$x+\frac{1}{x}$ type expressions — notes, question types & practice · Algebra

01

Overview

The whole subtopic is one machine. Because $x \times \frac{1}{x} = 1$ , the product of the two pieces is fixed, and every power sum becomes a short polynomial in $k$ :

x^2+\frac{1}{x^2} = k^2 - 2 \qquad x^3+\frac{1}{x^3} = k^3 - 3k \qquad x^4+\frac{1}{x^4} = (k^2-2)^2 - 2

There is also a multiply-down rung: $x^5+\frac{1}{x^5} = \left(x^2+\frac{1}{x^2}\right)\left(x^3+\frac{1}{x^3}\right) - k$ .

02

Watch the ladder work

Take $k = 3$ . The square rung gives $9 - 2 = 7$ . The cube rung gives $27 - 9 = 18$ . The fourth rung squares the second: $49 - 2 = 47$ . The fifth rung multiplies rung two by rung three and subtracts $k$ : $7 \times 18 - 3 = 123$ . Four answers, and $x$ itself never appeared.

Rule: Square for even rungs, use $k^3 - 3k$ for the cube, and chain rungs already computed instead of restarting.

03

The minus ladder

For $x - \frac{1}{x} = m$ , the cross term flips sign:

x^2+\frac{1}{x^2} = m^2 + 2 \qquad x^3-\frac{1}{x^3} = m^3 + 3m

With $m = 3$ : the square rung gives $9 + 2 = 11$ and the cube rung gives $27 + 9 = 36$ . The two ladders connect through $\left(x+\frac{1}{x}\right)^2 = \left(x-\frac{1}{x}\right)^2 + 4$ , so $k = \sqrt{m^2 + 4} = \sqrt{13}$ here.

Watch: The square rung of the minus ladder adds 2. Writing $m^2 - 2$ here is the single most common slip in this chapter.

04

Quadratics with equal end coefficients

An equation like $4x^2 - 9x + 4 = 0$ has matching first and last coefficients. Divide by $x$ : $4x - 9 + \dfrac{4}{x} = 0$ , so $x + \dfrac{1}{x} = \dfrac{9}{4}$ . Then the square rung gives $\dfrac{81}{16} - 2 = \dfrac{49}{16}$ . The same move on $3x^2 - 7x + 3 = 0$ gives $k = \dfrac{7}{3}$ and a square rung of $\dfrac{31}{9}$ .

05

Cyclic values

Two values of $k$ collapse the ladder entirely. If $k = 1$ , then $x^2 - x + 1 = 0$ ; multiplying by $(x+1)$ gives $x^3 = -1$ , so $x^6 = 1$ . If $k = -1$ , then $x^2 + x + 1 = 0$ and $x^3 = 1$ . Stems with giant exponents like $x^{30}$ reduce to small powers by taking the exponent modulo 3 or 6.

06

Mixed forms

For $2x + \dfrac{1}{2x} = 5$ , square carefully. The middle term is $2 \times 2x \times \dfrac{1}{2x} = 2$ , so $4x^2 + \dfrac{1}{4x^2} = 25 - 2 = 23$ . In general the middle term of $\left(px + \dfrac{1}{qx}\right)^2$ is $\dfrac{2p}{q}$ , not 2.

07

Combining the two ladders

Multiplying the sum ladder by the difference ladder kills the cross terms:

\left(x+\frac{1}{x}\right)\left(x-\frac{1}{x}\right) = x^2-\frac{1}{x^2}

From $m = 3$ above, $k = \sqrt{13}$ , so $x^2 - \dfrac{1}{x^2} = 3\sqrt{13}$ in one line. Any pair of $k$ and $m$ used together must first satisfy $k^2 - m^2 = 4$ ; an option that breaks this cannot be true for any real $x$ .

Tip: Plan the route before multiplying. For $x^4+\frac{1}{x^4}$ , square twice and never touch the cube rung; for the fifth rung, multiply rung two by rung three.

08

Question types you will see

Each type: how to recognise it, the method step by step, and one question to try.

Type 1very common4 practice Q

The ladder from x + 1/x = k

How to spot it:

A value of $x + \frac{1}{x}$ is given and a power sum of $x$ is asked.

x^3+\frac1{x^3} = k^3 - 3k

Method

Identify the wanted rung: square, cube, fourth or fifth.
Square rung: $k^2 - 2$ ; cube rung: $k^3 - 3k$ .
Fourth rung squares the square rung and subtracts 2.
Fifth rung multiplies rung two by rung three, minus $k$ .

Why it works:

The fixed product $x \times \frac{1}{x} = 1$ turns every rung into a polynomial in $k$ .

Try this

If $x + \dfrac{1}{x} = 5$ , find $x^3 + \dfrac{1}{x^3}$ .

Show solution

$k^3 = 125$ .
$3k = 15$ .
$125 - 15 = 110$ .

Answer

110

Type 2very common2 practice Q

The difference ladder from x - 1/x = m

How to spot it:

A value of $x - \frac{1}{x}$ is given and a power sum is asked.

x^3-\frac1{x^3} = m^3 + 3m

Method

Square rung: $m^2 + 2$ (the sign flips to plus).
Cube rung: $m^3 + 3m$ .
To reach the sum ladder, use $k^2 = m^2 + 4$ .

Why it works:

Squaring $x - \frac{1}{x}$ leaves a minus 2 cross term, so it moves across as plus 2.

Try this

If $x - \dfrac{1}{x} = 3$ , find $x^3 - \dfrac{1}{x^3}$ .

Show solution

$m^3 = 27$ .
$3m = 9$ .
$27 + 9 = 36$ .

Answer

36

Type 3very common2 practice Q

Quadratic with equal end coefficients

How to spot it:

An equation like $3x^2 - 8x + 3 = 0$ appears; a reciprocal power sum of its root is asked.

ax^2 - bx + a = 0 \Rightarrow x + \frac1x = \frac ba

Method

Check the first and last coefficients match.
Divide the whole equation by $x$ .
Read off $k = b/a$ .
Run the ladder for the asked rung.

Why it works:

Dividing by $x$ merges the end terms into $a\left(x + \frac{1}{x}\right)$ .

Try this

If $3x^2 - 8x + 3 = 0$ , find $x^2 + \dfrac{1}{x^2}$ .

Show solution

Divide by $3x$ : $x + \dfrac{1}{x} = \dfrac{8}{3}$ .
Square: $\dfrac{64}{9}$ .
Subtract 2: $\dfrac{64}{9} - \dfrac{18}{9} = \dfrac{46}{9}$ .

Answer

46/9

Type 4common2 practice Q

Cyclic values of the ladder

How to spot it:

$k$ equals 1 or -1, or the asked stem has huge exponents like $x^{18}$ or $x^{30}$ .

k = 1 \Rightarrow x^3 = -1;\qquad k = -1 \Rightarrow x^3 = 1

Method

Multiply $x^2 \mp x + 1 = 0$ by the complementary factor.
Conclude $x^3 = \mp 1$ .
Reduce every exponent modulo 3 or 6.
Substitute the reduced powers.

Why it works:

These $x$ values are cube roots of unity family, so powers repeat with period 3.

Try this

If $x + \dfrac{1}{x} = 1$ , find $x^3 + \dfrac{1}{x^3}$ .

Show solution

$k = 1$ , so $x^2 - x + 1 = 0$ and $x^3 = -1$ .
Then $\dfrac{1}{x^3} = -1$ too.
$x^3 + \dfrac{1}{x^3} = 1 - 3 = -2$ by the cube rung.

Answer

-2

Type 5common

Mixed reciprocal forms

How to spot it:

The given relation is $px + \frac{1}{qx}$ or similar, and the target is the squared form.

\left(px+\frac{1}{qx}\right)^2 = p^2x^2+\frac{1}{q^2x^2}+\frac{2p}{q}

Method

Square the given relation.
Compute the middle term as $2p/q$ .
Move it across to isolate the target.

Why it works:

Only the coefficients decide the middle term; the $x$ cancels inside the product.

Try this

If $2x + \dfrac{1}{2x} = 5$ , find $4x^2 + \dfrac{1}{4x^2}$ .

Show solution

Square: $4x^2 + \dfrac{1}{4x^2} + 2 \times 2x \times \dfrac{1}{2x} = 25$ .
The middle term is $2$ .
$4x^2 + \dfrac{1}{4x^2} = 25 - 2 = 23$ .

Answer

23

09

Formula sheet

Square rung

x^2+\frac{1}{x^2} = \left(x+\frac1x\right)^2-2 = \left(x-\frac1x\right)^2+2

Cube rung, sum

x^3+\frac{1}{x^3} = \left(x+\frac1x\right)^3 - 3\left(x+\frac1x\right)

Cube rung, difference

x^3-\frac{1}{x^3} = \left(x-\frac1x\right)^3 + 3\left(x-\frac1x\right)

Fourth rung

x^4+\frac{1}{x^4} = \left(x^2+\frac{1}{x^2}\right)^2 - 2

Fifth rung

x^5+\frac{1}{x^5} = \left(x^2+\tfrac1{x^2}\right)\left(x^3+\tfrac1{x^3}\right)-\left(x+\tfrac1x\right)

Ladders link

\left(x+\frac1x\right)^2 - \left(x-\frac1x\right)^2 = 4

Equal-end quadratic

ax^2 - bx + a = 0 \Rightarrow x+\frac1x = \frac ba

Mixed form

\left(px+\frac{1}{qx}\right)^2 = p^2x^2+\frac{1}{q^2x^2}+\frac{2p}{q}

10

Shortcuts that save time

⚡ Run the ladder in order

Square rung first, then cube rung, then reuse them for higher rungs. Never restart from k for each question part.

Example

If x + 1/x = 4, find x^2 + 1/x^2, x^4 + 1/x^4 and x^3 + 1/x^3.

Show solution

$x^2+\dfrac{1}{x^2} = 16 - 2 = 14$ .
$x^4+\dfrac{1}{x^4} = 14^2 - 2 = 194$ .
$x^3+\dfrac{1}{x^3} = 64 - 12 = 52$ .

Answer

14, 194, 52

⚡ Divide the quadratic by x

Equal first and last coefficients mean the quadratic is a k-value in disguise. Divide by x and read it off.

Example

If 3x^2 - 7x + 3 = 0, find x^2 + 1/x^2.

Show solution

Divide by $3x$ : $x + \dfrac{1}{x} = \dfrac{7}{3}$ .
Square rung: $\dfrac{49}{9} - 2$ .
$= \dfrac{31}{9}$ .

Answer

31/9

⚡ Switch ladders with one square

The two ladders differ by 4 under a square: k squared equals m squared plus 4. Convert once, then stay on the new ladder.

Example

If x - 1/x = 3, find the positive value of x + 1/x.

Show solution

$k^2 = m^2 + 4$ .
$k^2 = 9 + 4 = 13$ .
$k = \sqrt{13}$ (positive).

Answer

sqrt(13)

11

Mistakes to avoid

Where most students lose marks on this subtopic.

Mistake 01

Subtracting 2 on the minus ladder's square rung.

The minus ladder adds 2: $m^2 + 2$ .

Mistake 02

Writing the cube rung as $k^3 - 3$ .

It is $k^3 - 3k$ ; the k must be multiplied back.

Mistake 03

Dividing a quadratic by x when the end coefficients differ.

The equal-ends trick needs $ax^2 + bx + a$ exactly.

Mistake 04

Using a middle term of 2 when squaring $2x + 1/(2x)$ .

The middle term is $2p/q$ ; here it is 2 only because p equals q.

Mistake 05

Expanding $x^{30}$ -type stems directly.

Reduce exponents modulo 3 or 6 first when k is 1 or -1.

12

Quick revision

Read this the night before the exam.

$k = x + \dfrac1x$ : square rung $k^2 - 2$ , cube rung $k^3 - 3k$ .
$m = x - \dfrac1x$ : square rung $m^2 + 2$ , cube rung $m^3 + 3m$ .
$k^2 = m^2 + 4$ connects the two ladders.
$ax^2 - bx + a = 0$ gives $k = b/a$ after dividing by $x$ .
$k = 1$ gives $x^3 = -1$ ; $k = -1$ gives $x^3 = 1$ .
Mixed form $px + \dfrac{1}{qx}$ : middle term is $\dfrac{2p}{q}$ .

13

Practice: 14 questions

Sets of 10, mixed across the question types above. Every answer has a step-by-step explanation.

Topic test · 10 questions

Suggested time 6 min · wrong answers go to your mistake notebook automatically.