<s>[INST] Is $y^2+f(y)b+c$ a quadratic equation? The solution to the question:

Let $x, y, z \in R$ such that $x+y+z=6$ and $x y+y z+z x=7$. Then find the range of values of $x, y$, and $z$.

given in book is as follows:

$x, y, z \in R$
$x+y+z=6$ and $x y+y z+z x=7$
$\Rightarrow y(6-y-z)+y z+z(6-y-z)=7$
$\Rightarrow \quad-y^{2}+(6-z+z-z) y+z(6-z)-7=0$
$\Rightarrow \quad y^{2}+(z-6) y+7+z(z-6)=0$
.
Now, $y$ is real. Therefore,
$(z-6)^{2}-4[7+z(z-6)] \geq 0$...(1)
or $3 z^{2}-12 z-8 \leq 0$
or $\frac{12-\sqrt{144+96}}{6} \leq z \leq \frac{12+\sqrt{144+96}}{6}$
or $\frac{6-2 \sqrt{15}}{3} \leq z \leq \frac{6+2 \sqrt{15}}{3}$.
From symmetry, $x$ and $y$ have same range.

But I doubt if this is correct; since $y$ is function of $z$, hence the author cannot solve the question this way. Let me explain my point clearly.
Quadratic equations are defined as those which can be expressed in the form $ax^2+bx+c=0$, where $a \not=0$ and $a, b, c$ are constants. But clearly in this case $z$ is not constant, it is function of $y$, hence $(1)$ is not a quadratic equation so we cant the do the steps after $(1)$ hence the solution is wrong.
So the question is, is $y^2+f(y)b+c$ a quadratic equation?
 [/INST] Just go back and recall the derivation of the quadratic formula, it never uses the fact that $a$, $b$ and $c$ are constant, you can use it any time. Just as an example, consider:
$$\begin{align*}
2x+4&= 20\\
4+2x-20&= 0\\
(1)\color{blue}{2}^2+(x)\color{blue}{2}+(-20)&= 0\\
\color{blue}{2}&= \dfrac{-x\pm\sqrt{x^2-4(1)(-20)}}{2(1)}\\
4+x&=\pm\sqrt{x^2+80}\\
16+8x+x^2&=x^2+80\\
x&= 8
\end{align*}$$
Here, the coefficient was itself a variable and the variable was a constant, but the quadratic formula, as you see works. Note the coefficient of variable squared must be non-zero (here, $1\neq 0$).
$x^2+f(x)b+c=0$ may or may not be a quadratic equation, but you can always use the quadratic formula!
Hope this helps. Ask anything if not clear :)
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