Is y2+f(y)b+c a quadratic equation?

Question

firenzesunzc65 · Accepted Answer

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:K $\begin{aligned} 2 x + 4 & = 20 \\ 4 + 2 x - 20 & = 0 \\ (1) 2^{2} + (x) 2 + (- 20) & = 0 \\ 2 & = \frac{- x \pm \sqrt{x^{2} - 4 (1) (- 20)}}{2 (1)} \\ 4 + x & = \pm \sqrt{x^{2} + 80} \\ 16 + 8 x + x^{2} & = x^{2} + 80 \\ x & = 8 \end{aligned}$
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!

Marin Lowe · Accepted Answer

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

).

Is \(\displaystyle{y}^{{2}}+{f{{\left({y}\right)}}}{b}+{c}\) a quadratic equation?

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