You stand on a platform and throw a ball straight up. It leaves your hand 2 m above the ground, moving upwards at 15 m/s. When does it hit the ground? How high does it go?

If you did sixth-form mechanics, you met the constant-acceleration ("SUVAT") equations. The one that matters here is

where \(s\) is displacement, \(u\) initial velocity, \(a\) acceleration and \(t\) time. Measuring upwards, \(u = 15\) m/s and \(a = -9.8\) m/s² (gravity pulls down). Adding the 2 m starting height, the height after \(t\) seconds is

"When does it hit the ground?" becomes: solve \(2 + 15t - 4.9t^2 = 0\). The unknown appears both squared and unsquared. You cannot "subtract 2, then divide by 15" as you would for a linear equation, because the \(t^2\) term refuses to go away. You need a new idea.

You have 100 m of fencing and want a rectangular pen against a long wall, so you only need to fence three sides. If the two sides perpendicular to the wall are \(x\) metres each, the side parallel to the wall is \(100 - 2x\), and the area is

Which \(x\) gives the biggest pen? Again a squared term, and this time the question is not "where is it zero?" but "where is it largest?"

A small company finds that, selling \(q\) hundred units a month, its profit in thousands of pounds is

At low volume, fixed costs swamp revenue; at very high volume, it has to cut prices so far that it loses money again. For which volumes does it break even (\(P = 0\))? Which volume maximises profit?

All three are about an expression of the form

and all three ask the same two questions: where is it zero? and where is it largest (or smallest)? By the end of this chapter you will answer both for any such expression, by several methods, and know which method to reach for. (Answers: the ball lands after about 3.19 s and peaks at about 13.48 m; the best pen is 25 m by 50 m; the company breaks even at 1,000 and 5,000 units and does best at 3,000.)

Solve \(x^2 - 5x + 6 = 0\).

If you can write the quadratic as a product of two linear factors, the zero-product property (Chapter 7) finishes the job: a product is zero exactly when one of its factors is zero. For a monic quadratic \(x^2 + bx + c\), expanding \((x + m)(x + n) = x^2 + (m + n)x + mn\) shows you need two numbers with sum \(b\) and product \(c\).

(a) \(x^2 - 5x + 6 = 0\). Need sum \(-5\), product \(6\): \(-2\) and \(-3\).

Check: \(4 - 10 + 6 = 0\) ✓, \(9 - 15 + 6 = 0\) ✓.

(b) \(2x^2 + x = 6\).

(c) The break-even problem: \(-2q^2 + 120q - 1000 = 0\).

The company breaks even at 1,000 and 5,000 units (\(q\) counts hundreds).

(d) \(x^2 = 7x\). Do not divide by \(x\):

When to choose it. When the coefficients are small integers and you can see the factor pair within a few seconds. A quick test: a quadratic with integer coefficients factors over the integers exactly when its discriminant \(b^2 - 4ac\) is a perfect square (\(0, 1, 4, 9, 16, \ldots\)). For \(x^2 - 5x + 6\), \(\Delta = 25 - 24 = 1 = 1^2\): factor it.

Solve \(x^2 + 6x - 7 = 0\), but pretend you cannot see the factorisation. More to the point: find a method that works for every quadratic, including \(x^2 - 2x - 1 = 0\), which does not factor nicely.

Some quadratics are trivially easy. \(x^2 = 16\) gives \(x = \pm 4\). Equally easy is

What makes these easy is that \(x\) appears only once, inside a square, so you can undo the square with a square root and then undo the addition. The general quadratic \(x^2 + 6x - 7\) has \(x\) in two places (\(x^2\) and \(6x\)). Completing the square merges those two appearances into one.

Think of \(x^2 + 6x\) as an area: an \(x\)-by-\(x\) square plus a rectangle \(6\) by \(x\). Cut the rectangle into two halves, each \(3\) by \(x\), and attach one to the right side of the square and the other to the top. You now have an almost-square shape of side \(x + 3\), missing a \(3\)-by-\(3\) corner.

Add that corner, area \(3^2 = 9\), and you have completed the square:

In general, half the coefficient of \(x\), squared, is the missing corner:

(Expand the right side to check: \(x^2 + 2\cdot\tfrac{p}{2}x + \tfrac{p^2}{4} = x^2 + px + \tfrac{p^2}{4}\).) This is the perfect-square identity \((a + b)^2 = a^2 + 2ab + b^2\) from Chapter 7, read backwards.

To keep an equation balanced, whatever you add to one side you must add to the other.

Cross-check by factoring: \(x^2 + 6x - 7 = (x + 7)(x - 1)\) ✓.

Solve \(2x^2 - 8x + 3 = 0\).

Numerically \(\sqrt{10}/2 \approx 1.5811\), so \(x \approx 3.5811\) or \(x \approx 0.4189\). Check the first: \(2(3.5811)^2 - 8(3.5811) + 3 \approx 25.649 - 28.649 + 3 = 0.000\) ✓.

Solve \(x^2 - 2x + 3 = 0\). Try it: See the roots 1 ± i√2.

A real number squared is never negative, so there are no real solutions. Geometrically, the parabola \(y = (x - 1)^2 + 2\) has its lowest point at height 2 and never reaches the axis. (In Chapter 15 you will allow \(\sqrt{-2} = i\sqrt 2\), and this equation gets the two roots \(1 \pm i\sqrt{2}\).)

Completing the square turns "find the best value" into "read off the vertex", because \(a(x - h)^2 + k\) is smallest (for \(a > 0\)) when the square is zero.

Why bother, when there is a formula? Three reasons. Completing the square is how the formula is derived (next section). It produces vertex form, which tells you the maximum or minimum of a quadratic (Section 8.8). And it generalises: you will complete the square again for circles (Chapter 10), in integrals (Chapter 19), and for the Gaussian distribution (Chapter 30).

Completing the square works every time, but it means redoing the same steps for each new equation. As programmers know, when you find yourself repeating a procedure with different inputs, you write a function. The quadratic formula is completing the square done once, symbolically, for general \(a, b, c\).

Start from the general quadratic, with \(a \neq 0\):

Two steps deserve a comment.

That reading is worth pausing on. The formula says the roots are

which is exactly what the picture of the parabola shows: two roots, mirror images about the vertical line through the vertex.

\(a = 3\), \(b = -5\), \(c = -1\).

With \(\sqrt{37} \approx 6.0828\): \(x \approx \frac{11.0828}{6} \approx 1.8471\) or \(x \approx \frac{-1.0828}{6} \approx -0.1805\).

Check with Vieta (Section 8.12): sum of roots should be \(-b/a = 5/3 \approx 1.6667\); \(1.8471 + (-0.1805) = 1.6666\) ✓. Product should be \(c/a = -1/3\); \(1.8471 \times (-0.1805) \approx -0.3334\) ✓.

\(-4.9t^2 + 15t + 2 = 0\): \(a = -4.9\), \(b = 15\), \(c = 2\).

The ball hits the ground after about \(3.19\) s. The negative root is discarded: the model only applies after release. (Discarding roots for physical reasons is normal and is part of the modelling step, not a defect of the mathematics.)

The formula is for the quadratics that do not factor nicely, which is nearly all of them in practice.

Sometimes you do not need the roots, only to know whether they exist. Does a ray hit a sphere (Section 8.15)? Does a ball thrown at a given speed ever reach a given height? Is a trinomial factorable over the integers? All three are answered by the sign (or the squareness) of one number.

The derivation showed

The left side is a square, hence \(\ge 0\). So:

Graphically, the discriminant measures how far the vertex is from the \(x\)-axis, relative to how the parabola opens. The vertex height is \(k = -\Delta/(4a)\) (we prove this in Section 8.8). For an upward parabola (\(a > 0\)), \(\Delta > 0\) means \(k < 0\): the minimum is below the axis, so the curve must cross it twice.

How many real roots does each have?

Try it: Slide c from Δ = 16; See the repeated root 3/2; See the complex pair.

Does the ball reach 15 m? Solve \(2 + 15t - 4.9t^2 = 15\), i.e. \(-4.9t^2 + 15t - 13 = 0\). \(\Delta = 225 - 4(-4.9)(-13) = 225 - 254.8 = -29.8 < 0\). No: it never gets that high (consistent with the peak of 13.48 m). Does it reach 10 m? \(-4.9t^2 + 15t - 8 = 0\) has \(\Delta = 225 - 156.8 = 68.2 > 0\): yes, twice, once going up and once coming down.

The discriminant answers yes/no questions ("does it ever happen?") without solving anything.

Find the maximum height of the ball and when it occurs; find the best fence. These are questions about the vertex, not the roots. Each of the three standard ways of writing a quadratic answers a different question directly:

In all three, \(a\) is the same number and has the same meaning: direction (sign) and steepness (size).

Vertex form is easiest to understand as a sequence of moves applied to the basic parabola \(y = x^2\), whose vertex is the origin.

Combining them: \(y = a(x - h)^2 + k\) is \(y = x^2\) stretched by \(a\), then moved so that its vertex is at \((h, k)\).

These rules are not special to parabolas. Replacing \(x\) by \(x - h\) shifts any graph right by \(h\), and multiplying by \(a\) stretches any graph vertically. You will reuse this exact toolkit for every function family in Chapter 11.

Standard → vertex: complete the square (keeping the equation as a function, so we add and subtract rather than adding to both sides).

Worked example. Write \(y = 2x^2 - 12x + 13\) in vertex form.

Vertex \((3, -5)\), a minimum because \(a = 2 > 0\). The minimum value of \(2x^2 - 12x + 13\) is \(-5\), attained at \(x = 3\).

The general vertex. Doing the same symbolically:

So \(h = -b/(2a)\) and \(k = -\Delta/(4a)\), which is the fact we used in Section 8.7.

Vertex → standard: expand. \(2(x - 3)^2 - 5 = 2(x^2 - 6x + 9) - 5 = 2x^2 - 12x + 13\).

Standard → factored: find the roots. By any method; then \(ax^2 + bx + c = a(x - r_1)(x - r_2)\). For \(2x^2 - 12x + 13\): \(2(x - 3)^2 = 5\) gives \(x = 3 \pm \sqrt{5/2}\), so \(y = 2\left(x - 3 - \sqrt{5/2}\right)\left(x - 3 + \sqrt{5/2}\right)\). Do not forget the leading \(a\): the factored form of \(2x^2 - 12x + 13\) is not \((x - r_1)(x - r_2)\), which has leading coefficient 1.

Factored → vertex: average the roots. The vertex is midway between the roots: \(h = (r_1 + r_2)/2\); then \(k = f(h)\). For \(y = (x + 1)(x - 3)\): \(h = (-1 + 3)/2 = 1\), \(k = (2)(-2) = -4\). Vertex \((1, -4)\), matching the anatomy diagram.

For the ball: cap; intercept \((0, 2)\); vertex at \(t = -15/(2 \cdot (-4.9)) = 15/9.8 \approx 1.531\), \(h(1.531) = 2 + 225/19.6 \approx 13.48\); roots \(\approx -0.128\) and \(\approx 3.189\); mirror point \((3.061, 2)\).

Plot \(y = ax^2 + bx + c\) and read off where it crosses the axis. On paper this gives one or two significant figures; with graphing software or a zoomable plot, more. Its real value is not precision but understanding: a graph shows at a glance how many roots there are, roughly where, and whether the answer from a formula is plausible. Graphing is also the natural method when the "quadratic" is buried in a more complicated equation, such as \(x^2 = 2^x\), where you might first want to know how many solutions to look for (there are three).

A numerical method does not produce a formula; it produces a sequence of better and better approximations. For a quadratic this is usually unnecessary, because the formula exists. But understanding the numerical approach here, where you can check the answer exactly, prepares you for equations that have no formula: polynomials of degree 5 and above, equations mixing polynomials with \(e^x\) or \(\sin x\), and the equations inside every physics engine.

Bisection is the simplest. If a continuous function is negative at one point and positive at another, it must cross zero somewhere between (the intermediate value theorem, Chapter 16). Repeatedly halve the interval, keeping the half where the sign change is.

Try it: Run these 40 halvings.

Forty halvings to get twelve decimal places: steady but slow.

Newton’s method is much faster. Starting from a guess \(x_0\), it repeatedly slides down the tangent line to where the tangent meets the axis:

where \(f'(x)\) is the derivative, the slope of the tangent, which for \(f(x) = ax^2 + bx + c\) is \(2ax + b\). You will derive this properly in Chapter 17 and study Newton’s method, and how it can fail, in Chapter 18. For a preview, take \(f(x) = x^2 - 2\) (root \(\sqrt 2\)) and \(x_0 = 1\): the iteration is \(x_{n+1} = x_n - (x_n^2 - 2)/(2x_n)\), giving \(1 \to 1.5 \to 1.41\overline{6} \to 1.4142157 \to 1.41421356237469\). The number of correct digits roughly doubles each step: four steps where bisection would need about forty.

The quadratic formula is exact. Floating-point arithmetic is not. Solve Try it: See the 25% error.

on a computer using the formula exactly as written:

The roots multiply to \(c/a = 1\) (Vieta), so if one is \(10^8\) the other must be \(10^{-8}\) (to about 16 digits: the exact small root is \(1.0000000000000000 \times 10^{-8}\)). The computer reports \(7.45 \times 10^{-9}\): an error of 25%, from a formula you just derived with complete rigour. Nothing is wrong with the algebra. Something is wrong with how the computer carried it out.

A double-precision float carries about 16 significant decimal digits (Chapter 32). Here \(b^2 = 10^{16}\) and \(4ac = 4\). Then

That number needs 18 significant digits, but doubles near \(10^8\) are spaced \(2^{-26} \approx 1.49 \times 10^{-8}\) apart, so the nearest one the computer can store is \(10^8 - 1.49 \times 10^{-8}\). The small root is computed as

the difference of two nearly equal numbers. Their first 16 digits agree and cancel, leaving only the last one or two digits, which are mostly rounding error: the computer gets \((1.49 \times 10^{-8})/2 \approx 7.45 \times 10^{-9}\), exactly the wrong answer printed above. This is catastrophic cancellation: subtraction of nearly equal approximate quantities destroys relative accuracy. The large root, \((-b + \sqrt\Delta)/(2a)\), adds two numbers of the same sign and is perfectly accurate.

You can see the effect without a computer by pretending your calculator keeps only 4 significant digits. Solve \(x^2 + 62.10x + 1 = 0\) (true roots \(\approx -0.01611\) and \(\approx -62.08\)): Try it: The 4-digit calculator.

The answer \(-0.02000\) is 24% too large in magnitude. The other root, \((-62.10 - 62.06)/2 = -124.16/2 \to -124.2/2 = -62.10\), is off only in its last digit (true value \(-62.08\)).

The product of the roots is \(x_1 x_2 = c/a\) (Section 8.12). So compute the root that involves no cancellation, and get the other from the product. Concretely, choose the sign in front of the square root to match the sign of \(b\), so that \(b\) and \(\operatorname{sign}(b)\sqrt\Delta\) add rather than cancel:

Here \(\operatorname{sign}(b)\) is \(+1\) if \(b \ge 0\) and \(-1\) if \(b < 0\). Why is \(c/q\) a root? Multiply the two roots from the textbook formula: \(\frac{-b + \sqrt\Delta}{2a}\cdot\frac{-b - \sqrt\Delta}{2a} = \frac{b^2 - \Delta}{4a^2} = \frac{4ac}{4a^2} = \frac{c}{a}\). So if \(x_1 = q/a\) is one root, the other is \(\frac{c/a}{q/a} = \frac{c}{q}\).

In the 4-digit example: \(62.10 + 62.06 = 124.16 \to 124.2\), so \(q = -62.10\), \(x_2 = q/a = -62.10\) and \(x_1 = c/q = 1/(-62.10) = -0.01610\). The true value is \(-0.016107\ldots\), so the small root is now off by at most one unit in the fourth digit, instead of 24%.

math.copysign(d, b) returns \(d\) with the sign of \(b\): exactly the \(\operatorname{sign}(b)\sqrt\Delta\) term.

The plot tells the story: the textbook formula loses roughly two digits of accuracy for every factor of 10 in \(|b|\) (its relative error grows like \(b^2\) times the machine precision \(\approx 10^{-16}\), about 25% at \(|b| = 10^8\) and 100% from about \(1.34\times 10^8\)), while the stable version stays at full precision throughout. Try it: Slide b and watch digits go.

The stable algorithm relied on "the roots multiply to \(c/a\)". Where does that come from, and what else is true?

If \(r_1\) and \(r_2\) are the roots of \(ax^2 + bx + c\), the factor theorem says

Two polynomials are equal for every \(x\) only if their coefficients match, so

For which times is the ball above 10 m? For which production levels is the company profitable? These ask where a quadratic is positive (or negative), not where it is zero: inequalities such as \(-2q^2 + 120q - 1000 > 0\).

A parabola can only change sign where it crosses the axis, at its roots. Between consecutive roots its sign is constant. So: find the roots, then decide the sign on each interval. For an upward parabola (\(a > 0\)) with two roots, the quadratic is negative between the roots and positive outside them; for a downward one, the reverse.

(a) \(x^2 - x - 6 > 0\). Try it: Sign chart for x² − x − 6.

\(a = 1 > 0\), so the parabola is above the axis outside the roots: \(x < -2\) or \(x > 3\). A sign table confirms:

In interval notation the answer is \((-\infty, -2) \cup (3, \infty)\). Here round brackets mean the endpoint is excluded, square brackets would mean included, \(\infty\) is not a number but a marker for "no upper end", and \(\cup\) (union) means "or".

(b) \(-x^2 + 4x + 5 \ge 0\). Multiply by \(-1\) to make the leading coefficient positive, reversing the inequality: Try it: Sign chart for example (b).

(c) Profitable production: \(-2q^2 + 120q - 1000 > 0\) is \(q^2 - 60q + 500 < 0\) (divide by \(-2\), flip), i.e. \((q - 10)(q - 50) < 0\), so \(10 < q < 50\): between 1,000 and 5,000 units. Try it: Sign chart for the profit.

(d) The ball above 10 m: \(-4.9t^2 + 15t - 8 > 0\). Roots (formula, \(\Delta = 68.2\), \(\sqrt\Delta \approx 8.2583\)): \(t = (15 \mp 8.2583)/9.8 \approx 0.688\) and \(2.373\). Downward parabola, so positive between the roots: about \(0.69 < t < 2.37\) seconds, a window of \(\sqrt{\Delta}/|a| \approx 1.69\) s. Try it: When is the ball above 10 m?.

Try it: Try a rational inequality.

Every object moving under constant acceleration (and nothing else) has a position that is a quadratic function of time. That includes a thrown ball, a dropped phone, a braking car (constant deceleration), and a rocket in the short phase where thrust and gravity are roughly constant.

Worked example: maximum height. For \(h(t) = 2 + 15t - 4.9t^2\), the vertex is at

and the maximum height is

In SUVAT language you would have said: at the top the velocity is zero, \(v = u + at\) gives \(0 = 15 - 9.8t\), so \(t = 15/9.8\). Same answer, reached through physics instead of algebra, and a good illustration that the vertex of a quadratic is where its rate of change is zero, the idea that drives Chapter 18.

Braking distance. A car travelling at \(u\) m/s decelerates at \(d\) m/s². Its stopping distance is \(u^2/(2d)\): quadratic in speed. Doubling speed quadruples braking distance, which is why the UK Highway Code’s stopping distances grow so steeply.

Any quantity that is a product of two things, one increasing and one decreasing linearly in the same variable, is a downward quadratic and has a maximum at its vertex.

The fence. \(A(x) = x(100 - 2x) = -2x^2 + 100x\). Vertex at \(x = -100/(2 \cdot (-2)) = 25\). Area \(A(25) = 25 \times 50 = 1250\) m². Shortcut: the roots are \(0\) and \(50\), and the vertex is midway between them.

Revenue. A café’s simple demand model says it would sell 1000 coffees a week if they were free, and loses 20 weekly sales for every £1 on the price: demand \(q = 1000 - 20p\). Revenue is price times quantity:

Roots \(p = 0\) and \(p = 50\); vertex midway at \(p = 25\), with \(R(25) = 25 \times 500 = 12{,}500\), i.e. £12,500 a week. Raising the price beyond £25 loses more in volume than it gains in margin. (Real demand curves are not linear, but this is the first model economists draw, and the reasoning is the same with a better model plus calculus.)

The break-even company is at its best at the midpoint of its break-even volumes, \(q = 30\): \(P(30) = -1800 + 3600 - 1000 = 800\), i.e. £800,000 at 3,000 units.

Every parabola has a special point, its focus. For \(y = \dfrac{x^2}{4f}\) the focus is at \((0, f)\). Every ray travelling parallel to the axis that hits the inside of the parabola reflects through the focus. Spin the parabola about its axis to get a paraboloid, and every parallel ray striking the dish converges at one point. Try it: Rays meeting at the focus.

That is why:

Worked example: where to put the receiver. A dish is 60 cm across and 9 cm deep. Place its vertex at the origin, so its cross-section is \(y = x^2/(4f)\) and the rim is at \((30, 9)\): Try it: Place the receiver yourself.

The receiver sits 25 cm in front of the centre of the dish. The reason the property holds is a short calculation with distances: every point of the parabola is equally far from the focus and from a line called the directrix (\(y = -f\)); the proof uses tangent lines and is done in Chapter 10.

A ray tracer makes images by firing a ray from the camera through each pixel and asking: what does this ray hit first? For a sphere, that question is a quadratic equation, solved millions of times per frame. This is perhaps the most-executed quadratic formula in the world.

A ray starts at a point \(\mathbf{o}\) (the camera) and travels in direction \(\mathbf{d}\). Bold letters denote vectors, here just triples of numbers \((x, y, z)\) (Chapter 26). The points on the ray are

meaning: start at \(\mathbf{o}\) and move \(t\) units of \(\mathbf{d}\); component by component, \(p_x = o_x + t d_x\) and so on. The parameter \(t\) measures how far along the ray you are; smaller \(t\) is closer to the camera.

A sphere with centre \(\mathbf{c}\) and radius \(r\) is the set of points at distance \(r\) from \(\mathbf{c}\). Squared distance is computed by Pythagoras in 3D: for a vector \(\mathbf{v} = (v_x, v_y, v_z)\),

the dot product of \(\mathbf{v}\) with itself. More generally, \(\mathbf{u}\cdot\mathbf{v} = u_x v_x + u_y v_y + u_z v_z\); it distributes over addition just like ordinary multiplication.

The ray point \(\mathbf{p}(t)\) is on the sphere when its squared distance from the centre is \(r^2\). Write \(\mathbf{f} = \mathbf{o} - \mathbf{c}\) for the vector from the centre to the camera. Then \(\mathbf{p}(t) - \mathbf{c} = \mathbf{f} + t\mathbf{d}\), and

A quadratic in \(t\), and every piece of this chapter now has a geometric meaning:

Camera at \(\mathbf{o} = (0, 1, 0)\), looking along \(\mathbf{d} = (1, 0, 0)\); sphere centre \(\mathbf{c} = (5, 0, 0)\), radius \(r = 2\).

The visible point is at \(t \approx 3.268\): \(\mathbf{p} \approx (3.268, 1, 0)\). Check: its distance from the centre is \(\sqrt{(3.268 - 5)^2 + 1^2} = \sqrt{3.000 + 1} = 2\) ✓.

The small eps guards against a classic ray-tracing bug, "shadow acne": a ray bounced off a surface, starting on that surface, finds a spurious intersection at \(t \approx 10^{-15}\) due to rounding, so the surface shadows itself. The stable formula matters here too, though not where you might guess. A small sphere far away is harmless: both formulas agree to about 16 digits. The textbook formula loses digits when \(c\) is tiny compared with \(b^2\), which happens when the ray starts very close to the surface of a large sphere (a shadow or reflection ray leaving a big floor or planet). Then the near root comes from cancellation, which is exactly the shadow-acne situation above.