Why isn't $y=a(x^2-Sx+P)$ same with $x^2-Sx+P$

Question

If we have roots of the function $y=ax^2+bx+c$ we can calculate $S=\frac{-b}{a}$ and also $P=\frac{c}{a}$ . Then we know that we can form the function this way: $$x^2-Sx+P$$ So on the other side we know that we have the function f(x)=y in different ways: $$y=ax^2+bx+c$$ ($\alpha$ and $\beta$ are roots of the quadratic function) $$y=a(x-\alpha)(x-\beta)$$ And my question is here: $$y=a(x^2-Sx+P)$$ Actually know that how we can form the qudratic equation using $x^2-Sx+P$ ,but the function must be like $y=a(x^2-Sx+P)$. Actually I don't know that why we add $a$. I know it will be removed when $(a)(\frac{-b}{a})$ But I don't know that what is $y=a(x^2-Sx+P)$ different whitout a!

Answer 1

a is there just to include the possibility of a quadratic having a coefficient other than 1 for $2^{nd}$ degree term $x^2$.

Otherwise, how will you make a quadratic of the form: $$y=(ax^2+bx+c)$$(where a $\neq$ 1)
merely by taking it like: $$y=(x-\alpha)(x-\beta)?$$
Clearly, coefficient of $x^2$ is 1 in this.

So, while assuming a quadratic when its roots are known, we take it $$y=a(x-\alpha)(x-\beta),$$ just to be on the safer side.

Answer 2

$$a(x^2-Sx+P)$$

and

$$x^2-Sx+P$$

do coincide at the roots (of course, they are both $0$).

The presence of $a$ matters when you evaluate them at other values than the roots !

Answer 3

Let $f(x)=a(x^2-Sx+P)$ and $g(x)=x^2-Sx+P$, then these functions have the same roots, ( if $a \ne 0$) but , if $a \ne 1$, these functions are different. For example $f(0)=aP \ne P=g(0)$ (if $P \ne 1$).

Answer 4

Two quadratic equations $$ y_1= (x-b)(x-c)$$ and $$y_2= a(x-b)(x-c)$$ have the same roots but not the same values at any other points. They are different function which share common roots.

For example $$ y_1=x^2+5x+6$$ and $$ y_2=3x^2+15x+18$$ are different functions with the same roots.

Notice that $$y_1(2) = 20$$ while $$y_2(2) = 60$$

What you like to say is that the two quadratic equations, $$ x^2+5x+6=0$$ and $$ 3x^2+15x+18=0 $$ are equivalent because we can factor $3$ out and $3$ is not zero so it does not change the roots.

Answer 5

You have to distinguish between “$f_1$ and $f_2$ are the same functions” and “$f_1$ and $f_2$ have the same roots”.

Clearly $$f_1(x) = a(x^2-Sx+P) = 0 \iff f_2(x) = x^2-Sx+P = 0$$ for any $a \ne 0$. So $f_1$ and $f_2$ have the same roots.

But $f_1 = a \cdot f_2(x) \ne f_2(x)$ for $a \ne 1$. Thus they are not the same functions.

Edit: if you want to get a visual impression: in this Desmos Graph you can modify the value of $a$ (press the play button or modify it manually). Then you see how the function changes but the roots stay the same.