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2.3 — Quadratics, and Where the Formula Comes From
You have 40 metres of fencing and a straight wall to build against. What is the largest rectangular area you can enclose using the wall as one side?
Call the two equal sides x. The remaining side takes what is left: 40 - 2x. So the area is
A(x) = x(40 - 2x) = 40x - 2x^2
That is a quadratic, and the question "what is the largest area" is the question "where is the top of this curve". By the end of this chapter you will answer it in two lines without any calculus at all.
1. The parabola and its shape
A quadratic is a polynomial of degree two:
y = ax^2 + bx + c, \qquad a \neq 0
Its graph is always a parabola. Not sometimes — always. And a controls it entirely:
- a > 0: opens upward, has a lowest point.
- a < 0: opens downward, has a highest point.
- Larger |a|: narrower. Smaller |a|: wider.
The turning point is the vertex, and the parabola is perfectly symmetric about the vertical line through it — the axis of symmetry. That symmetry is not decorative; it is the key to everything in this chapter, and it comes from the fact that x^2 treats +d and -d identically.
2. Completing the square
Every quadratic can be rewritten so the variable appears exactly once. That single restructuring solves the equation, locates the vertex, and produces the famous formula. It is the most valuable technique in elementary algebra and it is barely taught.
Start with the picture. x^2 + 6x is a square of side x with two 3 \times x rectangles glued to two of its sides. Together they make an L-shape. To complete it into a full square of side x+3, you need to fill the missing corner, which is 3 \times 3 = 9.
So x^2 + 6x = (x+3)^2 - 9: we added 9 to complete the square, so we subtract 9 to keep the value unchanged.
The rule: halve the coefficient of x, square it, add and subtract it.
x^2 + bx = \left(x + \frac{b}{2}\right)^2 - \frac{b^2}{4}
Worked example. Rewrite x^2 - 8x + 3.
Half of -8 is -4; squared is 16.
x^2 - 8x + 3 = (x-4)^2 - 16 + 3 = (x-4)^2 - 11
Now read the answer off. The squared term is never negative and is zero when x = 4, so the minimum value is -11, occurring at x = 4. Vertex at (4, -11), found with no calculus and no formula.
When a \neq 1, factor it out of the first two terms first.
2x^2 + 12x + 5 = 2(x^2 + 6x) + 5 = 2\big[(x+3)^2 - 9\big] + 5 = 2(x+3)^2 - 13
Vertex at (-3, -13), opening upward since a = 2 > 0.
The resulting form y = a(x-h)^2 + k is called vertex form, and the vertex is (h, k). It is worth being able to move between vertex form and expanded form in both directions.
The fencing problem, solved
A(x) = -2x^2 + 40x = -2(x^2 - 20x) = -2\big[(x-10)^2 - 100\big] = -2(x-10)^2 + 200
Since a = -2 is negative, this opens downward, so the vertex is the maximum. The largest area is 200 square metres, at x = 10 — a 10 \times 20 rectangle. Two lines, as promised, and the calculus method of Chapter 5.4 will give the same answer with more machinery.
3. Deriving the quadratic formula
Now complete the square on the general quadratic and watch the formula appear. Every step is one you have already done.
Start from ax^2 + bx + c = 0. Divide by a, legal since a \neq 0:
x^2 + \frac{b}{a}x + \frac{c}{a} = 0
Move the constant across:
x^2 + \frac{b}{a}x = -\frac{c}{a}
Complete the square. Half of \frac{b}{a} is \frac{b}{2a}; squared it is \frac{b^2}{4a^2}. Add it to both sides:
x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} = \frac{b^2}{4a^2} - \frac{c}{a}
The left side is now a perfect square. Put the right side over a common denominator:
\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}
Take the square root of both sides, remembering from Chapter 1.4 that this gives two answers:
x + \frac{b}{2a} = \pm\frac{\sqrt{b^2-4ac}}{2a}
Subtract:
\boxed{\;x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\;}
That is the whole derivation, and it is worth doing once by hand. The formula is not handed down from anywhere; it is completing the square with letters instead of numbers. If you ever forget it, you can rebuild it in ninety seconds.
4. The discriminant tells you the answer before you compute it
The quantity under the root has its own name:
\Delta = b^2 - 4ac
It is the discriminant — from Latin discriminare, to distinguish — and it distinguishes the three cases.
| \Delta | Roots | Graph |
|---|---|---|
| \Delta \gt 0 | two distinct real | crosses axis twice |
| \Delta = 0 | one repeated real | touches axis once |
| \Delta \lt 0 | two complex | never touches axis |
The reason is visible in the formula: \pm\sqrt{\Delta} gives two different numbers when \Delta is positive, gives zero either way when \Delta is zero, and asks for the square root of a negative when \Delta is negative — which no real number provides.
Worked examples.
x^2 - 5x + 6 = 0: \Delta = 25 - 24 = 1 > 0. Two roots: x = \frac{5\pm1}{2} = 3 or 2.
x^2 - 4x + 4 = 0: \Delta = 16 - 16 = 0. One root: x = 2, repeated. Indeed (x-2)^2.
x^2 + x + 1 = 0: \Delta = 1 - 4 = -3 < 0. No real roots. The parabola floats entirely above the axis. Chapter 2.5 finds the two complex roots.
5. Vieta's relations: the roots without solving
If a quadratic has roots r_1 and r_2, then it factors as a(x - r_1)(x - r_2). Expand:
a(x^2 - (r_1+r_2)x + r_1 r_2) = ax^2 + bx + c
Match coefficients:
r_1 + r_2 = -\frac{b}{a}, \qquad r_1 r_2 = \frac{c}{a}
You now know the sum and product of the roots without finding them. For x^2 - 5x + 6, the roots sum to 5 and multiply to 6 — and indeed 2 and 3 do both.
This is what you are really doing when you factor a quadratic by inspection: hunting for two numbers with a known sum and product. Naming the technique makes it deliberate instead of a guess.
It also builds quadratics to order. Want a quadratic whose roots are 3 and -7? Sum is -4, product is -21, so x^2 + 4x - 21.
Named after François Viète, who around 1590 was among the first to use letters for known quantities as well as unknowns — the notational step that made general formulas expressible at all.
6. Where quadratics come from in the physical world
Projectile motion. Anything thrown under gravity, ignoring air resistance, has height
h(t) = h_0 + v_0 t - \tfrac{1}{2}g t^2
with g \approx 9.8\ \text{m/s}^2. The path is a parabola. Galileo established this around 1638, and it was the first time a curve of nature had been identified with a curve of geometry.
Worked example. A ball is thrown upward at 20\ \text{m/s} from a height of 1.5\ \text{m}. When does it hit the ground?
1.5 + 20t - 4.9t^2 = 0
Multiply by -1 to get 4.9t^2 - 20t - 1.5 = 0, then apply the formula:
t = \frac{20 \pm \sqrt{400 + 29.4}}{9.8} = \frac{20 \pm 20.72}{9.8}
That gives t = 4.16 or t = -0.073. Time cannot be negative here, so the answer is about 4.16 seconds. The negative root is not a mistake — it is where the parabola would have crossed the ground had the ball been launched from below and passed the thrower's hand at t=0. The mathematics does not know the ball did not exist then; you do. Discarding physically meaningless roots is a judgement the formula cannot make for you.
Area and optimisation, as in the fencing problem: any situation where a total is fixed and a product is to be maximised gives a quadratic, and the answer is always at the symmetric point.
Satellite dishes and headlights. Rotate a parabola about its axis and you get a shape with a remarkable property: every ray arriving parallel to the axis reflects to a single point, the focus. That is why a dish concentrates a weak satellite signal onto one receiver, and why a headlight with a bulb at the focus emits a parallel beam. Chapter 3.4 proves this.
Bridge cables — the near miss. A suspension bridge's main cable, carrying a uniformly heavy deck, hangs in a parabola. A chain or cable hanging under its own weight alone hangs in a catenary, which is y = \cosh(x) and is not a parabola, though it looks like one. Galileo thought it was a parabola; Bernoulli, Huygens and Leibniz settled it in 1691. The distinction depends on where the weight sits, and getting it wrong is the sort of thing that matters when you build.
7. The cubic, and a story worth knowing
Quadratics were solved in Babylon, in India and by al-Khwārizmī. The cubic held out until sixteenth-century Italy, and its resolution is the strangest episode in the history of mathematics.
Mathematicians then held public problem-solving duels, with posts and salaries at stake, so a method that nobody else had was a career. Scipione del Ferro solved one type of cubic around 1515 and told nobody but a student. Niccolò Tartaglia rediscovered it and won a famous contest in 1535. Gerolamo Cardano coaxed the method out of Tartaglia under an oath of secrecy, then found del Ferro's earlier work and published everything in Ars Magna (1545) — reasoning, defensibly, that the oath covered Tartaglia's discovery and not del Ferro's. Tartaglia never forgave him. Cardano's student Lodovico Ferrari went on to solve the quartic.
The formula exists and is genuinely horrible, involving nested cube roots of square roots. Its historical importance is not that anyone uses it — nobody does — but that solving cubics forced mathematicians to write down square roots of negative numbers as intermediate steps that cancelled later, producing correct real answers by way of impossible objects. Complex numbers were not invented to solve x^2 + 1 = 0; they were forced on people by cubics whose answers were plainly real. That is the story Chapter 2.5 picks up.
And then it stops. Abel proved in 1824, and Galois explained in 1832, that no formula in radicals exists for the general fifth-degree equation. Not undiscovered — impossible. Galois' explanation, invented in the months before he died in a duel at twenty, created group theory. Chapter 11.4 tells that story.
Every formula above, built from scratch
None of the results in this chapter are worth memorising, because each one can be rebuilt in under a minute from something simpler. What follows is that rebuilding, one result at a time, so the formula and the reason for it sit on the same page as the explanation that needed them.
Quadratics
The quadratic formula, built by completing the square
ax^2+bx+c = 0 \quad \Longrightarrow \quad x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}
Read it aloud. "x equals minus b, plus or minus the square root of b squared minus four a c, all over two a."
Where it comes from. The only equation anybody can solve instantly is (\text{something})^2 = \text{number}, because you just take the square root. So the whole derivation is one long effort to force the quadratic into that shape. That effort is called completing the square.
Start with the equation and divide through by a, which is allowed because a \ne 0 (otherwise it is not a quadratic):
x^2 + \frac{b}{a}x + \frac{c}{a} = 0
Move the constant out of the way:
x^2 + \frac{b}{a}x = -\frac{c}{a}
Now look at the left side and ask: what would make it a perfect square? Comparing with (x+k)^2 = x^2 + 2kx + k^2, the middle terms match if 2k = \frac{b}{a}, so k = \frac{b}{2a}. The square would then need a constant term of k^2 = \frac{b^2}{4a^2}, which is not there. Add it to both sides — adding the same thing to both sides keeps the equation true:
x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} = -\frac{c}{a} + \frac{b^2}{4a^2}
The left side is now exactly the square we designed it to be:
\left(x + \frac{b}{2a}\right)^2 = \frac{b^2}{4a^2} - \frac{c}{a}
Tidy the right side over the common denominator 4a^2, remembering \frac{c}{a} = \frac{4ac}{4a^2}:
\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}
Take the square root of both sides. Both a positive and a negative root square to the same thing, which is where the \pm enters and why a quadratic has two solutions:
x + \frac{b}{2a} = \pm\frac{\sqrt{b^2-4ac}}{2a}
Subtract \frac{b}{2a}:
x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}
Every symbol in that formula now has a home. The -b/2a is the axis of symmetry — the parabola's midline. The square root is how far the two roots sit either side of it. And b^2-4ac, the discriminant, decides what happens under the root sign.
What the discriminant tells you
\Delta = b^2 - 4ac
| \Delta | Roots | Graph |
|---|---|---|
| > 0 | two real | crosses the axis twice |
| = 0 | one repeated | touches the axis |
| < 0 | two complex | never touches |
The reason is visible in the formula: \Delta sits under the square root, and the square root of a negative number is not real.
Vieta's formulas: the roots without solving
x_1 + x_2 = -\frac{b}{a}, \qquad x_1 x_2 = \frac{c}{a}
Where they come from. If x_1 and x_2 are the roots, the quadratic can be written in factored form:
ax^2+bx+c = a(x-x_1)(x-x_2)
Expand the right side:
a\left(x^2 - (x_1+x_2)x + x_1x_2\right) = ax^2 - a(x_1+x_2)x + ax_1x_2
Two polynomials are equal for all x only if their matching coefficients are equal. Comparing the x terms: b = -a(x_1+x_2), so x_1+x_2 = -b/a. Comparing the constants: c = ax_1x_2, so x_1x_2 = c/a.
Why this is worth having. It gives you both roots' sum and product from a glance at the coefficients, which is how you factorise by inspection: to factor x^2 - 7x + 12, look for two numbers multiplying to 12 and adding to 7 — that is 3 and 4, so the roots are 3 and 4 and the factorisation is (x-3)(x-4).
Vertex form
ax^2+bx+c = a\left(x + \frac{b}{2a}\right)^2 + \left(c - \frac{b^2}{4a}\right)
This falls straight out of the completing-the-square work above, and it hands you the turning point directly: the parabola's vertex is at x = -\frac{b}{2a}, with value c - \frac{b^2}{4a}. Since a square is never negative, that value is the minimum when a>0 and the maximum when a<0. This is optimisation without calculus, and it is the fastest route to "what price maximises revenue" problems.
8. Where this shows up in your life
Every thrown, kicked, or fired object. The arc is a parabola, and the range/height calculations are this chapter.
Every satellite dish, torch reflector and solar concentrator.
Every optimisation where a total is fixed. Maximum area for a given perimeter, maximum revenue when raising the price loses customers proportionally, the best speed to drive for fuel economy — all quadratics with the answer at the vertex.
Every collision-detection routine in a game engine. Asking whether a moving circle hits another moving circle reduces to asking whether a quadratic in time has a real root — so the code computes a discriminant and checks its sign.
Every image resize on your phone. Bilinear and bicubic interpolation fit low-degree polynomials between pixels.
Quadratics are fully solved when the discriminant is not negative. When it is, we need numbers that do not yet exist. Before building them, the next chapter finishes the real-number toolkit: inequalities and systems of equations.