Quadratic equations and functions

MotivationH

A garden against a wall

Students at a secondary school in Kisumu are fencing a vegetable garden against the long wall of their classroom block. They have $20$ metres of fencing for the other three sides. If the two sides that stick out from the wall are $x$ metres long, the side parallel to the wall is $20 - 2x$ metres, and the garden's area is

A(x) = x(20 - 2x) = 20x - 2x^2 \quad \text{square metres}.

Two questions come up at once. The agriculture teacher wants exactly $48\ \text{m}^2$ : what should $x$ be? And what is the largest garden the fencing can enclose?

The first question is an equation, $-2x^2 + 20x = 48$ , in which the unknown appears squared: a quadratic equation. The second is about the highest point of the graph of $A$ , which turns out to be a parabola. By the end of this lesson you will be able to answer both, and to prove that your answers are right.

IntuitionH

Completing the square: a picture from Baghdad

Around 820 CE, the scholar Muḥammad ibn Mūsā al-Khwārizmī, working in Baghdad, explained how to solve "a square and ten roots equal thirty-nine": in our notation, $x^2 + 10x = 39$ . His explanation was a picture.

Completing the square for x² + 10x = 39. The square x² and two strips of 5x make the L-shaped area 39. The missing corner is a 5 × 5 square of area 25, so the completed square has area 39 + 25 = 64 and side x + 5 = 8.

Split the $10x$ into two strips of $5x$ and lay them along two sides of the square $x^2$ . The L-shape has area $39$ , and it is a large square with a $5 \times 5$ corner missing. Add that corner to both sides:

x^2 + 10x + 25 = 64, \qquad (x + 5)^2 = 64, \qquad x + 5 = 8, \qquad x = 3.

Every quadratic equation can be solved this way. Doing it once and for all, with letters instead of numbers, produces the quadratic formula.

DefinitionsH

Definitions

Definition— quadratic equation and function

A quadratic equation is an equation $ax^2 + bx + c = 0$ where $a, b, c$ are real numbers and $a \neq 0$ . The function $f(x) = ax^2 + bx + c$ is a quadratic function, and its graph is called a parabola. A solution of the equation is a root of $f$ (or a zero).

Definition— discriminant

The discriminant of $ax^2 + bx + c$ is the number $\Delta = b^2 - 4ac$ .

Definition— vertex and axis of symmetry

The vertex of a parabola is its lowest point (if $a > 0$ ) or highest point (if $a < 0$ ). The axis of symmetry is the vertical line through the vertex: the parabola is its own mirror image in this line.

We will use one property of the real numbers again and again.

Theorem— Zero-product property

For real numbers $u$ and $v$ : if $uv = 0$ , then $u = 0$ or $v = 0$ .

Proof

Suppose $uv = 0$ and $u \neq 0$ . Multiplying both sides by $1/u$ gives $v = \tfrac{1}{u} \cdot 0 = 0$ .

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Worked examplesH

Worked examples

Example— By factorising: the garden

The garden needs $20x - 2x^2 = 48$ . Rearranging and dividing by $2$ : $x^2 - 10x + 24 = 0$ . We look for two numbers with product $24$ and sum $-10$ : $-4$ and $-6$ . So

(x - 4)(x - 6) = 0,

and by the zero-product property $x = 4$ or $x = 6$ . Both work: a $4 \times 12$ garden or a $6 \times 8$ garden. Factorising is quick when the roots are whole numbers, but most quadratics do not factorise so nicely.

Example— By completing the square

Solve $x^2 + 6x - 7 = 0$ . Half of $6$ is $3$ , and $(x + 3)^2 = x^2 + 6x + 9$ , so

x^2 + 6x - 7 = (x + 3)^2 - 16 = 0 \quad\Longrightarrow\quad (x + 3)^2 = 16 \quad\Longrightarrow\quad x + 3 = \pm 4.

So $x = 1$ or $x = -7$ . Do not forget the $\pm$ : both $4^2$ and $(-4)^2$ are $16$ .

Example— By the formula

Solve $2x^2 - 4x - 3 = 0$ . Here $a = 2$ , $b = -4$ , $c = -3$ and $\Delta = 16 + 24 = 40 > 0$ , so there are two real roots:

x = \frac{4 \pm \sqrt{40}}{4} = 1 \pm \frac{\sqrt{10}}{2} \approx 2.58 \text{ or } -0.58.

Example— What the discriminant says

$x^2 - 4x + 4$ : $\Delta = 16 - 16 = 0$ . One root, $x = 2$ , because $x^2 - 4x + 4 = (x - 2)^2$ . The parabola just touches the $x$ -axis.
$x^2 + x + 1$ : $\Delta = 1 - 4 = -3 < 0$ . No real roots: the parabola stays above the axis. In the complex numbers there are two roots, $x = \tfrac{-1 \pm i\sqrt 3}{2}$ .

Example— A quadratic with given roots

Find a quadratic with roots $2 + \sqrt 3$ and $2 - \sqrt 3$ . Their sum is $4$ and their product is $4 - 3 = 1$ . By Vieta's formulas (proved below), $x^2 - 4x + 1$ works. Check with the formula: $x = \tfrac{4 \pm \sqrt{12}}{2} = 2 \pm \sqrt 3$ .

Example— The largest garden

$A(x) = -2x^2 + 20x$ has $a = -2 < 0$ , so its vertex is a highest point, at $x = -\tfrac{b}{2a} = -\tfrac{20}{-4} = 5$ . The largest garden is $5$ m by $10$ m, with area $A(5) = 50\ \text{m}^2$ . Completing the square shows it directly: $A(x) = 50 - 2(x - 5)^2$ , and $2(x - 5)^2 \ge 0$ .

Example— A ball kicked upwards

A footballer kicks a ball straight up from a height of $1$ m at $14$ m/s. Ignoring air resistance (a modelling assumption), its height after $t$ seconds is about $h(t) = 1 + 14t - 4.9t^2$ metres. The vertex is at $t = \tfrac{14}{9.8} \approx 1.43$ s, and the maximum height is

h_{\max} = c - \frac{b^2}{4a} = 1 - \frac{196}{4 \cdot (-4.9)} = 1 + 10 = 11 \text{ m}.

The ball lands when $h(t) = 0$ : $t = \tfrac{14 + \sqrt{196 + 19.6}}{9.8} \approx 2.93$ s. The other root, about $-0.07$ s, is a valid solution of the equation but means nothing for the ball. Always ask which roots make sense in the situation.

ProofH

Proofs: the formula, the discriminant and the vertex

Theorem— Completing the square

If $a \neq 0$ , then for every $x$ :

ax^2 + bx + c = a\left(x + \frac{b}{2a}\right)^2 - \frac{b^2 - 4ac}{4a}.

Proof

Expand the right-hand side: $a\left(x^2 + \tfrac{b}{a}x + \tfrac{b^2}{4a^2}\right) - \tfrac{b^2}{4a} + \tfrac{4ac}{4a} = ax^2 + bx + \tfrac{b^2}{4a} - \tfrac{b^2}{4a} + c = ax^2 + bx + c$ .

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Multiplying by $4a$ clears every fraction and gives an identity that holds even when $a = 0$ :

4a\,(ax^2 + bx + c) = (2ax + b)^2 - (b^2 - 4ac). \tag{$\ast$}

Theorem— Quadratic formula

Let $a \ne 0$ and $\Delta = b^2 - 4ac \ge 0$ . Then $ax^2 + bx + c = 0$ if and only if

x = \frac{-b + \sqrt{\Delta}}{2a} \quad \text{or} \quad x = \frac{-b - \sqrt{\Delta}}{2a}.

Proof

By $(\ast)$ , and because $4a \neq 0$ , the equation $ax^2 + bx + c = 0$ is equivalent to $(2ax + b)^2 = \Delta$ . Since $\Delta \ge 0$ , it has the square root $s = \sqrt\Delta \ge 0$ , and $(2ax + b)^2 = s^2$ means $(2ax + b - s)(2ax + b + s) = 0$ . By the zero-product property, $2ax + b = s$ or $2ax + b = -s$ , that is, $x = \tfrac{-b \pm s}{2a}$ . Every step is reversible, so these two numbers are roots, and there are no others.

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Theorem— The discriminant decides

For $a \neq 0$ , the equation $ax^2 + bx + c = 0$ has

two different real roots if $\Delta > 0$ ;
exactly one real root, $x = -\tfrac{b}{2a}$ , if $\Delta = 0$ ;
no real roots if $\Delta < 0$ .

Proof

If $\Delta > 0$ then $\sqrt\Delta > 0$ , and the two roots given by the formula differ by $\tfrac{2\sqrt\Delta}{2a} \ne 0$ . If $\Delta = 0$ the formula gives $-\tfrac{b}{2a}$ twice. If $\Delta < 0$ , a root $x$ would satisfy $(2ax + b)^2 = \Delta < 0$ by $(\ast)$ . But the square of a real number is never negative.

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Theorem— Vieta's formulas

If $r$ and $s$ are the roots of $ax^2 + bx + c$ (with $r = s$ allowed when $\Delta = 0$ ), then

r + s = -\frac{b}{a}, \qquad rs = \frac{c}{a}.

Proof

From the formula: $r + s = \tfrac{-b + \sqrt\Delta - b - \sqrt\Delta}{2a} = -\tfrac{b}{a}$ and $rs = \tfrac{b^2 - \Delta}{4a^2} = \tfrac{4ac}{4a^2} = \tfrac{c}{a}$ .

Without the formula, for two different roots: subtracting $as^2 + bs + c = 0$ from $ar^2 + br + c = 0$ gives $(r - s)\big(a(r + s) + b\big) = 0$ . Since $r - s \neq 0$ , $r + s = -b/a$ . Then $c = -ar^2 - br = -ar^2 + a(r + s)r = ars$ .

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Equivalently, $ax^2 + bx + c = a(x - r)(x - s)$ . This is the factor theorem of the polynomials lesson at work: each root gives a factor.

Theorem— Vertex and axis of symmetry

Let $f(x) = ax^2 + bx + c$ with $a \neq 0$ and put $h = -\tfrac{b}{2a}$ . Then:

$f(h + t) = f(h - t)$ for every $t$ : the line $x = h$ is an axis of symmetry;
$f(h) = c - \tfrac{b^2}{4a} = -\tfrac{\Delta}{4a}$ ;
if $a > 0$ then $f(x) \ge f(h)$ for all $x$ (the vertex $(h, f(h))$ is the lowest point); if $a < 0$ then $f(x) \le f(h)$ (it is the highest point).

Proof

By completing the square, $f(x) = a(x - h)^2 + f(h)$ with $f(h) = -\tfrac{\Delta}{4a}$ . Then $f(h + t) = at^2 + f(h) = f(h - t)$ , which proves 1. For 3, $f(x) - f(h) = a(x - h)^2$ has the sign of $a$ , because $(x - h)^2 \ge 0$ .

∎

Note

There is a pleasant shortcut, popularised by Po-Shen Loh in 2019 (and, as he points out, rooted in very old ideas). Divide by $a$ to get $x^2 + Bx + C = 0$ . By Vieta, the roots have average $-B/2$ , so write them as $-\tfrac{B}{2} \pm u$ . Their product is $\tfrac{B^2}{4} - u^2 = C$ , so $u = \sqrt{\tfrac{B^2}{4} - C}$ — the quadratic formula again.

Formal mathematicsF

The proofs, checked by Lean

In Lean the real numbers are ℝ and Real.sqrt is the square root. The tactic field_simp clears denominators (it needs to know $a \neq 0$ ), and ring checks polynomial identities. Completing the square, and the fraction-free identity $(\ast)$ :

complete_squareLean 4Proved here from basic factsFormally verified

/-- **Completing the square:** for `a ≠ 0`,
`ax² + bx + c = a(x + b/(2a))² − (b² − 4ac)/(4a)`.
`field_simp` clears the denominators `2a` and `4a`, and `ring` checks the resulting identity. -/
theorem complete_square {a b c : ℝ} (ha : a ≠ 0) (x : ℝ) :
    a * x ^ 2 + b * x + c = a * (x + b / (2 * a)) ^ 2 - (b ^ 2 - 4 * a * c) / (4 * a) := by
  field_simp
  ring