A trick in algebra is not magic and not a rule to memorise. Almost every one of them is one of two harmless moves: multiply by 1, or add 0. Neither changes the value of anything, so both are allowed anywhere, at any time. What they change is the shape, and the right shape makes a root disappear, a fraction split, or a hard product easy to do in your head. This chapter takes the most useful tricks of that kind one by one, starting with multiplying by the conjugate, and for each one says what it is, why it works, when to reach for it and where it can go wrong.
By the end you can
- Explain why multiplying by 1 or adding 0 is always allowed, and what it is good for.
- Multiply by the conjugate to remove square roots, and say why squaring would not do the same job.
- Rationalise a denominator with one root, with a sum or difference of roots, and with a cube root.
- Add and subtract the same term to complete a square, split a fraction or factor x⁴ + 4.
- Use the notable products backwards to multiply and square numbers in your head.
Every trick in this chapter, and in the other tricks chapters of the track, is described in the same four parts. What it is: the move, in one line. Why it works: a short proof, so you can rebuild the trick if you forget it, and trust it. When to use it: the sign in the problem that tells you this trick will help, because knowing a trick is useless if you never notice the moment for it. The trap: the one way it usually goes wrong.
| Part | The question it answers |
|---|---|
| What it is | What exactly do I write? |
| Why it works | Why is the result equal to what I started with? |
| When to use it | What in the problem tells me to try this? |
| The trap | How does it usually go wrong? |
1 is the number that leaves everything alone under multiplication, and 0 is the one that leaves everything alone under addition. So you may multiply any expression by 1 and add 0 to it, and it stays the same expression. The point is to write the 1 and the 0 in a useful disguise: 1 as c/c for some well-chosen c that is not zero, and 0 as c − c. This is different from the rule for equations, "do the same to both sides". Here nothing is done to the other side, because nothing has changed. That is why these moves also work on a single expression that is not part of any equation.
What it is
The conjugate of a sum of two terms, a + b, is the same two terms with the sign between them flipped, a − b. The conjugate of a − b is a + b. The terms themselves are not changed, only the sign that joins them. The trick is to multiply by the conjugate when a square root is in the way, because the product of a pair of conjugates has no square root left in it. In a fraction, you multiply the top and the bottom by it, so you are multiplying by 1.
| Expression | Its conjugate | The product |
|---|---|---|
| 3 − √2 | 3 + √2 | 9 − 2 = 7 |
| √x + 1 | √x − 1 | x − 1 |
| √5 + √3 | √5 − √3 | 5 − 3 = 2 |
| 2√7 − 5 | 2√7 + 5 | 28 − 25 = 3 |
| a + bi | a − bi | a² + b² (complex numbers, since i² = −1) |
Why it works
Multiply out (a + b)(a − b) term by term. You get a² − ab + ab − b². The two cross terms, −ab and +ab, are opposites and cancel, leaving a² − b², the difference of squares. Now let one of the terms be a square root. The root only appears in the cross terms, and they are gone. What is left are the squares, and squaring a square root removes it: (√2)² = 2.
Why not simply square the expression? Because the square of a sum keeps its cross term: (a + b)² = a² + 2ab + b². The cross term 2ab still holds the root. Only the conjugate makes the two cross terms opposite instead of equal. The comparison below uses 3 + √2.
First panel: a square of side a with a corner of side b cut out has area a² − b². Press ▶: the L shape is cut into two strips and one is turned and moved, giving a rectangle (a + b) by (a − b) with the same area. Second panel: (a + √n) is multiplied by itself and by its conjugate (a − √n) in a 2 × 2 grid, one cell per pair of terms. The root lives only in the two red cross cells. Squared, they add up; with the conjugate they are opposite and cancel, leaving a² − n with no root.
When to use it
| The sign in the problem | What to multiply by |
|---|---|
| a sum or difference with a root in a denominator, 1/(3 − √2) | top and bottom by the conjugate of the denominator |
| a difference of two roots in an equation, √(x + 5) − √x = 1 | both sides by the sum of the roots |
| a limit that gives 0/0 and has a root, (√(x + 1) − 1)/x at x = 0 | top and bottom by the conjugate of the part with the root |
| two nearly equal roots subtracted, √10001 − √10000 | top and bottom by their sum, to turn the difference into a quotient |
| a complex number in a denominator, 1/(2 + i) | top and bottom by the conjugate, 2 − i |
Use 1: rationalising a denominator
A fraction with a square root in its denominator is usually rewritten without it; the step is called rationalising, because the denominator becomes a rational number. Without the root below, fractions are easier to compare, add and estimate: 1/√2 means "1 divided by 1.41421…", while √2/2 is simply half of 1.41421…. A single root needs only itself: 1/√2 = √2/2, because √2 · √2 = 2. A sum or difference with a root needs its conjugate.
Use 2: an equation with a difference of roots
In √(x + 5) − √x = 1, squaring right away leaves a cross term with a root in it, exactly as in the comparison above. The conjugate gives a cleaner route. Multiply both sides by √(x + 5) + √x. This is an equation, so here the same thing is done to both sides, and that is only safe if the factor is not zero (the Equation Toolkit chapter explains why). This sum is positive for every allowed x, at least √5, so the move is safe. On the left, the difference of squares leaves (x + 5) − x = 5, and the x disappears.
Use 3: two nearly equal roots
How big is √10001 − √10000? Both roots are about 100, so the difference is tiny, and subtracting two nearly equal numbers is where a calculator, or a computer, loses digits: if each root is known to 6 digits, the difference may keep only one or two of them right. The conjugate turns the subtraction into a division, which loses nothing. It also gives a quick estimate: when n is large, √(n + 1) and √n are both about √n, so their sum is about 2√n.
Use 4: cube roots need a different partner
The plain conjugate does not remove a cube root: (∛2 − 1)(∛2 + 1) = ∛4 − 1, still a root. The idea is the same, but the identity has to produce cubes instead of squares. The difference of cubes is a³ − b³ = (a − b)(a² + ab + b²), which you can check by multiplying out: a³ + a²b + ab² − a²b − ab² − b³, and four terms cancel. So the partner of a − b, for cubes, is a² + ab + b². For a sum, a³ + b³ = (a + b)(a² − ab + b²).
The traps
Where else the conjugate appears
| Where | What it removes | Chapter |
|---|---|---|
| Dividing complex numbers, (5 + 5i)/(1 + 2i) | the i in the denominator, since (1 + 2i)(1 − 2i) = 1 + 4 = 5 | Complex Numbers |
| A limit that gives 0/0, such as (√(x + 1) − 1)/x at x = 0 | the root that keeps the 0/0, so a factor x can cancel | Limits & Continuity |
| The derivative of √x from its definition | the difference √(x + h) − √x on top | The Derivative |
| Trigonometry: (1 − cos x)(1 + cos x) | the 1 − cos x: the product is 1 − cos²x = sin²x | Identities & Rotation |
The conjugate is only one way to write 1. Two others come up all the time. The first clears fractions that sit inside a fraction: multiply the top and the bottom by the inner denominator. When you see a small fraction inside a big one, this is the move. The trap is the same as before: every term on top and every term below gets multiplied, and the inner denominator must not be zero.
The second is unit conversion. 1000 m and 1 km are the same length, so the fraction 1000 m / 1 km is equal to 1, and so is 1 h / 3600 s. Multiplying by them changes the units, not the speed. Choose each fraction so the unit you want to remove appears once on top and once below, and cancels.
Adding c and taking c away again adds 0. The art is choosing c so that one copy of it completes a pattern you know and the other copy is left over, harmless. The pattern can be a perfect square, a copy of the denominator, or a difference of squares.
Completing the square
x² + 6x + 5 is not a perfect square, but x² + 6x + 9 is: it is (x + 3)². So add 9 to complete it, and subtract 9 to stay honest. The number to add is always (half of the x coefficient)². The sign that tells you to try this: you need the smallest or largest value of a quadratic, its vertex, or you want to solve it without the formula. The Quadratics chapter builds this step as a picture and uses it to prove the quadratic formula.
Splitting a fraction
In (x + 3)/(x + 1), x appears twice, on top and below, and it is hard to see how the value behaves. Write the top as the denominator plus what is left: x + 3 = (x + 1) + 2, which is adding and subtracting 1. Now the fraction splits into 1 + 2/(x + 1), and x appears once. At a glance, the value is a little more than 1 when x is large and positive, and it is never exactly 1. This is the step that later turns many integrals into easy ones.
Factoring x⁴ + 4
x⁴ + 4 looks impossible to factor: it is a sum, and it is never zero for a real x. Adding and subtracting 4x² makes it a perfect square minus a square, and a difference of squares factors at once. This is called the Sophie Germain identity. It shows that n⁴ + 4 is never a prime number for a whole number n > 1, because both factors are then bigger than 1: n² − 2n + 2 = (n − 1)² + 1 > 1.
The notable products are usually read from left to right, to expand brackets. Read from right to left they become tools for mental arithmetic. The sign: two numbers at the same distance from a round number, or a number close to a round one that has to be squared. Write the numbers as round ± small, then use the identity; the round part is easy to square and the small part is small.
| Identity | In your head |
|---|---|
| (a + b)(a − b) = a² − b² | 47 · 53 = 50² − 3² = 2500 − 9 = 2491 |
| (a − b)² = a² − 2ab + b² | 98² = 100² − 2 · 100 · 2 + 2² = 10000 − 400 + 4 = 9604 |
| (a + b)² = a² + 2ab + b² | 203² = 40000 + 1200 + 9 = 41209 |
| a² − b² = (a + b)(a − b) | 1001² − 999² = 2000 · 2 = 4000 |
| (10a + 5)² = 100 · a(a + 1) + 25 | 65² = 100 · 6 · 7 + 25 = 4225 |
| When you see | Try | Why it works |
|---|---|---|
| one root below, 6/√3 | multiply top and bottom by the root | √a · √a = a |
| a sum or difference with a root, 1/(√5 − 2) | multiply by the conjugate | (a + b)(a − b) = a² − b² |
| a cube root in a difference, 1/(∛3 − 1) | multiply by a² + ab + b² | (a − b)(a² + ab + b²) = a³ − b³ |
| a fraction inside a fraction | multiply top and bottom by the inner denominator | it is c/c = 1 |
| the smallest value, or the vertex, of a quadratic | complete the square | + (b/2)² − (b/2)² = 0; a square is ≥ 0 |
| x on top and below, (x + 3)/(x + 1) | make a copy of the denominator on top | add and subtract, then split the sum |
| a sum of two squares that will not factor | add and subtract the missing middle term | it becomes a difference of squares |
| numbers close to a round number | round ± small, then a notable product | the round part is easy, the small part is small |
A round of 10 questions
Answer as many as you can. Mistakes come back at the end for a review round.
Knowing a trick is half of it; the other half is seeing when it applies. Each question is a problem and what it asks for. Pick the trick that opens it. A wrong answer shows the right trick and one sentence on why it works there. The questions come in a new order every round.
| Wrong | Right | Why |
|---|---|---|
| 1/(3 − √2) → 1/(3 − √2) · (3 + √2) | · (3 + √2)/(3 + √2) | multiplying only the bottom, or only the top, changes the value; c/c is what equals 1 |
| conjugate of √(x − 1) + 2 is √(x + 1) + 2 | √(x − 1) − 2 | flip the sign between the terms, never inside them |
| (√a + √b)² = a + b | a + 2√(ab) + b | squaring keeps the cross term; only the conjugate cancels it |
| 1/(√a + √b) = 1/√a + 1/√b | multiply by the conjugate | a fraction does not split over a sum in its denominator: 1/(1 + 1) = 1/2, not 2 |
| (∛2 − 1)(∛2 + 1) = 1 | ∛4 − 1 | for cube roots the partner is a² + ab + b², not a + b |
| x² + 6x + 5 = (x + 3)² + 5 | (x + 3)² − 4 | the 9 that completes the square must also be taken away |
A round of 10 questions
Answer as many as you can. Mistakes come back at the end for a review round.
Every exercise is made on the spot, so a new round is never the same. The computer picks the answer first and builds the question around it, working with exact fractions, so the answer key is always right. Answers are read exactly: 0.75, 3/4 and 6/8 count as the same number (where lowest terms are asked for, 6/8 just gets a reminder). A wrong answer shows the worked solution line by line, in the order the lesson does it; the new number in each line is coloured.
Key ideas
- Multiplying by 1 (as c/c, with c ≠ 0) and adding 0 (as c − c) never change a value, only its shape; they work on any expression.
- The conjugate of a + b is a − b. Their product a² − b² has no cross term, so a square root in a or b disappears; squaring would keep the cross term 2ab.
- In a fraction, multiply top and bottom by the conjugate; in an equation, multiply both sides, and check the factor is not zero.
- A difference of nearly equal roots is better written as 1 over their sum: no digits are lost, and it gives the estimate 1/(2√n).
- For cube roots the partner of a − b is a² + ab + b², from a³ − b³ = (a − b)(a² + ab + b²).
- Adding and subtracting the same term completes a square, makes a copy of a denominator, or turns a sum like x⁴ + 4 into a difference of squares.
- The notable products read backwards are mental arithmetic: 47 · 53 = 50² − 3², 98² = (100 − 2)².