Consider the equality $x^2 + ax + by = z$ whereby $a, b$ and $z$ are constants and one should find values for $x$ and $y$ (that will involve z) s.t. the equality holds. How can I find such solutions for $x$ and $y$ that depend on $z, a, b$? Also, $x, y, z> 0$. It would be interesting to find all solutions but finding one solution would also be enough for me.
Asked
Active
Viewed 38 times
0
-
Solve $x$ in terms of $y$ (or vice versa) – fleablood Apr 12 '21 at 19:13
-
@fleablood thanks, that answered my question, how can I close the question? – Jacob Apr 12 '21 at 19:15
-
Did you mean $x^2 + ax + by = z$ or did you mean $x^2 + ax + by^2 = z$. If $b\ne 0$ then $y = \frac {-z -x^2 -ax}b$. Just set $x$ to anything and $y$ follows. But if $b=0$ then $y$ can be absolutely anything and $x$ are solutions to $x^2 + ax = z$. – fleablood Apr 12 '21 at 19:18
1 Answers
0
If $b=0$, you obtain $x^2+ax-z=0$. You may solve this using the quadratic formula. You will find $x=\frac{-a+\sqrt{a^2+4z}}{2}$ (the negative branch will give you a negative $x$), and $y$ may be arbitrary.
Next, consider the case $b>0$. You find $y=\frac{-x^2-ax+z}{b}$. As you want the numerator to be positive, you can argue that $0\leq x\leq \frac{-a+\sqrt{a^2+4z}}{2}$.
Lastly, you may consider the case $b<0$. You will find the same $y$ as in the previous case, but this time $x\geq \frac{-a+\sqrt{a^2+4z}}{2}$.
Zuy
- 4,656