Many problems ask for less than they seem to. "Find r² + q², where r and q are the roots" does not need the roots. "Find the sum of the coefficients of (2x − 1)⁵" does not need the expansion. The tricks in this chapter all notice what the question really needs and go straight for it: the sum and the product of the roots, a symmetry, one well-chosen value of x. Each one is described as in the previous chapter: what it is, why it works, when to use it and the trap.
By the end you can
- Find symmetric expressions of the roots, such as r² + q² or 1/r + 1/q, from the coefficients alone.
- Solve a symmetric system by changing to s = x + y and p = xy.
- Use t = x + 1/x to get x² + 1/x² and to solve an equation whose coefficients read the same both ways.
- Get coefficient sums and remainders by plugging in x = 1, −1 or 0.
- Test an identity with numbers, and know when a few values are a proof and when they are not.
What it is, and why it works
If r and q are the roots of x² − s·x + p = 0, then r + q = s and r·q = p. You can read both straight off the equation. The reason: a quadratic with roots r and q is (x − r)(x − q), and multiplying it out gives x² − (r + q)x + rq. Comparing with x² − s·x + p, the middle coefficient is minus the sum and the last one is the product. For a·x² + b·x + c = 0, divide by a first: the sum is −b/a and the product is c/a. The Quadratics chapter proves this in full; here the point is what it lets you skip.
When to use it
The sign is a question about the roots that does not change if you swap them. r² + q² is the same as q² + r²; so are 1/r + 1/q, (r − q)² and r³ + q³. Such an expression is called symmetric, and every symmetric expression in r and q can be written using s and p alone. So you never need the roots, which may be ugly fractions with square roots, or not even real.
| You want | From s and p | How |
|---|---|---|
| r² + q² | s² − 2p | (r + q)² = r² + 2rq + q², take away 2rq |
| (r − q)² | s² − 4p | (r − q)² = r² + q² − 2rq; this is the discriminant (divided by a²) |
| 1/r + 1/q | s / p | common denominator: (q + r)/(rq) |
| r³ + q³ | s³ − 3ps | (r + q)³ = r³ + q³ + 3rq(r + q) |
| r²q + rq² | ps | take rq out: rq(r + q) |
| expression | from s and p | value | from the roots |
|---|---|---|---|
| r² + q² | s² − 2p | 17 | 17 |
| (r − q)² | s² − 4p | 9 | 9 |
| 1/r + 1/q | s / p | 1.25 | 1.25 |
| r³ + q³ | s³ − 3sp | 65 | 65 |
The purple curve is y = x² − s·x + p, whose roots are r and q, with s = r + q and p = rq. In the first mode drag the two dots along the axis; in the second set s and p yourself. The table works out four expressions in two ways: from s and p alone (amber formula), and directly from the roots (green). They always match, even when s² − 4p is negative and the curve misses the axis: then the roots are complex numbers α ± βi, and the expressions still come out real.
Two more uses. If you know one root, the product gives the other at once: 2x² − 7x + 3 = 0 has the root 3, and the product of the roots is 3/2, so the other root is (3/2)/3 = 1/2. And to build an equation with given roots, write x² − (sum)x + (product) = 0: the roots 2 and −5 come from x² + 3x − 10 = 0.
x + y = 5 and x² + y² = 13 is a system that does not change if you swap x and y. That symmetry is the sign. Change the unknowns to s = x + y and p = xy: the system becomes simple, and at the end x and y are the two roots of t² − s·t + p = 0, by trick 1. Why does this always work? Because any expression that does not change when x and y swap can be written using s and p; this fact is called the fundamental theorem of symmetric polynomials.
When x and 1/x appear together, call their sum t. Squaring t gives x² + 2 + 1/x², because the cross term is 2 · x · (1/x) = 2: the x cancels out of it. So x² + 1/x² = t² − 2, without ever finding x. Cubing in the same way gives x³ + 1/x³ = t³ − 3t.
The same t solves equations whose coefficients read the same from both ends, called palindromic: x⁴ + x³ − 4x² + x + 1 = 0 has coefficients 1, 1, −4, 1, 1. That is the sign. Divide by x², which is safe here because x = 0 is not a solution (the constant term is 1, not 0), so no solution is lost. The terms pair up as x² + 1/x² and x + 1/x, and a degree-4 equation becomes a quadratic in t.
Write a polynomial as P(x) = c₀ + c₁x + c₂x² + … + cₙxⁿ, where c₀, c₁, … are its coefficients. Every power of 1 is 1, so P(1) is the sum of all the coefficients. The powers of −1 alternate between +1 and −1, so P(−1) is the alternating sum c₀ − c₁ + c₂ − …. Every power of 0 above the zeroth is 0, so P(0) = c₀, the constant term. The sign to use it: a question about the coefficients of a big expansion, not the expansion itself.
The same idea appears elsewhere in the track. The remainder of P(x) divided by x − 1 is P(1), the remainder theorem of the polynomial-division chapter. And (1 + x)ⁿ at x = 1 says that the binomial coefficients C(n, 0) + C(n, 1) + … + C(n, n) add up to 2ⁿ.
Before trusting an identity or an answer, put numbers in. One value where the two sides differ is enough to show the identity is false: (a + b)² = a² + b² fails at a = b = 1, since 4 ≠ 2. Agreement at a few values usually only makes it likely. But for polynomials it can be a proof. The difference of the two sides, D(x), is again a polynomial. A polynomial of degree at most n that is not zero has at most n roots, since each root gives a factor (the factor theorem). So if both sides have degree at most n and agree at n + 1 different values, D has n + 1 roots, D must be the zero polynomial, and the identity holds for every x.
Which is bigger, 2³⁰⁰ or 3²⁰⁰? Neither fits on a calculator. Make the exponents equal: 300 and 200 share the factor 100, so 2³⁰⁰ = (2³)¹⁰⁰ = 8¹⁰⁰ and 3²⁰⁰ = (3²)¹⁰⁰ = 9¹⁰⁰. For positive numbers, raising to the same positive power keeps the order, and 8 < 9, so 3²⁰⁰ is bigger. When the exponents share no useful factor, take logarithms: the logarithm also keeps the order, and it turns a power into a product, log(aⁿ) = n · log a.
| Trick | The idea in one line | Chapter |
|---|---|---|
| Gauss pairing | 1 + 2 + … + 100: pair the first with the last, 1 + 100, 2 + 99, …: 50 pairs of 101 = 5050 | Sequences & Series |
| Telescoping | 1/(k(k + 1)) = 1/k − 1/(k + 1): in the sum every middle term cancels its neighbour, leaving 1 − 1/(n + 1) | Summation & Its Rules |
| Substitution | x⁴ − 13x² + 36 = 0: name the repeated block, u = x², and it is a quadratic | Equation Toolkit |
| When you see | Try | Why it works |
|---|---|---|
| an expression of the roots that does not change if they swap | sum and product | (x − r)(x − q) = x² − (r + q)x + rq |
| a system unchanged when x and y swap | s = x + y, p = xy | symmetric expressions are expressions in s and p |
| x and 1/x together, or coefficients that read the same both ways | t = x + 1/x | x · (1/x) = 1, so the cross terms are numbers |
| a question about the coefficients, not the expansion | plug in 1, −1 or 0 | 1ᵏ = 1, (−1)ᵏ = ±1, 0ᵏ = 0 |
| an identity or an answer to trust | test with numbers such as 2 and 3 | one failure disproves; for degree n, n + 1 agreements prove |
| two huge powers to compare | a common exponent, or logarithms | both keep the order of positive numbers |
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 |
|---|---|---|
| x² − 5x + 6: the roots add up to −5 | +5 | the sum is −b/a, and here b = −5 |
| 2x² − 6x + 4: the roots multiply to 4 | 4/2 = 2 | divide by the leading coefficient a |
| r² + q² = s² | s² − 2p | (r + q)² also contains the cross term 2rq |
| x² + 1/x² = t² | t² − 2 | the cross term 2 · x · (1/x) is 2, not 0 |
| checking only with x = 0 and x = 1 | also try 2 or 3 | x² = x holds at 0 and 1 but nowhere else |
| dividing a palindromic equation by x² without a thought | first check that x = 0 is not a root | dividing by an expression that can be 0 can lose a solution |
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
- The roots of a·x² + b·x + c add up to −b/a and multiply to c/a; any expression that does not change when they swap can be written with these two numbers.
- r² + q² = s² − 2p, (r − q)² = s² − 4p, 1/r + 1/q = s/p, r³ + q³ = s³ − 3ps; they hold for complex roots too.
- A system unchanged when x and y swap becomes simple in s = x + y and p = xy; at the end x and y are the roots of t² − st + p = 0.
- With t = x + 1/x: x² + 1/x² = t² − 2. Equations whose coefficients read the same both ways become quadratics in t, after dividing by x².
- P(1) is the sum of the coefficients, P(−1) the alternating sum and P(0) the constant term.
- One counterexample disproves an identity. For polynomials of degree at most n, agreement at n + 1 values proves it.
- Compare huge powers by making the exponents equal, or by comparing their logarithms.