-8=0$$ we can write
$$z = 8 - x^2 -6y - \sqrt{z-y^2-6x-26}$$

we see that for real solutions $$z-y^2-6x-26 \ge 0$$ or $$z \ge y^2+6x+26$$

thus $$z = 8 - x^2 -6y - \sqrt{z-y^2-6x-26} \ge y^2+6x+26$$ or

$$- \sqrt{z-y^2-6x-26} \ge y^2+6x+26 +x^2 + 6y -8$$ the RHS can be condensed to

$$x^2 + y^2 + 6(x+y) +18 = 0$$ which can be written as

$$(x+3)^2 + (y+3)^2 =0$$ finally giving

$$- \sqrt{z-y^2-6x-26} \ge (x+3)^2 + (y+3)^2$$

This has only one real root ##(x,y)= (-3,-3)## when the LHS is zero then solving for ##z## gives ##z=17##
so ##(x,y,z) →(-3,-3,17)##
[\spoiler]
 
Back
Top