Quadratic Equation

A school carpets its rectangular office. It takes 80 m² of carpet.

a. What are the room’s length and breadth? From the area alone you can’t tell: 10 m by 8 m, 16 m by 5 m and 20 m by 4 m all make 80 m², and so do endless others with decimals. One fact about two unknowns isn’t enough.

b. Now add a second fact: the room is 2 m longer than it is wide. Let the breadth be x m; the length is then (x + 2) m, and

$$\begin{aligned} (x + 2)x &= 80 \\ x^2 + 2x - 80 &= 0 \end{aligned}$$

You factorized expressions like this in Grade 9. Look for two numbers that multiply to −80 and add up to 2: they are 10 and −8.

$$\begin{aligned} x^2 + 10x - 8x - 80 &= 0 \\ x(x + 10) - 8(x + 10) &= 0 \\ (x + 10)(x - 8) &= 0 \end{aligned}$$

So x + 10 = 0, giving x = −10, or x − 8 = 0, giving x = 8. A room can’t be −10 m wide, so x = 8.

The room is 10 m long and 8 m wide. Check: 10 × 8 = 80 ✓, and 10 is 2 more than 8 ✓.

An equation in one variable whose highest power is 2 is a quadratic equation (Latin quadratus, a square: the variable is squared). Every quadratic equation can be written in the standard form

$$ax^2 + bx + c = 0 \qquad (a \ne 0)$$

where a, b and c are numbers. If a were 0 there would be no $x^2$ term, and the equation would be linear. The values of x that make the equation true are its roots (or solutions). A quadratic equation has at most two roots, and the equation itself can’t tell which of them a problem wants: the room’s equation is just as happy with −10 as with 8.

There are three ways to solve a quadratic equation, each built on the one before: factorization, completing the square, and the formula.

What a root means in a picture: draw the graph of $y = x^2 - 5x + 6$ and the roots of $x^2 - 5x + 6 = 0$ are where the curve meets the X-axis (Figure 7.2). A curve shaped like ∪ (or ∩, when a is negative) can cross the axis twice, touch it once, or miss it altogether, which is why a quadratic equation has two roots, one, or none. Optional Mathematics draws these graphs in detail.

a. Factorization method

The whole method rests on one fact about zero:

If the product of two numbers is 0, at least one of them is 0. If pq = 0, then p = 0 or q = 0.

Nothing like it is true for other numbers: pq = 6 could be 2 × 3, 1 × 6, $12 \times \tfrac12$ or a thousand other pairs. That is why the equation must have 0 on one side before you factorize.

Let the breadth be x m. Then the length is (2x + 1) m, and length × breadth = area:

$$\begin{aligned} (2x + 1)x &= 300 \\ 2x^2 + x - 300 &= 0 \end{aligned}$$

This time a = 2. To split the middle term, look for two numbers whose product is ac = 2 × (−300) = −600 and whose sum is b = 1. They are 25 and −24.

$$\begin{aligned} 2x^2 + 25x - 24x - 300 &= 0 \\ x(2x + 25) - 12(2x + 25) &= 0 \\ (2x + 25)(x - 12) &= 0 \end{aligned}$$

So 2x + 25 = 0, giving x = −12.5, or x − 12 = 0, giving x = 12. A breadth can’t be negative, so x = 12.

The playground is 12 m wide and 2 × 12 + 1 = 25 m long. Check: 25 × 12 = 300 ✓.

Why ac? If $ax^2 + bx + c$ factorizes as (px + q)(rx + s), multiplying out gives $prx^2 + (ps + qr)x + qs$. The two parts of the middle term, ps and qr, multiply to pqrs, which is (pr)(qs) = ac. So the search is always among the factor pairs of ac, a short list:

Worked examples: factorization

b. Completing the square

Some quadratics don’t factorize with whole numbers: try $x^2 + 6x - 2 = 0$, and no pair of whole numbers multiplies to −2 and adds up to 6. The second method works for every quadratic. It starts from the easiest kind of all.

a. $x^2 - 9$ is a difference of two squares, $x^2 - 3^2 = (x + 3)(x - 3)$, so x = −3 or 3. Or more quickly: $x^2 = 9 = 3^2$, so x = ±3 (“plus or minus 3”). Both 3 and −3 square to 9.

If $x^2 = a^2$, then x = ±a. The same is true of a bracket: if $(x - 5)^2 = 9$, then x − 5 = ±3.

b. So if the left side were a perfect square like $(x - \text{something})^2$, the equation would be easy. Move the 6 across, then add the square of half the coefficient of x to both sides. Half of −5 is $-\tfrac52$, and its square is $\tfrac{25}{4}$:

$$\begin{aligned} x^2 - 5x &= -6 \\ x^2 - 5x + \left(\tfrac52\right)^2 &= \left(\tfrac52\right)^2 - 6 \\ \left(x - \tfrac52\right)^2 &= \tfrac{25}{4} - \tfrac{24}{4} \\ \left(x - \tfrac52\right)^2 &= \left(\tfrac12\right)^2 \end{aligned}$$

The first line used $(a - b)^2 = a^2 - 2ab + b^2$: $x^2 - 5x$ is $x^2 - 2 \cdot x \cdot \tfrac52$, so it needs $\left(\tfrac52\right)^2$ to become a square. Now take square roots:

$$\begin{aligned} x - \tfrac52 &= \pm \tfrac12 \\ x &= \tfrac52 \pm \tfrac12 \end{aligned}$$

Taking +, x = 3; taking −, x = 2. The roots are 2 and 3, the same as by factorization.

Why “completing the square”? Picture $x^2 + 6x = 16$: the term $x^2$ is a square of side x, and 6x can be cut into two strips, each x by 3, laid along two of its sides (Figure 7.4). The L-shape is a square of side x + 3 with one small corner missing, 3 by 3. Add that corner, 9, to both sides and the square is complete:

$$\begin{aligned} x^2 + 6x + 9 &= 16 + 9 \\ (x + 3)^2 &= 25 \\ x + 3 &= \pm 5 \end{aligned}$$

so x = 2 or x = −8. (The picture shows only the positive root: a side can’t be negative, but an equation’s root can.)

The steps: make the coefficient of $x^2$ equal to 1 (divide by a); keep the x terms on one side; add the square of half the coefficient of x to both sides; write the left side as a square; take square roots, with ±; solve the two equations.

Worked examples: completing the square

c. Solving by formula

Completing the square works every time, so let us do it once and for all, with letters. The result is a formula that gives the roots of any quadratic equation at once.

Move c across, then divide by a:

$$\begin{aligned} ax^2 + bx &= -c \\ x^2 + \frac{b}{a}x &= -\frac{c}{a} \end{aligned}$$

Half of $\dfrac{b}{a}$ is $\dfrac{b}{2a}$. Add its square to both sides:

$$x^2 + 2 \cdot x \cdot \frac{b}{2a} + \left(\frac{b}{2a}\right)^2 = \left(\frac{b}{2a}\right)^2 - \frac{c}{a}$$

The left side is now a square. On the right, put everything over $4a^2$:

$$\begin{aligned} \left(x + \frac{b}{2a}\right)^2 &= \frac{b^2}{4a^2} - \frac{4ac}{4a^2} \\ \left(x + \frac{b}{2a}\right)^2 &= \frac{b^2 - 4ac}{4a^2} \end{aligned}$$

Take square roots (the square root of $4a^2$ is 2a), with ±:

$$\begin{aligned} x + \frac{b}{2a} &= \pm \frac{\sqrt{b^2 - 4ac}}{2a} \\ x &= -\frac{b}{2a} \pm \frac{\sqrt{b^2 - 4ac}}{2a} \end{aligned}$$

So the roots of $ax^2 + bx + c = 0$ are

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

that is, $\dfrac{-b + \sqrt{b^2 - 4ac}}{2a}$ and $\dfrac{-b - \sqrt{b^2 - 4ac}}{2a}$. This is the quadratic formula.

Worked examples: the formula

What $b^2 - 4ac$ tells you

The number under the square root, $b^2 - 4ac$, decides how many roots there are before you finish:

• positive: two different roots ($+\sqrt{\ }$ and $-\sqrt{\ }$ give different answers);
• zero: one root, repeated (±0 changes nothing). $x^2 - 6x + 9 = 0$ is $(x - 3)^2 = 0$: only x = 3, which the graph touches without crossing;
• negative: no real root, since no real number squares to a negative. $x^2 + x + 1 = 0$ has $b^2 - 4ac = 1 - 4 = -3$, and the graph of $y = x^2 + x + 1$ never reaches the X-axis.

Back to the road

The fastest speed at which the car stops before the ball is about 50.6 km/h.

Here is where that comes from. The car’s stopping distance has two parts. While the driver reacts (0.7 s for an alert driver), the car keeps going at full speed: at v meters per second it covers 0.7v m. Then it brakes. Braking at 6.5 m/s², a typical rate on a dry road, it covers $v^2/13$ m before it stops (twice the speed, four times the braking distance). These are the numbers in the opening’s widget, and they give close to the Highway Code’s own distances: at 30 miles an hour (about 48 km/h) it lists 9 m of thinking and 14 m of braking.

To stop in exactly 25 m:

$$\begin{aligned} 0.7v + \frac{v^2}{13} &= 25 \\ 9.1v + v^2 &= 325 \\ v^2 + 9.1v - 325 &= 0 \end{aligned}$$

No whole numbers multiply to −325 and add up to 9.1, so use the formula with a = 1, b = 9.1, c = −325:

$$\begin{aligned} v &= \frac{-9.1 \pm \sqrt{9.1^2 + 4 \times 325}}{2} \\ &= \frac{-9.1 \pm \sqrt{1382.81}}{2} \\ &= \frac{-9.1 \pm 37.19}{2} \end{aligned}$$

So $v \approx 14.04$ m/s or $v \approx -23.14$ m/s. A negative speed means nothing here, so $v \approx 14.04$ m/s, and $14.04 \times 3.6 \approx 50.6$ km/h.

What it means: in the graph of the opening’s widget, 50.6 km/h is where the curve crosses the 25 m line. At 60 km/h, under a fifth faster, the car needs about 33 m, because the braking distance grows with the square of the speed. On a wet road, or with a driver looking at a phone, the safe speed is much lower: Going Further’s widget lets you try both.

Question 1

Which of these are quadratic equations? Give a reason.

Question 2

Solve by factorization.

Question 3

Solve by completing the square.

Question 4

Solve by using the formula.

Questions 5 and 6

Questions to discuss

Most quadratic equations in real life arrive as words. Turning the words into an equation is the real skill; solving it is then routine.

The table finds the answer by trying, which works only because it is small. Algebra finds it directly. In x years Sumitra will be 12 + x and her sister 18 + x:

$$\begin{aligned} (12 + x)(18 + x) &= 280 \\ 216 + 30x + x^2 &= 280 \\ x^2 + 30x - 64 &= 0 \\ (x + 32)(x - 2) &= 0 \end{aligned}$$

x = −32 or x = 2. Thirty-two years ago neither sister was born, so x = 2: in 2 years the product of their ages will be 280, as the table showed.

Steps for a word problem:

1. Say what the letter stands for, with its unit (“let the breadth be x m”).
2. Write every other quantity in terms of it.
3. Write the equation the words give, and put it in standard form.
4. Solve it.
5. Keep only the roots that fit the story, and say why you reject the others.
6. Answer the question in a sentence, and check it against the words.

Worked examples: word problems

In the SEE’s Compulsory Mathematics paper, Algebra carries 15 marks in three questions, with knowledge, understanding, application and higher-ability parts spread among them (CDC’s specification grid). In CDC’s model paper the quadratic equation question is about a rectangular field:

Part	Level	Marks	What it asks
a.	knowledge	1	the standard form of a quadratic equation
b.	application	3	solve it: the length and breadth
c.	application	1	use the answer: how many smaller plots fit

Part a asks only for the general form; the equation from the story is part of b. Here is a question in that shape, and one to practice with, whose first part asks for the story’s own equation instead.

Questions 1 to 7: squares

Questions 8 to 12: numbers

Questions 13 to 15: ages

Question 16

Question 17

Questions 18 and 19