Showing that a quadratic cannot have 3 roots using Determinants Consider the equation $px^2 + qx + r = 0$ and let us assume that a, b and c satisfies the above equation. 
So,
$$a^2p + aq + r = 0$$
$$b^2p + bq + r = 0 $$
$$c^2p + cq + r = 0$$
They can be represented using a matrix.
\begin{bmatrix}a^2&a&1\\b^2&b&1\\c^2&c&1\end{bmatrix}
From the equation,the matrix sends a point $(p, q, r)$ to the origin $(0, 0, 0)$.
Now my book states the following:
There are two possibilities:


*

*$(p,q,r)$ is at the origin

*$(p,q,r)$ is some distinct point


$(1)$ is not possible as for that value of $(p,q,r)$, the equation is no more an equation.
So the latter is the case. According to the book, for $(2)$ to be true, the matrix must be singular. I don't quite get this.
 A: $$px^2+qx+r = 0$$
Imagine that $x$ has $3$ roots $a,b,c$
Representing in matrix
$$\begin{bmatrix}a^2&a&1\\b^2&b&1\\c^2&c&1\end{bmatrix}\begin{bmatrix}p\\q\\r\\\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}$$
Since the matrix product is $$\begin{bmatrix}0\\0\\0\end{bmatrix}$$
Then it's  determinant must be zero
$$\begin{bmatrix}a^2&a&1\\b^2&b&1\\c^2&c&1\end{bmatrix}$$
$$\Delta = (a-b)(a-c)(b-c)$$
$$ (a-b)(a-c)(b-c) = 0$$
This equation here mean the the quadratic polynomial is cursed to always have only $2$ roots
$$ (a-b)(a-c)(b-c) = 0$$
If we image that $a,b$ are two of it's root and we solve the quadratic for $c$, the result of $c$ would be $c=a$ or $c=b$
$$ (a-b)(a-c)(b-c) = 0$$
Now if we take only $a$ as one of its root, related $b$ and $c$ linearly and solving for them, it would also turn out that the last root is either one of the first two
A: If the matrix equation
$$\begin{bmatrix}a^2&a&1\\b^2&b&1\\c^2&c&1\end{bmatrix}\begin{bmatrix}p\\q\\r\\\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}$$
has a non-trivial solution $(p,q,r)\neq (0,0,0)$, then the matrix is singular by definition.
