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?

  • $\begingroup$ To be precise no one says "solve the equation". If you rewrite the "quadratic" in $y^2$ in terms of $y-\frac{z-6}{2}$ you'll be able to see that the appropriate function of $z$ [the "constant"] is indeed positive and then etc etc $\endgroup$ Jul 13 '21 at 15:32
  • $\begingroup$ One we have more than one variable in the polynomial, we say that it's "quadratic in $x$" if the highest power of $x$ that appears is $2$. Then we can treat the other variables as constants, whose value is unknown. $\endgroup$
    – saulspatz
    Jul 13 '21 at 15:40
  • $\begingroup$ @saulspatz But why? It will be helpful if you provide some reference. Thanks. $\endgroup$
    – Osmium
    Jul 13 '21 at 15:44
  • $\begingroup$ Reference for what? It's just the quadratic formula, which you already know. $\endgroup$
    – saulspatz
    Jul 13 '21 at 15:46
  • $\begingroup$ Let's say $f(x) =2x^2$ then will your claim still hold? $\endgroup$
    – Osmium
    Jul 13 '21 at 15:48

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 :)

  • $\begingroup$ Thanks. Just to be sure you're saying that we can use quadratic formula even if the equation is not an quadratic equation, so for example I can use quadratic formula on: $x^2+cos(x)x+2=0$ (though that many not give solution!)? Also I am not able to follow how you removed \pm. Thanks $\endgroup$
    – Osmium
    Jul 14 '21 at 1:36
  • $\begingroup$ @IDKWTD: Yes, and as for the \pm, I had something else in my mind, in this equation, we don't even need to remove \pm. I have edited my post. $\endgroup$ Jul 14 '21 at 1:54

It sounds like you are asking: If you have a quadratic in the variable $y$, does the quadratic formula hold, if the coefficients of the quadratic depend on some other variable $z$?

In your case, if $$y^2 + (z-6)y + (7+z(z-6)) = 0,$$ can you still treat the coefficients $z-6$ and $7+z(z-6)$ as constant and use the usual quadratic formula to find an expression for $y$? The answer is yes. If you go through the derivation of the quadratic formula, you will see that it is not important that the coefficients are constants: They are allowed to vary with other variables, and the key is that the coefficients are constant with respect to y.

So if you have an expression of the form $$a(z)y^2 + b(z)y + c(z) = 0,$$ where $a$, $b$, and $c$ are functions of $z$, then I would indeed call this a quadratic in $y$, and I would gladly apply the quadratic formula to find an expression for $y$ in terms of $z$ (and here you of course need to take care if $a(z)\neq 0$).

  • $\begingroup$ But $y$ and $z$ are related to each other by equation $x+y+z=6$. $\endgroup$
    – Osmium
    Jul 14 '21 at 0:37

First, the quadratic formula deals with polynomials. If you introduce some unspecified function $f$, you muddy the waters, since $f$ needn't be a polynomial. For example, if $f(x)=\sin x$, then $x^2+f(x)$ is not quadratic in $x$.

I understand that you saw the quadratic formula derived, it was assumed that $a,b,c$ are constants, but the proof works so long as they aren't functions of $x$.

When we say that the roots of $p(x)=ax^2+bx+c$ are $$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$ this is true so long as $a\neq0$ and $a,b,$ and $c$ do not depend on $x$. You can verify this by substituting them into the formula for $p$.

It doesn't matter what $a,b,$ and $c$ are, so long as they don't depend on $x$. They could be trigonometric functions of some other variable, for example.

  • $\begingroup$ I agree with you on every point your wrote but clearly in this case $y$ and $z$ are related to each other by the equationz: $x+y+z=6 $ and $x y+y z+z x=7 $. $\endgroup$
    – Osmium
    Jul 14 '21 at 0:49
  • $\begingroup$ Yes, but that's completely irrelevant, as I and others have explained. $\endgroup$
    – saulspatz
    Jul 14 '21 at 4:04
  • 1
    $\begingroup$ The formula works even if $a$, $b$ and $c$ depend upon $x$. $\endgroup$ Jul 14 '21 at 6:42

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