determination coefficients of quadratic equation using sum and product of its solutions? let $x_1$ and $x_2$ are real solutions of the quadratic equation $ f(x)=ax^2+bx+c=0$ with $a, b, c$ are real such that its solutions  satisfy the following system :
$$\begin{cases}x_1+x_2=16, \\ x_1 x_2=55.\end{cases}$$
Now my question here is : what are $a, b, c$ ?
I have got $a=1,b=-16, c=55$ but someone said me that I am wrong , $a$ should  be arbitrary as a reason the system have infinitely many solutions 
 A: If $x_1$ and $x_2$ are solutions of the quadratic equation $x^2-16x+55=0$, then they are also solutions of the quadratic equation $ax^2-16ax+55a=0$ for any non-zero $a\in\Bbb R$.
So it is correct that the system has an infinite number of solutions.
A: Let's assume the solutions are $x_1$ and $x_2$. Now, we can write:
\begin{align*}
f(x) &= ax^2 + bx + c \qquad (1)\\
f(x) &= (x - x_1)(x - x_2) = x^2 - x(x_1 + x_2) + x_1 x_2 \qquad (2)
\end{align*}
Since the first term of $(1)$ has a highest power of $1$, we scale equation $(2)$ into:
$$
f(x) = a(x - x_1)(x - x_2) = ax^2 - xa(x_1 + x_2) + ax_1 x_2 \qquad (3)
$$
We can use the relations $x_1 + x_2 = 16$, $x_1x_2 = 55$ in equation $(3)$ to arrive at:
$$
f(x) = ax^2 -16ax + 55a
$$
which means we have an (infinite) family of polynomials $f(x)$ for different choices of $a$.
EDIT: showing that the roots of $f(x)$ are the same as asked:
Note that we can find the roots of $x^2 - 16x + 55$ using (say) the quadratic equation:
$$
x_{1, 2} = \frac{16 \pm \sqrt{16^2 - 4 \cdot 1 \cdot 55}}{2 \cdot 1} = \frac{16 \pm 6}{2} = 8\pm3 = \{ 11, 5\} 
$$
Hence, $$f(x) = a(x^2 - 16x + 55) = a(x - 11)(x - 5)$$
which means that the roots are $\{11, 5\}$
EDIT: An attempt to clarify the confusion between roots and solutions
We have two "solutions" floating around in this question.


*

*The collection of polynomials of the form $f_a(x) = a(x - 11)(x - 5)$

*The roots of each $f_{a_0}(x)$, for each $a$. which are $x_1 = 11, x_2 = 5$.


Given a fixed $a = a_0$, the solution of $f_{a_0}(x)$ is $x_1 = 11, x_2 = 5$.
Given that the roots obey the equation $x_1 + x_2 = 16, x_1 x_2 = 55$. Equivalently, given that the roots are $x_1 = 11, x_2 = 5$, .there are an infinite number of polynomials $f_a(x)$ which have the roots, one for each choice of $a$.
So the fact that, for example, $f_1(x) = (x - 11)(x - 5)$ and $f_2(x) = 2(x- 11)(x - 5)$ both have the same roots ($x_1 = 11, x_2 = 5$) should not contradict the fact that there are two such polynomials, one called $f_1(x)$ and one called $f_2(x)$.
Indeed, this is analogous to this situation which might be easier to think about:
$$
a(x) \equiv x - 10 \quad
b(x) \equiv 2x - 20 \quad
c(x) \equiv 3x - 30
$$
All of $a(x), b(x), c(x)$ have the root $(x = 10)$, but $a(x) \neq b(x) \neq c(x)$.
Similarly, all of the $f_a(x)$ have roots $(x_1 = 11, x_2 = 5$), but they are different polynomials, one for each choice of $a$.
