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

Alexander Lewis 2022-10-24 Answered
Why isn't $y=a\left({x}^{2}-Sx+P\right)$ same with ${x}^{2}-Sx+P$
If we have roots of the function $y=a{x}^{2}+bx+c$ 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 $f\left(x\right)=y$ in different ways:
$y=a{x}^{2}+bx+c$
($\alpha$ and $\beta$ are roots of the quadratic function)
$y=a\left(x-\alpha \right)\left(x-\beta \right)$
And my question is here:
$y=a\left({x}^{2}-Sx+P\right)$
Actually know that how we can form the qudratic equation using ${x}^{2}-Sx+P$, but the function must be like $y=a\left({x}^{2}-Sx+P\right)$. Actually I don't know that why we add a. I know it will be removed when $\left(a\right)\left(\frac{-b}{a}\right)$. But I don't know that what is $y=a\left({x}^{2}-Sx+P\right)$ different whitout a!
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Dana Simmons
Step 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:
$y=\left(a{x}^{2}+bx+c\right)$
(where $a\ne 1$)
merely by taking it like:
$y=\left(x-\alpha \right)\left(x-\beta \right)?$
Step 2
Clearly, coefficient of ${x}^{2}$ is 1 in this.
So, while assuming a quadratic when its roots are known, we take it
$y=a\left(x-\alpha \right)\left(x-\beta \right),$
just to be on the safer side.
###### Did you like this example?
Christopher Saunders
Step 1
${y}_{1}=\left(x-b\right)\left(x-c\right)$
and
${y}_{2}=a\left(x-b\right)\left(x-c\right)$
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
${y}_{2}=3{x}^{2}+15x+18$
are different functions with the same roots.
Step 2
Notice that
${y}_{1}\left(2\right)=20$
while
${y}_{2}\left(2\right)=60$
What you like to say is that the two quadratic equations,
${x}^{2}+5x+6=0$
and
$3{x}^{2}+15x+18=0$
are equivalent because we can factor 3 out and 3 is not zero so it does not change the roots.