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We know that the quadratic equation $$f(x)=ax^2+bx+c=0$$ has roots $$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=-\frac b{2a}\pm \frac 1a\sqrt{-\left(ac-\frac {b^2}4\right)}$$

Also, $f(x)$ can be written in matrix notation as follows: $$f(x)= \left(\begin{matrix}x&1\\\end{matrix}\right) \left(\begin{matrix}a&\frac b2\\\frac b2&c\end{matrix}\right) \left(\begin{matrix}x\\1\end{matrix}\right)=\mathbf{x^T Q x}$$ where the determinant of $\mathbf Q$ is $\left(ac-\frac {b^2}4\right)=-\frac 14\left(b^2-4ac\right)$, where coincidentally the familiar $(b^2-4ac)$ is the discriminant of the quadratic $f(x)$.

Hence the roots of the quadratic $f(x)=0$ may be written as $$x=-\frac b{2a}\pm \frac 1a\sqrt{-\det(\mathbf Q)}$$ This is equivalent to $$\left(x+\frac b{2a}\right)^2=\frac {-\det(\mathbf Q)}{a^2}$$ Or in neater form, $$\left(ax+\frac b{2}\right)^2={-\det(\mathbf Q)}$$

Question
Can the roots of $f(x)=0$ be derived and written completely in matrix notation, given the link between the determinant and discriminant as shown above?

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    $\begingroup$ One of notation propositions - the last equation is equivalent to: $(\mathbf{x}^T\mathbf{w})^2+\det(Q)=0$ where $\mathbf {x}^T= \begin{bmatrix} x & 1 \end{bmatrix}$ and $\mathbf{w}=Q\mathbf{e_1}$. $\endgroup$ – Widawensen Apr 24 '18 at 10:26
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    $\begingroup$ Maybe somewhat relevant/related: eigenvalues of a 2x2 are $T/2 \pm \sqrt{T^2/4-D}$ and are roots of $x^2-Tx+D=0$ where $T$ is the trace and $D$ is the determinant. $\endgroup$ – jdods Apr 24 '18 at 10:42
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I have obtained some formula with a little changed notation comparing the notation used in the question.

I've interchanged the components of $\mathbf x$ (the new vector is denoted as $\mathbf{ \hat{x}}$ ) and I've interchanged the entries on the main diagonal of $ \mathbf Q$ (new matrix is denoted as $ \mathbf M$ - interchanging entries on the diagonal for $2 \times 2$ matrices doesn't change the determinant) .

Then $f(x)$ can be written down as

$$f(x)= \begin{bmatrix}1&x\\\end{bmatrix} \begin{bmatrix}c&\frac b2\\\frac b2&a\end{bmatrix} \begin{bmatrix}1\\x\end{bmatrix} =\mathbf{\hat{x}^T M \hat{x}}=0$$

It can be calculated that

$ \begin{bmatrix}1 & x\\0 & 1\end{bmatrix} \begin{bmatrix}c & \frac{b}{2}\\\frac{b}{2} & a\end{bmatrix} \begin{bmatrix}1 & 0\\x & 1\end{bmatrix} = \begin{bmatrix}\frac{b x}{2} + c + x \left(a x + \frac{b}{2}\right) & a x + \frac{b}{2}\\a x + \frac{b}{2} & a\end{bmatrix} = \begin{bmatrix} 0 & a x + \frac{b}{2}\\a x + \frac{b}{2} & a \end{bmatrix} $

We know also that determinant of a matrix product is equal to the product of determinants of multiplied matrices.

Additionally we have $\det( \mathbf Q)=\det(\mathbf M)$.

From this follows that $ \det( \mathbf Q)=-\left( ax+ \frac{b}{2}\right)^2$.

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  • $\begingroup$ Used calculations {{1,x} }* {{c,b/2},{b/2,a}} *{{1 }, {x }} at Wolphram Alpha $\endgroup$ – Widawensen Apr 24 '18 at 12:56
  • $\begingroup$ and {{1,x},{0,1}} * {{c,b/2},{b/2,a}} * {{1,0 }, {x,1 }} $\endgroup$ – Widawensen Apr 24 '18 at 13:02
  • $\begingroup$ Interesting results. Can you explain further the basis for your conclusion? It is clear that the determinant of the last matrix is $-(ax+\frac b2)^2$ but it is not clear how that links to $\det(\mathbf Q)$. $\endgroup$ – hypergeometric Apr 24 '18 at 16:35
  • $\begingroup$ @hypergeometric Determinant of matrix $M$ is the same as determinant of $Q$ ( see their forms or calculate them if you don't believe) , they have only changed $c$ and $a$ on the main diagonal , and taking into account that $\det(AMC)=\det(A)\det(M)\det(C)$ ( see multiplication of three matrices) and $\det(A)=1$ and $\det(C)=1$ ( upper triangular with 1's on main diagonal) the result follows.. Anyway I have a feeling that I solve your problem, is something more unclear in the solution? $\endgroup$ – Widawensen Apr 25 '18 at 8:30
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    $\begingroup$ Thanks for your solution (+1). $\endgroup$ – hypergeometric May 5 '18 at 15:54
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Let us switch to homogeneous coordinates and use $2b$ instead of $b$ for convenience.

$$ax^2+2bxy+cy^2=\begin{pmatrix}x&y\end{pmatrix}\begin{pmatrix}a&b\\b&c\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\mathbf{p^TQp}=0.$$

Assuming $Q$ diagonalizable, we have (the transformation $P$ can be taken orthonormal)

$$\mathbf{p^TQp}=\mathbf{p^TP^TDPp}=\mathbf{q^TDq}=0.$$

The last expression is of the form

$$\lambda_0u^2+\lambda_1v^2=0$$ where the lambdas are the Eigenvalues of $\mathbf Q$. For the equation to have real solutions, the Eigenvalues must have opposite signs, or $\lambda_0\lambda_1\le0$, which is precisely $-\det(\mathbf Q)=b^2-ac\ge0$.

The conic factors as

$$\left(\sqrt{|\lambda_0|}u+\sqrt{|\lambda_1|}v\right)\left(\sqrt{|\lambda_0|}u-\sqrt{|\lambda_1|}v\right)=0$$

and the solutions are given by plugging $u=\mathbf{e_0p},v=\mathbf{e_1p}$ in the two factors, setting $y=1$ and solving the linear equations for $x$.

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    $\begingroup$ Thanks - interesting solution. (+1). $\endgroup$ – hypergeometric May 5 '18 at 15:54

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