Notebooks · posts/the-quadratic-formula-derived

The quadratic formula, derived

Adam Zettel · September 22, 2026 · written with an agent

A garden has a perimeter of 10 m and an area of 6 m². How long are its sides? The answer needs $x^{2} - 5 x + 6 = 0$ , and solving it by completing the square gives the quadratic formula.

A rectangular garden has a perimeter of 10 m and an area of 6 m². How long are its sides?

Call one side $x$ . The two sides add up to half the perimeter, so the other side is $5 - x$ , and the area says

x (5 - x) = 6, that is, x^{2} - 5 x + 6 = 0.

You could guess $x = 2$ here, but for most areas a guess won't work, so we solve the general equation $a x^{2} + b x + c = 0$ instead, in three steps.

Look inside this program

Atlas
The program’s map: every function and which calls which.
There’s no recorded run of this program yet.

Shift+Enter in a cell runs it and moves to the next. Running a cell runs every cell above it too, as one program.

[1]lines 20–75

rung "Complete the square" {
    #| Factor $a$ out of the first two terms, then add and subtract $(b/2a)^2$ inside:
    #|
    #| $$ax^2 + bx + c = a\left(x + \frac{b}{2a}\right)^2 + c - \frac{b^2}{4a}.$$
    #|
    #| That is $a(x - h)^2 + k$, with $h = -b/2a$ and $k = c - b^2/4a$. The point $(h, k)$ is the vertex of the parabola.
    #|
    #| For the garden, $h = 2.5$ and $k = -0.25$: a square garden, 2.5 m on a side, would have an area of $2.5^2 = 6.25$ m², which is 0.25 m² too much.
    def vertex(a, b, c) {
        return -b / (2 * a), c - b^2 / (4 * a)
    }
    test "the vertex of $x^2 - 5x + 6$ is $(2.5, -0.25)$" {
        h, k = vertex(1, -5, 6)
        assert_near(h, 2.5, 1.0e-12)
        assert_near(k, -0.25, 1.0e-12)
    }
}

rung "Solve from the vertex" {
    #| Now $a(x - h)^2 + k = 0$ is one step from solved. Move $k$ across and take the square root:
    #|
    #| $$(x - h)^2 = -\frac{k}{a}, \qquad x = h \pm \sqrt{-\frac{k}{a}}.$$
    #|
    #| For the garden, $\sqrt{0.25} = 0.5$, so $x = 2.5 \pm 0.5$ and the sides are 2 m and 3 m.
    #|
    #| Write $h$ and $k$ out in terms of $a$, $b$ and $c$: $-k/a = (b^2 - 4ac)/4a^2$, and the roots become the quadratic formula,
    #|
    #| $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.$$
    def roots(a, b, c) {
        h, k = vertex(a, b, c)
        s = sqrt(-k / a)
        return h - s, h + s
    }
    test "the roots of $x^2 - 5x + 6$ are 2 and 3" {
        x1, x2 = roots(1, -5, 6)
        assert_near(x1, 2.0, 1.0e-12)
        assert_near(x2, 3.0, 1.0e-12)
    }
}

rung "When there are no real roots" {
    #| The formula takes the square root of the discriminant,
    #|
    #| $$D = b^2 - 4ac.$$
    #|
    #| If $D > 0$ there are two roots, if $D = 0$ one, and if $D < 0$ none: the parabola never reaches the axis. A garden with a perimeter of 10 m and an area of 7 m² gives $x^2 - 5x + 7 = 0$ and $D = 25 - 28 = -3$. No such garden exists. The square, at 6.25 m², is the largest area that perimeter can enclose.
    def discriminant(a, b, c) {
        return b^2 - 4 * a * c
    }
    test "$x^2 - 5x + 7$ has no real roots" {
        assert_true(discriminant(1, -5, 7) < 0)
    }
}

# SPDX-License-Identifier: MIT OR Apache-2.0
# SPDX-FileCopyrightText: 2026 The Viv authors

rung "Complete the square" {
    #| Factor $a$ out of the first two terms, then add and subtract $(b/2a)^2$ inside:
    #|
    #| $$ax^2 + bx + c = a\left(x + \frac{b}{2a}\right)^2 + c - \frac{b^2}{4a}.$$
    #|
    #| That is $a(x - h)^2 + k$, with $h = -b/2a$ and $k = c - b^2/4a$. The point $(h, k)$ is the vertex of the parabola.
    #|
    #| For the garden, $h = 2.5$ and $k = -0.25$: a square garden, 2.5 m on a side, would have an area of $2.5^2 = 6.25$ m², which is 0.25 m² too much.
    def vertex(a, b, c) {
        return -b / (2 * a), c - b^2 / (4 * a)
    }
    test "the vertex of $x^2 - 5x + 6$ is $(2.5, -0.25)$" {
        h, k = vertex(1, -5, 6)
        assert_near(h, 2.5, 1.0e-12)
        assert_near(k, -0.25, 1.0e-12)
    }
}

rung "Solve from the vertex" {
    #| Now $a(x - h)^2 + k = 0$ is one step from solved. Move $k$ across and take the square root:
    #|
    #| $$(x - h)^2 = -\frac{k}{a}, \qquad x = h \pm \sqrt{-\frac{k}{a}}.$$
    #|
    #| For the garden, $\sqrt{0.25} = 0.5$, so $x = 2.5 \pm 0.5$ and the sides are 2 m and 3 m.
    #|
    #| Write $h$ and $k$ out in terms of $a$, $b$ and $c$: $-k/a = (b^2 - 4ac)/4a^2$, and the roots become the quadratic formula,
    #|
    #| $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.$$
    def roots(a, b, c) {
        h, k = vertex(a, b, c)
        s = sqrt(-k / a)
        return h - s, h + s
    }
    test "the roots of $x^2 - 5x + 6$ are 2 and 3" {
        x1, x2 = roots(1, -5, 6)
        assert_near(x1, 2.0, 1.0e-12)
        assert_near(x2, 3.0, 1.0e-12)
    }
}

rung "When there are no real roots" {
    #| The formula takes the square root of the discriminant,
    #|
    #| $$D = b^2 - 4ac.$$
    #|
    #| If $D > 0$ there are two roots, if $D = 0$ one, and if $D < 0$ none: the parabola never reaches the axis. A garden with a perimeter of 10 m and an area of 7 m² gives $x^2 - 5x + 7 = 0$ and $D = 25 - 28 = -3$. No such garden exists. The square, at 6.25 m², is the largest area that perimeter can enclose.
    def discriminant(a, b, c) {
        return b^2 - 4 * a * c
    }
    test "$x^2 - 5x + 7$ has no real roots" {
        assert_true(discriminant(1, -5, 7) < 0)
    }
}

# SPDX-License-Identifier: MIT OR Apache-2.0
# SPDX-FileCopyrightText: 2026 The Viv authors