Symmetric system of $3$ equations: $x+y+z=23$, $xy+yz+xz=144$, $xyz=252$

Question

I need help solving the following system for all real ordered triples $(x,y,z)$ without guessing and checking:

$$x+y+z=23$$

$$xy+yz+xz=144$$

$$xyz=252$$

Preferably, the solution should use methods taught in high school.

Thanks.

Answer 1

By Vieta's formula, $x,y,z$ are the roots of the following cubic equation: $$x^3-23x^2+144x-252=0$$ Now, use the cubic formula to solve.

You could also try guessing and checking for integer roots of $x$. Using the rational roots theorem, roots must be factors of $252$. Then, once you find a root, use synthetic division to get a quadratic and then use the quadratic formuka.

Answer 2

The second equation is,

$$y(x+z)+xz=144$$

From the first equation, $x+z=23-y$ From the third equation, $xz=\frac{252}{y}$

So we have:

$$y(23-y)+\frac{252}{y}=144$$

$$-y^3+23y^2-144y+252=0$$

$$y^3-23y^2+144y-252=0$$

Which can be solved by the cubic formula. And perhaps the rational root theorem and checking roots, or graphing and checking is simple way to go if you are willing to accept checking only a few amount of hypothesized answers.

$y=3$

Synthetic division and factoring gives:

$y=6$

$y=14$

Furthermore, by the symmetry of your equations we have:

$$x^3-23x^2+144x-252=0$$

$$z^3-23z^2+144z-252=0$$

Answer 3

We have three equations

$$x + y + z = 23$$

$$xy + yz + xz = 144$$

$$xyz = 252$$

Multiplying the 2nd equation by $z$, we obtain

$$\underbrace{xyz}_{=252} + yz^2 + xz^2 = 144z$$

which can be rewritten as

$$252 + (x + y) z^2 = 144z$$

From the 1st equation, we have $x+y=23-z$. Hence,

$$252 + (23-z) z^2 = 144 z$$

and, thus,

$$z^3 - 23 z^2 + 144 z - 252 = 0$$

Plot this cubic polynomial. Find one root. Use, say, Ruffini's rule to find the other two. For each of the three roots, $\bar{z}$, we have a system of two equations in $x$ and $y$, one linear and one bilinear,

$$x + y = 23 - \bar{z}$$

$$x y = \frac{252}{\bar{z}}$$

Multiplying the 1st equation by $y$,

$$\underbrace{x y}_{= \frac{252}{\bar{z}}} + y^2 = (23 - \bar{z}) y$$

This is a quadratic equation in $y$, so we can use the quadratic formula. Once we have solved for $y$, we can find the corresponding $x$.