# 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?

• 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 Jul 13 '21 at 15:32
• 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. Jul 13 '21 at 15:40
• @saulspatz But why? It will be helpful if you provide some reference. Thanks. Jul 13 '21 at 15:44
• Reference for what? It's just the quadratic formula, which you already know. Jul 13 '21 at 15:46
• Let's say $f(x) =2x^2$ then will your claim still hold? Jul 13 '21 at 15:48

## 3 Answers

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 :)

• 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 Jul 14 '21 at 1:36
• @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. 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 $$a(z)\neq 0$$).

• But $y$ and $z$ are related to each other by equation $x+y+z=6$. 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.

• 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 $x y+y z+z x=7$. Jul 14 '21 at 0:49
• Yes, but that's completely irrelevant, as I and others have explained. Jul 14 '21 at 4:04
• The formula works even if $a$, $b$ and $c$ depend upon $x$. Jul 14 '21 at 6:42