x + 1/x: Cycles, Limits and Several Variables

SSC CGL · CHSL · CPO — method by method

4 Part30 worked examples101 practice questions

The previous chapter (“x + 1/x Type Expressions”) taught climbing the ladder: $x + \frac{1}{x}$ to $x^{2} + \frac{1}{x ^{2}}$ , $x^{3} + \frac{1}{x ^{3}}$ … but climbing to $x^{100}$ is hopeless. For a few special values ( $x + \frac{1}{x} = 1, - 1, 0, 2, 3$ ) the ladder values start repeating, and then any huge power is just a counting question inside the cycle. This chapter covers that cycle, then limit cases such as $x + \frac{1}{x} = 2$ and minimum values, and finally $\frac{a}{b} + \frac{b}{a}$ and three-variable questions.

Before thisMake the ladder of that chapter (square, cube, multiply) solid first; here we take it as known.

Cycles

Special values · huge powers ·

x^{2} + x + 1 = 0

The repeating of powers at special values of x + 1/x.

Three methods: spotting the cycle, bringing a huge power into the cycle, and finding the value of a polynomial from an equation like $x^{2} + x + 1 = 0$ .

Special values and the cycle of powers

a_{n} = \pm 2 \Rightarrow x^{n} = \pm 1

The cycle in which powers go round for $x + \frac{1}{x} = 1, - 1, 0, 2, 3$ .

The idea

Take $x + \frac{1}{x} = 1$ . Multiply by $x$ : $x^{2} + 1 = x$ , i.e. $x^{2} - x + 1 = 0$ . Now multiply by $(x + 1)$ : $(x + 1) (x^{2} - x + 1) = x^{3} + 1 = 0$ . So $x^{3} = - 1$ , and then $x^{6} = 1$ : after $6$ steps the powers are back where they started. That is the story of every special value. The direct rule: if some rung has $a_{n} = 2$ then $x^{n} + \frac{1}{x ^{n}} - 2 = 0$ , i.e. $(x^{n} - 1)^{2} = 0$ and $x^{n} = 1$ ; if $a_{n} = - 2$ then $x^{n} = - 1$ ; and if $a_{n} = 0$ then $x^{2 n} = - 1$ .

∑ Formula

a_{n} = x^{n} + \frac{1}{x ^{n}} : a_{n} = 2 \Rightarrow x^{n} = 1, a_{n} = - 2 \Rightarrow x^{n} = - 1, a_{n} = 0 \Rightarrow x^{2 n} = - 1

Reading the cycle off the ladder

$a_{n}$: the $n$ -th rung of the ladder
$n$: the power

◭ Diagram

x + \frac{1}{x} = 1

: the powers of

x

go round a circle in

6 0^{\circ}

steps and are back after

6

steps;

x^{n} + \frac{1}{x ^{n}}

is twice the horizontal position of that point. (The picture only explains the cycle.)

◭ Diagram

x + \frac{1}{x} = 3

: the step is

3 0^{\circ}

, so a cycle of

12

steps;

x^{6} = - 1

and

x^{12} = 1

$x + \frac{1}{x}$	Equation	Key relation	Cycle	$a_{0}, a_{1}, a_{2}, \dots$ (one cycle)
$1$	$x^{2} - x + 1 = 0$	$x^{3} = - 1$	$6$	$2, 1, - 1, - 2, - 1, 1$
$- 1$	$x^{2} + x + 1 = 0$	$x^{3} = 1$	$3$	$2, - 1, - 1$
$0$	$x^{2} + 1 = 0$	$x^{2} = - 1$	$4$	$2, 0, - 2, 0$
$2$	$x^{2} - 2 x + 1 = 0$	$x^{4} = - 1$	$8$	$2, 2, 0, - 2, - 2, - 2, 0, 2$
$3$	$x^{2} - 3 x + 1 = 0$	$x^{6} = - 1$	$12$	$2, 3, 1, 0, - 1, - 3, - 2, - 3, - 1, 0, 1, 3$
$\frac{5 - 1}{2}$	$x^{2} - \frac{5 - 1}{2} x + 1 = 0$	$x^{5} = 1$	$5$	$2, \frac{- 1 + 5}{2}, \frac{- 1 - 5}{2}, \frac{- 1 - 5}{2}, \frac{- 1 + 5}{2}$

Here: the first column is the given value, the second the equation it makes, the third the power at which $\pm 1$ first appears, the fourth the length of the cycle (after how many powers everything repeats) and the last column the $a_{n}$ values of one cycle. $a_{0} = 2$ always.

$x + \frac{1}{x} = 2$ and $- 2$ are special too, but they come in later methods ( $x = 1$ and $x = - 1$ ). The four commonest special values in exams are $1$ (cycle $6$ ), $- 1$ (cycle $3$ ), $3$ (cycle $12$ ) and $0$ or $2$ (cycle $4$ or $8$ ).

Spot itSpot it: the value of

x + \frac{1}{x}

lies between

- 2

and

2

(like

\pm 1, 0, \pm 2, \pm 3

) and the power is large. Count the ladder values first; the cycle shows at the first

\pm 2

Steps

Compute the ladder values $a_{1}, a_{2}, \dots$ in order ( $a_{n + 1} = s a_{n} - a_{n - 1}$ ).
At the first $n$ with $a_{n} = \pm 2$ write $x^{n} = \pm 1$ (if a zero appears, $x^{2 n} = - 1$ ).
Take powers of that relation and carry on.

Example 1 · Basic

x + \frac{1}{x} = 1

, find

x^{3}

and

x^{6}

Solution, step by step

1The ladder: $a_{1} = 1$ , $a_{2} = 1^{2} - 2 = - 1$ , $a_{3} = 1^{3} - 3 \times 1 = - 2$ .
2 $a_{3} = - 2$ means $x^{3} + \frac{1}{x ^{3}} = - 2$ , i.e. $(x^{3} + 1)^{2} = 0$ , so $x^{3} = - 1$ .
3Squaring gives $x^{6} = (- 1)^{2} = 1$ .

Answer -1 ; 1

Alternative: multiplying

x^{2} - x + 1 = 0

(x + 1)

also gave

x^{3} + 1 = 0

Example 2 · Standard

x + \frac{1}{x} = 3

, find the cycle and then

x^{12}

Solution, step by step

1 $a_{1} = 3$ , $a_{2} = 3 - 2 = 1$ , $a_{3} = 3 \times 1 - 3 = 0$ .
2 $a_{3} = 0$ means $x^{3} + \frac{1}{x ^{3}} = 0$ , i.e. $x^{6} + 1 = 0$ and $x^{6} = - 1$ .
3So $x^{12} = (x^{6})^{2} = (- 1)^{2} = 1$ ; the cycle is $12$ powers long.

Answer 1

Note: the first

\pm 2

appears at

a_{6} = - 2

, which is exactly

x^{6} = - 1

Example 3 · Standard

x + \frac{1}{x} = 0

, find

x^{2}

and

x^{4} + \frac{1}{x ^{4}}

Solution, step by step

1 $a_{1} = 0$ , so $a_{2} = 0^{2} - 2 = - 2$ .
2 $a_{2} = - 2$ means $x^{2} = - 1$ .
3 $x^{4} = 1$ , so $x^{4} + \frac{1}{x ^{4}} = 1 + 1 = 2$ .

Answer -1 ; 2

No real

x

has

x^{2} = - 1

, but the algebra formulas still work.

Example 4 · Advanced

x + \frac{1}{x} = 2

, find

x^{4}

x^{8}

and

x^{12} + \frac{1}{x ^{12}}

Solution, step by step

1 $a_{1} = 2$ , $a_{2} = 2 - 2 = 0$ . A zero appeared, so $x^{2} + \frac{1}{x ^{2}} = 0$ , i.e. $x^{4} = - 1$ .
2 $x^{8} = (x^{4})^{2} = 1$ .
3 $x^{12} = x^{8} \cdot x^{4} = 1 \times (- 1) = - 1$ , and $\frac{1}{x ^{12}} = - 1$ , so $x^{12} + \frac{1}{x ^{12}} = - 2$ .

Answer -1 ; 1 ; -2

12 = 8 + 4

: you can reach any power by adding powers of the cycle.

Example 5 · Advanced

x + \frac{1}{x} = \frac{5 - 1}{2}

, find

x^{5}

Solution, step by step

1 $a_{1} = \frac{5 - 1}{2}$ ; $a_{2} = a_{1}^{2} - 2 = \frac{3 - 5}{2} - 2 = \frac{- 1 - 5}{2}$ .
2 $a_{3} = a_{1} a_{2} - a_{1} = \frac{- 1 - 5}{2}$ , $a_{4} = a_{1} a_{3} - a_{2} = \frac{- 1 + 5}{2}$ .
3 $a_{5} = a_{1} a_{4} - a_{3} = 2$ . This is $2$ , so $x^{5} = 1$ : the cycle is $5$ long.

Answer 1

This five-cycle is rare; the ladder method still identifies it every time.

Practice

BasicTry it yourself first

x + \frac{1}{x} = 1

, then

x^{3}

equals:

BasicTry it yourself first

x + \frac{1}{x} = - 1

, then

x^{3}

equals:

StandardTry it yourself first

x + \frac{1}{x} = 3

, then

x^{6}

equals:

StandardTry it yourself first

x + \frac{1}{x} = 0

, then

x^{6} + \frac{1}{x ^{6}}

equals:

StandardTry it yourself first

x + \frac{1}{x} = 2

, then

x^{4}

equals:

StandardTry it yourself first

x + \frac{1}{x} = 3

, then

x^{5} + \frac{1}{x ^{5}}

equals:

AdvancedTry it yourself first

x + \frac{1}{x} = 3

, then

x^{18}

equals:

AdvancedTry it yourself first

x + \frac{1}{x} = 1

, then

x^{4} + \frac{1}{x ^{4}}

equals:

AdvancedTry it yourself first

x + \frac{1}{x} = 0

, then

x^{6}

equals:

Huge powers: bring them into the cycle

x^{N} \to x^{N mod P}

Questions like $x^{100} + \frac{1}{x ^{100}}$ without climbing the ladder.

The idea

When $x^{P} = 1$ (cycle length $P$ ), everything repeats after every $P$ powers. So write $N = P \times q + r$ ( $r$ is the remainder of $N \div P$ ); then $x^{N} = (x^{P})^{q} x^{r} = x^{r}$ . A huge power $N$ is done with the small power $r$ : $x^{N} + \frac{1}{x ^{N}} = a_{r}$ . It is the same “clock arithmetic” in which $13$ o’clock means $1$ o’clock. The cycle length comes from the table of the previous method.

∑ Formula

x^{P} = 1 \Rightarrow x^{N} + \frac{1}{x ^{N}} = a_{N mod P}

Bringing the power into the cycle

$N$: the given (large) power
$P$: the length of the cycle
$N mod P$: the remainder of $N \div P$
$a_{r}$: the $r$ -th rung of the ladder

$x + \frac{1}{x}$	cycle $P$
$1$	$6$
$- 1$	$3$
$3$	$12$
$0$	$4$
$2$	$8$

Here: the given value is on the left and its cycle on the right. Divide by $P$ , take the remainder, then read that $a_{r}$ in the table of the previous method.

The second form: a bare $x^{N}$ (without $\frac{1}{x ^{N}}$ ) is reduced by the cycle too. If $x^{3} = - 1$ then $x^{3 k} = (- 1)^{k}$ and $x^{n + 3} = - x^{n}$ ; so $x^{16} + x^{13} = x^{3} x^{13} + x^{13} = - x^{13} + x^{13} = 0$ . If $x^{3} = 1$ then $x^{n + 3} = x^{n}$ and a whole cycle of consecutive powers adds to $1 + x + x^{2} = 0$ .

Squaring roadA second road when

N

is a power of two (

64

512

): square again and again,

a_{2 n} = a_{n}^{2} - 2

. The value gets stuck at

- 1

(

(- 1)^{2} - 2 = - 1

) and at

2

(

2^{2} - 2 = 2

Steps

Get the cycle length $P$ from the ladder (first time $a_{P} = 2$ ).
Divide $N$ by $P$ and take the remainder $r$ .
The answer is $a_{r}$ (for a bare $x^{N}$ it is $x^{r}$ , with the sign of the cycle).

Example 6 · Basic

x + \frac{1}{x} = 1

, find

x^{100} + \frac{1}{x ^{100}}

Solution, step by step

1The cycle is $6$ powers long: $a_{0}, \dots, a_{5} = 2, 1, - 1, - 2, - 1, 1$ and then $a_{6} = 2$ .
2 $100 = 6 \times 16 + 4$ , so $x^{100} = x^{4}$ .
3 $a_{4} = - 1$ , so the answer is $- 1$ .

Answer -1

The

96

powers (a multiple of

6

) make no difference.

Example 7 · Standard

x + \frac{1}{x} = - 1

, find

x^{50} + \frac{1}{x ^{50}}

Solution, step by step

1 $a_{1} = - 1$ , $a_{2} = 1 - 2 = - 1$ , $a_{3} = (- 1) (- 1) - (- 1) = 2$ . So $x^{3} = 1$ and the cycle is $3$ long.
2 $50 = 3 \times 16 + 2$ , so $x^{50} = x^{2}$ .
3 $a_{2} = - 1$ , so the answer is $- 1$ .

Answer -1

For a cycle of

3

look at

N mod 3

: remainder

0

gives

2

, otherwise

- 1

Example 8 · Standard

x + \frac{1}{x} = 3

, find

x^{150} + \frac{1}{x ^{150}}

and

x^{200} + \frac{1}{x ^{200}}

Solution, step by step

1The cycle is $12$ long: $a_{0}, \dots, a_{11} = 2, 3, 1, 0, - 1, - 3, - 2, - 3, - 1, 0, 1, 3$ .
2 $150 = 12 \times 12 + 6$ , so $a_{150} = a_{6} = - 2$ .
3 $200 = 12 \times 16 + 8$ , so $a_{200} = a_{8} = - 1$ .

Answer -2 ; -1

a_{6} = - 2

means

x^{150} = - 1

: at half a cycle the sign flips.

Example 9 · Advanced

x + \frac{1}{x} = 1

, find

x^{25} + x^{26} + x^{27} + x^{28} + x^{29} + x^{30}

Solution, step by step

1 $x^{3} = - 1$ . So $x^{28} = x^{3} x^{25} = - x^{25}$ , i.e. $x^{25} + x^{28} = 0$ .
2In the same way $x^{26} + x^{29} = 0$ and $x^{27} + x^{30} = 0$ .
3All three pairs are zero, so the total is $0$ .

Answer 0

Powers

3

apart make pairs, because

x^{3} = - 1

Example 10 · Advanced

x + \frac{1}{x} = 2

, find

x^{64} + \frac{1}{x ^{64}}

Solution, step by step

1Squaring road: $a_{2} = (2)^{2} - 2 = 0$ .
2 $a_{4} = 0^{2} - 2 = - 2$ , $a_{8} = (- 2)^{2} - 2 = 2$ .
3The value sticks at $2$ : $a_{16} = 2^{2} - 2 = 2$ , and likewise $a_{32} = a_{64} = 2$ .

Answer 2

Cycle road:

64 = 8 \times 8

, so

x^{64} = 1

and the answer is

2

Practice

BasicTry it yourself first

x + \frac{1}{x} = 1

, then

x^{150} + \frac{1}{x ^{150}}

equals:

BasicTry it yourself first

x + \frac{1}{x} = - 1

, then

x^{101} + \frac{1}{x ^{101}}

equals:

StandardTry it yourself first

x + \frac{1}{x} = 3

, then

x^{47} + \frac{1}{x ^{47}}

equals:

StandardTry it yourself first

x + \frac{1}{x} = 0

, then

x^{102} + \frac{1}{x ^{102}}

equals:

StandardTry it yourself first

x + \frac{1}{x} = 2

, then

x^{19} + \frac{1}{x ^{19}}

equals:

StandardTry it yourself first

x + \frac{1}{x} = 1

, then

x^{99} + x^{96}

equals:

AdvancedTry it yourself first

x + \frac{1}{x} = 1

, then

x^{2025} + \frac{1}{x ^{2025}}

equals:

AdvancedTry it yourself first

x + \frac{1}{x} = 3

, then

x^{1000} + \frac{1}{x ^{1000}}

equals:

AdvancedTry it yourself first

x + \frac{1}{x} = - 1

, then

x^{100} + x^{200}

equals:

Expressions with x² + x + 1 = 0

x^{3} = 1

x^{3} = - 1

When the equation $x^{2} \pm x + 1 = 0$ is given and a polynomial value is asked.

The idea

This is the same special value, with the equation given in place of $x + \frac{1}{x}$ : $x + \frac{1}{x} = - 1 ⟺ x^{2} + x + 1 = 0$ and $x + \frac{1}{x} = 1 ⟺ x^{2} - x + 1 = 0$ . So the same cycle: $x^{2} + x + 1 = 0$ gives $x^{3} = 1$ and $x^{2} - x + 1 = 0$ gives $x^{3} = - 1$ . Bring every power $x^{e}$ down to $x^{0}$ , $x^{1}$ or $x^{2}$ (with its sign), then put $x^{2} = - x - 1$ (or $x^{2} = x - 1$ ): the $x$ terms cancel and a number remains.

∑ Formula

x^{2} + x + 1 = 0 \Rightarrow x^{3} = 1, x^{2} = - x - 1, 1 + x + x^{2} = 0; x^{2} - x + 1 = 0 \Rightarrow x^{3} = - 1, x^{2} = x - 1

What the two equations give

$x^{3}$: the cycle relation
$1 + x + x^{2} = 0$: three consecutive powers add to zero

Spot itSpot it: big powers like

x^{20}, x^{100}

in the polynomial and the given equation

x^{2} \pm x + 1 = 0

. Handy:

x^{2} + 1 = - x

and

x + 1 = - x^{2}

(when

x^{2} + x + 1 = 0

) collapse many products in one line.

Steps

Write $x^{3} = \pm 1$ (multiply by $(x \mp 1)$ ).
Bring every power down to $x^{0}, x^{1}, x^{2}$ .
Replace $x^{2}$ using the equation and cancel the $x$ terms.

Example 11 · Basic

x^{2} + x + 1 = 0

, find

x^{10} + x^{8} + 1

Solution, step by step

1Multiply by $(x - 1)$ : $x^{3} - 1 = 0$ , so $x^{3} = 1$ .
2 $x^{10} = x^{9} \cdot x = x$ and $x^{8} = x^{6} \cdot x^{2} = x^{2}$ .
3Sum: $x + x^{2} + 1 = 0$ , which is exactly the given equation.

Answer 0

Three consecutive powers adding to zero is the most frequent outcome of this method.

Example 12 · Standard

x^{2} + x + 1 = 0

, find

\frac{( x ^{2} + 1 ) ^{3}}{x ^{3}}

Solution, step by step

1From the equation $x^{2} + 1 = - x$ .
2So the numerator is $(x^{2} + 1)^{3} = (- x)^{3} = - x^{3}$ .
3The value is $\frac{- x ^{3}}{x ^{3}} = - 1$ .

Answer -1

We did not even need

x^{3} = 1

here, only

x^{2} + 1 = - x

Example 13 · Standard

x^{2} - x + 1 = 0

, find

x^{12} + x^{6} + 1

Solution, step by step

1Multiply by $(x + 1)$ : $x^{3} + 1 = 0$ , so $x^{3} = - 1$ .
2 $x^{6} = (- 1)^{2} = 1$ and $x^{12} = (- 1)^{4} = 1$ .
3Sum: $1 + 1 + 1 = 3$ .

Answer 3

x^{3} = - 1

the power

3 k

has the sign

(- 1)^{k}

Example 14 · Advanced

x^{2} + x + 1 = 0

, find

1 + x + x^{2} + x^{3} + \dots + x^{2025}

Solution, step by step

1The terms run from $x^{0}$ to $x^{2025}$ , $2026$ in all. Split the first $2025$ into groups of three: $2025 = 3 \times 675$ groups.
2Each group is $x^{3 k} (1 + x + x^{2}) = 0$ , because $1 + x + x^{2} = 0$ .
3The last term left is $x^{2025} = (x^{3})^{675} = 1$ . Total $0 + 1 = 1$ .

Answer 1

The counting: divide the number of terms by

3

and look at the remainder; the leftover terms make the answer.

Example 15 · Advanced

x^{2} + x + 1 = 0

, find

(x + 2) (x^{2} + 2)

Solution, step by step

1Expand: $(x + 2) (x^{2} + 2) = x^{3} + 2 x + 2 x^{2} + 4$ .
2 $x^{3} = 1$ and $2 x + 2 x^{2} = 2 (x + x^{2}) = 2 \times (- 1) = - 2$ (because $x + x^{2} = - 1$ ).
3Sum: $1 - 2 + 4 = 3$ .

Answer 3

Keeping

x + x^{2} = - 1

as a ready-made pair is quickest.

Practice

BasicTry it yourself first

x^{2} + x + 1 = 0

, then

x^{20} + x^{10} + 1

equals:

BasicTry it yourself first

x^{2} + x + 1 = 0

, then

x^{3} + x^{6} + x^{9}

equals:

StandardTry it yourself first

x^{2} - x + 1 = 0

, then

x^{9} + x^{3}

equals:

StandardTry it yourself first

x^{2} + x + 1 = 0

, then

x^{4} + x^{5} + 3 + x^{6}

equals:

StandardTry it yourself first

x^{2} + x + 1 = 0

, then

x^{2} + x + 5

equals:

StandardTry it yourself first

x^{2} - x + 1 = 0

, then

x^{18} + x^{12} + x - x^{2}

equals:

AdvancedTry it yourself first

x^{2} + x + 1 = 0

, then

x^{100} + x^{200} + x^{300}

equals:

AdvancedTry it yourself first

x^{2} - x + 1 = 0

, then

x^{2025}

equals:

AdvancedTry it yourself first

x^{2} + x + 1 = 0

, then

x^{30} + x^{31} + x^{32} + 2

equals:

Limits

x + \frac{1}{x} = 2

· minimum values

How small $x + \frac{1}{x}$ can be and what happens at $2$ .

Both methods come from one fact: for real $x$ , $x + \frac{1}{x}$ never goes between $- 2$ and $2$ .

The limit of x + 1/x and what = 2 means

∣ x + \frac{1}{x} ∣ \geq 2

At $x + \frac{1}{x} = \pm 2$ we get $x = \pm 1$ , and which values are impossible.

The idea

From $(x - 1)^{2} \geq 0$ we get $x^{2} + 1 \geq 2 x$ . For $x > 0$ , dividing by $x$ gives $x + \frac{1}{x} \geq 2$ , and for $x < 0$ the direction reverses: $x + \frac{1}{x} \leq - 2$ . So for real $x$ the value of $x + \frac{1}{x}$ never lies between $- 2$ and $2$ . Equality happens only when $(x - 1)^{2} = 0$ , i.e. $x = 1$ (or $x = - 1$ for $- 2$ ). Hence $x + \frac{1}{x} = 2 \Rightarrow x = 1$ and $x + \frac{1}{x} = - 2 \Rightarrow x = - 1$ . Every power is then a plain number and the question opens in one line. In the picture the two branches of $y = x + \frac{1}{x}$ stay outside the band $- 2 < y < 2$ .

◭ Diagram

y = x + \frac{1}{x}

: the branches lie only in

y \geq 2

and

y \leq - 2

. The line

y = 3

meets the curve twice (two values of

x

, reciprocals of each other);

y = 1

never meets it (no real

x

). (The pale pink band is the forbidden middle zone.)

∑ Formula

x + \frac{1}{x} = 2 \Rightarrow x = 1, x + \frac{1}{x} = - 2 \Rightarrow x = - 1, x real \Rightarrow x + \frac{1}{x} \geq 2

The limit rules

$x$: a real number, $x \neq = 0$
$∣ ∣$: absolute value (sign removed)

Similarly $x^{2} + \frac{1}{x ^{2}} = 2$ gives $(x^{2} - 1)^{2} = 0$ , i.e. $x = \pm 1$ , and $x^{n} + \frac{1}{x ^{n}} = 2$ gives $x^{n} = 1$ . Trap: in a question with $∣ s ∣ < 2$ (like $x + \frac{1}{x} = 1$ ) there is no real $x$ , yet the algebra formulas still work; exam questions use exactly this (previous method).

Spot itSpot it:

x + \frac{1}{x} = 2

- 2

is given, or the question asks “which value is possible”. First write

x = \pm 1

and then solve by putting in numbers.

Steps

Multiply by $x$ to make a perfect square: $(x \mp 1)^{2} = 0$ .
Get $x = \pm 1$ .
Put the number into every power; the sign of $(- 1)^{n}$ depends on $n$ being even or odd.

Example 16 · Basic

x + \frac{1}{x} = 2

, find

x^{15} + \frac{2}{x ^{8}} - 3 x

Solution, step by step

1Multiply by $x$ : $x^{2} - 2 x + 1 = 0$ , i.e. $(x - 1)^{2} = 0$ , so $x = 1$ .
2Every power of $1$ is $1$ : $1 + 2 \times 1 - 3 \times 1 = 0$ .

Answer 0

In such questions the powers are only there to scare you; at

x = 1

everything flattens.

Example 17 · Standard

x + \frac{1}{x} = - 2

, find

x^{25} - 3 x^{40} + \frac{1}{x ^{9}}

Solution, step by step

1 $x^{2} + 2 x + 1 = 0$ , i.e. $(x + 1)^{2} = 0$ , so $x = - 1$ .
2 $(- 1)^{25} = - 1$ (odd power), $(- 1)^{40} = 1$ (even power), $\frac{1}{( - 1 ) ^{9}} = - 1$ .
3Value $= - 1 - 3 \times 1 + (- 1) = - 5$ .

Answer -5

An odd power flips the sign, an even power does not.

Example 18 · Standard

x^{2} + \frac{1}{x ^{2}} = 2

and

x > 0

, find

x + \frac{1}{x}

and

x^{50} + \frac{1}{x ^{50}}

Solution, step by step

1 $x^{4} + 1 = 2 x^{2}$ , i.e. $(x^{2} - 1)^{2} = 0$ , so $x^{2} = 1$ and $x = \pm 1$ ; as $x > 0$ , $x = 1$ .
2 $x + \frac{1}{x} = 1 + 1 = 2$ and $x^{50} + \frac{1}{x ^{50}} = 2$ .

Answer 2 ; 2

Without

x > 0

we could also have

x = - 1

and the first answer could be

- 2

Example 19 · Advanced

Which of

1.9, - 1.5, 0.5, - 2.2

can be the value of

x + \frac{1}{x}

for a real

x

Solution, step by step

1 $x + \frac{1}{x} = s$ gives $x^{2} - s x + 1 = 0$ ; a real root needs $s^{2} - 4 \geq 0$ , i.e. $∣ s ∣ \geq 2$ .
2 $1.9, - 1.5, 0.5$ have absolute value below $2$ , so they are impossible; $∣ - 2.2∣ = 2.2 \geq 2$ is fine.

Answer −2.2

Check: at

s = - 2.2

x = \frac{- 2.2 \pm 0.84}{2}

is real.

Practice

BasicTry it yourself first

x + \frac{1}{x} = 2

, then

x^{7} + x^{9}

equals:

BasicTry it yourself first

x + \frac{1}{x} = - 2

, then

x^{7} + x^{9}

equals:

StandardTry it yourself first

x + \frac{1}{x} = 2

, then

x^{50} - \frac{1}{x ^{20}} + x

equals:

StandardTry it yourself first

For real

x (x \neq = 0)

which of these is a possible value of

x + \frac{1}{x}

StandardTry it yourself first

x + \frac{1}{x} = - 2

, then

2 x^{10} + x^{3} - \frac{1}{x ^{4}}

equals:

AdvancedTry it yourself first

x + \frac{1}{x} = - 2

, then

x^{11} + \frac{1}{x ^{11}} + 3 x^{2}

equals:

AdvancedTry it yourself first

x + \frac{1}{x} = 2

, then

5 x^{100} - 2 \frac{1}{x ^{3}}

equals:

AdvancedTry it yourself first

For real

x (x \neq = 0)

which of these is a possible value of

x + \frac{1}{x}

Minimum and maximum values

k x + \frac{m}{x} \geq 2 km

The smallest value of expressions like $x + \frac{1}{x}$ for $x > 0$ .

The idea

We saw $x + \frac{1}{x} \geq 2$ for $x > 0$ . The same with coefficients: expand $(k x - \frac{m}{x})^{2} \geq 0$ and you get $k x + \frac{m}{x} \geq 2 km$ . The minimum occurs when the two terms are equal: $k x = \frac{m}{x}$ , i.e. $x = \frac{m}{k}$ . In the picture the curve comes down to its lowest point and rises again. The maximum (for $x < 0$ , or for a subtracted term) is the mirror image.

◭ Diagram

y = 2 x + \frac{8}{x}

(

x > 0

): the lowest point is at

x = \frac{8}{2} = 2

with value

216 = 8

∑ Formula

k x + \frac{m}{x} \geq 2 km (x > 0)

The minimum-value formula (equality at

x = m / k

)

$k, m$: positive coefficients
$2 km$: the minimum value
$m / k$: the $x$ at which it occurs

Other forms: $\frac{( x + p ) ( x + q )}{x} = x + (p + q) + \frac{pq}{x} \geq p + q + 2 pq$ , and $x^{2} + \frac{1}{x ^{2}} \geq 2$ , $(x + \frac{1}{x})^{2} \geq 4$ , $\frac{( x + 1 ) ^{2}}{x} \geq 4$ . All have their minimum at $x = 1$ (where the terms are equal). For $x < 0$ , $k x + \frac{m}{x} \leq - 2 km$ (a maximum).

Spot itSpot it: “minimum for

x > 0

” and an expression with

x

and

\frac{1}{x}

terms. Trap: if the product of the two terms depends on

x

(

x^{2} + \frac{4}{x}

) the formula does not apply; the product must be constant. And for

x < 0

the sign flips.

Steps

Write the expression as $k x + \frac{m}{x}$ plus a constant (dividing by $x$ if needed).
Check that the product $km$ of the two terms is constant.
Minimum $= 2 km$ + constant, at $x = \frac{m}{k}$ .

Example 20 · Basic

For

x > 0

find the minimum value of

x + \frac{9}{x}

and the

x

where it occurs.

Solution, step by step

1 $k = 1$ , $m = 9$ , $km = 9$ . Minimum $= 29 = 6$ .
2Equality when $x = \frac{9}{x}$ , i.e. $x^{2} = 9$ , $x = 3$ .

Answer 6 ; x = 3

Check: at

x = 3

3 + 3 = 6

✓; at

x = 1

1 + 9 = 10

and at

x = 9

9 + 1 = 10

, both larger.

Example 21 · Standard

For

x > 0

find the minimum value of

3 x + \frac{12}{x}

Solution, step by step

1 $k = 3$ , $m = 12$ , $km = 36$ , and the product is constant (the $x$ cancels).
2Minimum $= 236 = 12$ , at $x = 12/3 = 2$ .

Answer 12

Check: at

x = 2

6 + 6 = 12

✓.

Example 22 · Standard

For

x > 0

find the minimum value of

\frac{( x + 4 ) ( x + 9 )}{x}

Solution, step by step

1Expand the numerator: $x^{2} + 13 x + 36$ . Divide by $x$ : $x + 13 + \frac{36}{x}$ .
2 $x + \frac{36}{x} \geq 236 = 12$ , equality at $x = 6$ .
3Minimum $= 13 + 12 = 25$ .

Answer 25

Note:

x = 4 \times 9 = 6

, the geometric mean of the two constants.

Example 23 · Advanced

Find the maximum of

10 - 3 x - \frac{12}{x}

for

x > 0

and the maximum of

x + \frac{16}{x}

for

x < 0

Solution, step by step

1First: $3 x + \frac{12}{x} \geq 236 = 12$ , so $10 - (3 x + \frac{12}{x}) \leq 10 - 12 = - 2$ . The maximum is $- 2$ (at $x = 2$ ).
2Second: $x < 0$ , let $x = - y$ ( $y > 0$ ). $x + \frac{16}{x} = - (y + \frac{16}{y}) \leq - 216 = - 8$ .
3The maximum is $- 8$ (at $x = - 4$ ).

Answer -2 ; -8

For negative

x

the minimum turns into a maximum in the mirror.

Practice

BasicTry it yourself first

For

x > 0

the minimum value of

x + \frac{25}{x}

is:

BasicTry it yourself first

For

x > 0

the minimum value of

4 x + \frac{1}{x}

is:

StandardTry it yourself first

For

x > 0

the minimum value of

8 x + \frac{18}{x}

is:

StandardTry it yourself first

For

x > 0

the minimum value of

\frac{( x + 2 ) ( x + 8 )}{x}

is:

StandardTry it yourself first

For

x > 0

, at which

x

5 x + \frac{20}{x}

smallest?

StandardTry it yourself first

For

x > 0

the minimum value of

\frac{( x + 3 ) ( x + 12 )}{x}

is:

AdvancedTry it yourself first

For

x > 0

the minimum value of

\frac{( x + 5 ) ( x + 20 )}{x}

is:

AdvancedTry it yourself first

For

x > 0

the maximum value of

10 - 3 x - \frac{12}{x}

is:

AdvancedTry it yourself first

For

x > 0

the maximum value of

4 - x - \frac{16}{x}

is:

AdvancedTry it yourself first

For

x > 0

, at which

x

4 x + \frac{9}{x}

smallest?

Two and three variables

\frac{a}{b} + \frac{b}{a}

· cyclic three variables

The same pair when two or three letters replace $x$ .

Calling $\frac{a}{b}$ by $x$ turns a two-variable question into the old ladder; with three variables the cyclic move is different.

Questions with a/b + b/a

Put

x = \frac{a}{b}

When $a^{2} + b^{2} = kab$ or $a : b$ is given and $\frac{a}{b} + \frac{b}{a}$ is asked.

The idea

Call $\frac{a}{b}$ by the name $x$ ; then $\frac{b}{a} = \frac{1}{x}$ . So $\frac{a}{b} + \frac{b}{a}$ is exactly the pair $x + \frac{1}{x}$ , and the whole ladder of the chapter runs here. Dividing the given $a^{2} + b^{2} = kab$ by $ab$ gives $\frac{a}{b} + \frac{b}{a} = k$ straight away. The squares of the sum and difference become multiples of $ab$ : $(a + b)^{2} = (k + 2) ab$ and $(a - b)^{2} = (k - 2) ab$ , whose quotient gives $(\frac{a + b}{a - b})^{2} = \frac{k + 2}{k - 2}$ .

∑ Formula

a^{2} + b^{2} = kab \Rightarrow \frac{a}{b} + \frac{b}{a} = k, \frac{( a + b ) ^{2}}{ab} = k + 2, \frac{( a - b ) ^{2}}{ab} = k - 2, (\frac{a + b}{a - b})^{2} = \frac{k + 2}{k - 2}

What

a^{2} + b^{2} = kab

gives

$a, b$: two non-zero numbers
$k$: the given constant

∑ Formula

a + b = S, ab = P \Rightarrow \frac{a}{b} + \frac{b}{a} = \frac{S ^{2} - 2 P}{P}, a : b = p : q \Rightarrow \frac{a}{b} + \frac{b}{a} = \frac{p ^{2} + q ^{2}}{pq}

The sum-product and ratio forms

$S, P$: sum and product
$p, q$: the terms of the ratio

Spot itSpot it: a relation between

a^{2} + b^{2}

and

ab

is given, or

\frac{a}{b}

\frac{b}{a}

is asked. Divide by

ab

at once, then the old ladder. Trap:

\frac{a + b}{a - b}

comes through a square root; read the sign condition (

a > b > 0

Steps

Divide the given condition by $ab$ .
Read $\frac{a}{b} + \frac{b}{a} = k$ as $x + \frac{1}{x} = k$ .
Apply the ladder (square, cube) or a square root.

Example 24 · Basic

a^{2} + b^{2} = 7 ab

, find

\frac{a}{b} + \frac{b}{a}

and

\frac{a ^{2}}{b ^{2}} + \frac{b ^{2}}{a ^{2}}

Solution, step by step

1Divide by $ab$ : $\frac{a ^{2}}{ab} + \frac{b ^{2}}{ab} = 7$ , i.e. $\frac{a}{b} + \frac{b}{a} = 7$ .
2With $x = \frac{a}{b}$ : $x + \frac{1}{x} = 7$ ; square: $x^{2} + \frac{1}{x ^{2}} = 7^{2} - 2 = 47$ .

Answer 7 ; 47

\frac{a ^{2}}{b ^{2}} + \frac{b ^{2}}{a ^{2}} = \frac{a ^{4} + b ^{4}}{a ^{2} b ^{2}}

is the same thing.

Example 25 · Standard

a : b = 2 : 5

, find

\frac{a}{b} + \frac{b}{a}

Solution, step by step

1 $\frac{a}{b} = \frac{2}{5}$ and $\frac{b}{a} = \frac{5}{2}$ .
2Sum: $\frac{2}{5} + \frac{5}{2} = \frac{4 + 25}{10} = \frac{29}{10}$ .

Answer 29/10

The formula

\frac{p ^{2} + q ^{2}}{pq} = \frac{4 + 25}{10}

is the same.

Example 26 · Standard

a + b = 9

and

ab = 14

, find

\frac{a}{b} + \frac{b}{a}

Solution, step by step

1 $\frac{a}{b} + \frac{b}{a} = \frac{a ^{2} + b ^{2}}{ab}$ .
2 $a^{2} + b^{2} = (a + b)^{2} - 2 ab = 81 - 28 = 53$ .
3Value $= \frac{53}{14}$ .

Answer 53/14

Here

a = 2

b = 7

\frac{2}{7} + \frac{7}{2} = \frac{4 + 49}{14}

✓.

Example 27 · Advanced

a^{2} + b^{2} = 10 ab

and

a > b > 0

, find

\frac{a + b}{a - b}

Solution, step by step

1 $(a + b)^{2} = a^{2} + b^{2} + 2 ab = 12 ab$ and $(a - b)^{2} = a^{2} + b^{2} - 2 ab = 8 ab$ .
2Divide: $(\frac{a + b}{a - b})^{2} = \frac{12}{8} = \frac{3}{2}$ .
3 $a > b > 0$ ⟹ positive: $\frac{a + b}{a - b} = \frac{3}{2} = \frac{6}{2}$ .

Answer √6/2

3/2 = 6 /2

(rationalising the denominator).

Practice

BasicTry it yourself first

a^{2} + b^{2} = 11 ab

, then

\frac{a}{b} + \frac{b}{a}

equals:

BasicTry it yourself first

a^{2} + b^{2} = 6 ab

, then

\frac{( a + b ) ^{2}}{ab}

equals:

StandardTry it yourself first

a : b = 3 : 7

, then

\frac{a}{b} + \frac{b}{a}

equals:

StandardTry it yourself first

a + b = 11

and

ab = 18

, then

\frac{a}{b} + \frac{b}{a}

equals:

StandardTry it yourself first

a^{2} + b^{2} = 18 ab

, then

\frac{( a - b ) ^{2}}{ab}

equals:

StandardTry it yourself first

a^{2} + b^{2} = 14 ab

and

a > b > 0

, then

\frac{a + b}{a - b}

equals:

AdvancedTry it yourself first

a^{2} + b^{2} = 5 ab

, then

\frac{a ^{3}}{b ^{3}} + \frac{b ^{3}}{a ^{3}}

equals:

AdvancedTry it yourself first

a^{2} + b^{2} = 34 ab

and

a > b > 0

, then

\frac{a + b}{a - b}

equals:

AdvancedTry it yourself first

a^{2} + b^{2} = 9 ab

, then

\frac{a ^{2}}{b ^{2}} + \frac{b ^{2}}{a ^{2}}

equals:

AdvancedTry it yourself first

a + b = 13

and

ab = 12

, then

\frac{a}{b} + \frac{b}{a}

equals:

Three variables: cyclic pairs

x yz = - p

and the

x yz = 1

fractions

Cyclic questions such as $x + \frac{1}{y}$ , $y + \frac{1}{z}$ , $z + \frac{1}{x}$ .

The idea

Now instead of $x + \frac{1}{x}$ three variables are tied in a cycle: $x$ with $\frac{1}{y}$ , $y$ with $\frac{1}{z}$ and $z$ with $\frac{1}{x}$ . The ladder does not run here, but the same style of move does: write everything in one variable (first result) or multiply and combine the three equations cleverly to get $x yz$ . Two famous results: (1) if $x + \frac{1}{y} = 1$ and $y + \frac{1}{z} = 1$ then $z + \frac{1}{x} = 1$ and $x yz = - 1$ . (2) if $x + \frac{1}{y} = y + \frac{1}{z} = z + \frac{1}{x} = p$ with $x, y, z$ distinct, then $p^{2} = 1$ and $x yz = - p$ . A third, from the same family: (3) if $ab c = 1$ then $\frac{1}{1 + a + ab} + \frac{1}{1 + b + b c} + \frac{1}{1 + c + c a} = 1$ .

◭ Diagram

Read each arrow as: the letter at the tail

+ \frac{1}{the letter at the head} = 1

. The three arrows form a cycle, which is why the third equation comes out true by itself and

x yz = - 1

∑ Formula

x + \frac{1}{y} = y + \frac{1}{z} = z + \frac{1}{x} = p, x, y, z distinct \Rightarrow p^{2} = 1, x yz = - p

The cyclic-pair rule

$x, y, z$: three distinct non-zero numbers
$p$: the shared value

∑ Formula

ab c = 1 \Rightarrow \frac{1}{1 + a + ab} + \frac{1}{1 + b + b c} + \frac{1}{1 + c + c a} = 1

The product-

1

fractions

$a, b, c$: three numbers with product $1$

Spot itSpot it: three letters tied in a cycle by reciprocals, or three variables with

ab c = 1

. Move: use

ab c = 1

in place of a “

1

” so all denominators match; in cyclic equations multiply by

z

(or

x

) and use the third equation.

Steps

Clear the fractions ( $x y + 1 = p y$ and so on).
Multiply one equation by the third letter and combine with another.
By the cycle the result is the same for every letter; subtract to get $p^{2} = 1$ and $x yz = - p$ .

Example 28 · Basic

x + \frac{1}{y} = 1

and

y + \frac{1}{z} = 1

, find

z + \frac{1}{x}

and

x yz

Solution, step by step

1From $x + \frac{1}{y} = 1$ we get $x y + 1 = y$ , so $x = \frac{y - 1}{y}$ .
2From $y + \frac{1}{z} = 1$ we get $z = \frac{1}{1 - y}$ .
3A number check: take $y = 3$ , then $x = \frac{2}{3}$ , $z = \frac{1}{1 - 3} = - \frac{1}{2}$ . Then $z + \frac{1}{x} = - \frac{1}{2} + \frac{3}{2} = 1$ and $x yz = \frac{2}{3} \times 3 \times (- \frac{1}{2}) = - 1$ .
4In general: $z + \frac{1}{x} = \frac{1}{1 - y} + \frac{y}{y - 1} = \frac{1 - y}{1 - y} = 1$ and $x yz = \frac{y - 1}{y} \cdot y \cdot \frac{1}{1 - y} = - 1$ .

Answer 1 ; -1

Whatever

y

is (except

0

and

1

), the result is the same.

Example 29 · Standard

x, y, z

are distinct and

x + \frac{1}{y} = y + \frac{1}{z} = z + \frac{1}{x} = p

, find

p

and

x yz

(with

p > 0

Solution, step by step

1Clear fractions: $x y + 1 = p y$ , $yz + 1 = p z$ , $z x + 1 = p x$ .
2Multiply the first by $z$ : $x yz + z = p yz = p (p z - 1)$ , so $x yz + p = (p^{2} - 1) z$ . Likewise $x yz + p = (p^{2} - 1) x$ .
3Subtracting: $(p^{2} - 1) (z - x) = 0$ ; since $z \neq = x$ , $p^{2} = 1$ , and $p > 0$ gives $p = 1$ .
4Then $x yz + p = 0$ , i.e. $x yz = - 1$ .

Answer 1 ; -1

This is the earlier (

= 1

) result; with

p = - 1

we would get

x yz = 1

Example 30 · Standard

ab c = 1

, find

\frac{1}{1 + a + ab} + \frac{1}{1 + b + b c} + \frac{1}{1 + c + c a}

Solution, step by step

1Aim: make all three denominators alike. Multiply top and bottom of the second by $a$ : $\frac{a}{a + ab + ab c} = \frac{a}{a + ab + 1}$ ( $ab c = 1$ ).
2Multiply top and bottom of the third by $ab$ : $\frac{ab}{ab + ab c + a \cdot ab c} = \frac{ab}{ab + 1 + a}$ .
3Sum $= \frac{1 + a + ab}{1 + a + ab} = 1$ .

Answer 1

Check with numbers: at

a = 2

b = 3

c = \frac{1}{6}

\frac{1}{9} + \frac{2}{9} + \frac{6}{9} = 1

✓.

Practice

BasicTry it yourself first

x + \frac{1}{y} = 1

and

y + \frac{1}{z} = 1

, then

x yz

equals:

BasicTry it yourself first

x + \frac{1}{y} = 1

and

y + \frac{1}{z} = 1

, then

z + \frac{1}{x}

equals:

StandardTry it yourself first

x, y, z

are distinct and

x + \frac{1}{y} = y + \frac{1}{z} = z + \frac{1}{x} = p

, then

p^{2}

equals:

StandardTry it yourself first

x, y, z

are distinct and

x + \frac{1}{y} = y + \frac{1}{z} = z + \frac{1}{x} = 1

, then

x yz

equals:

StandardTry it yourself first

ab c = 1

, then

\frac{1}{1 + a + ab} + \frac{1}{1 + b + b c} + \frac{1}{1 + c + c a}

equals:

StandardTry it yourself first

ab c = 1

, then

\frac{a}{1 + a + ab} + \frac{b}{1 + b + b c} + \frac{c}{1 + c + c a}

equals:

AdvancedTry it yourself first

x + \frac{1}{y} = 1

and

y + \frac{1}{z} = 1

, then

x^{3} y^{3} z^{3}

equals:

AdvancedTry it yourself first

x, y, z

are distinct and

x + \frac{1}{y} = y + \frac{1}{z} = z + \frac{1}{x} = - 1

, then

x yz

equals:

AdvancedTry it yourself first

ab c = 1

with

a = 2, b = 3

, then

\frac{1}{1 + a + ab} + \frac{1}{1 + b + b c} + \frac{1}{1 + c + c a}

equals:

AdvancedTry it yourself first

x + \frac{1}{y} = 1

and

y + \frac{1}{z} = 1

, then

x yz + \frac{1}{x yz}

equals:

Summary, mixed practice, revision

All methods in one place

One table of all methods, exam-style mixed questions and revision cards.

Here is the essence of this chapter and exam-style mixed questions: first spot the method, then solve.

Summary of all methods

Which method when

One page to pick the method as soon as you read a question.

Method	Use when	Key formula / move
Special values	$x + \frac{1}{x} = 1, - 1, 0, 2, 3$	first $a_{n} = \pm 2$ : $x^{n} = \pm 1$ ; $a_{n} = 0$ : $x^{2 n} = - 1$
Huge power	like $x^{100} + \frac{1}{x ^{100}}$	$a_{N mod P}$ ; cycles $6, 3, 12, 4, 8$
$x^{2} \pm x + 1 = 0$	a polynomial value	$x^{3} = \pm 1$ , $x^{2} = \mp x - 1$
Limit	$x + \frac{1}{x} = \pm 2$ , or which value is possible	$x = \pm 1$ ; $∣ x + \frac{1}{x} ∣ \geq 2$
Minimum value	$x > 0$ , $k x + \frac{m}{x}$	$2 km$ at $x = m / k$
$\frac{a}{b} + \frac{b}{a}$	$a^{2} + b^{2} = kab$ , $a : b$ , $a + b$ and $ab$	$k$ ; $\frac{p ^{2} + q ^{2}}{pq}$ ; $\frac{S ^{2} - 2 P}{P}$
Three variables	cyclic $x + \frac{1}{y}$ , or $ab c = 1$	$p^{2} = 1$ , $x yz = - p$ ; fractions add to $1$

Here: $a_{n} = x^{n} + \frac{1}{x ^{n}}$ , $P$ is the cycle length, $N mod P$ the remainder of $N \div P$ , $k, m$ positive coefficients and $S, P$ (second last row) the sum and product. First ask: is the value between $- 2$ and $2$ (a cycle), exactly $\pm 2$ (a limit), or are there several letters?

Two trapsTwo traps: (1) at

x + \frac{1}{x} = \pm 2

x

itself is

\pm 1

and the ladder is not needed; (2) in a cycle a remainder

0

means

a_{0} = 2

(that is

x^{N} = 1

Mixed practice — set 1

Basic to standard

All kinds mixed together; spot the method first, then solve.

These questions are not in method order, just like the exam.

BasicTry it yourself first

x + \frac{1}{x} = 2

, then

x^{3} + \frac{1}{x ^{3}}

equals:

BasicTry it yourself first

x + \frac{1}{x} = 1

, then

x^{12}

equals:

BasicTry it yourself first

x + \frac{1}{x} = - 1

, then

x^{200} + \frac{1}{x ^{200}}

equals:

BasicTry it yourself first

x^{2} + x + 1 = 0

, then

x^{13} + x^{14} + x^{15}

equals:

BasicTry it yourself first

x + \frac{1}{x} = 2

, then

2 x^{5} + \frac{1}{x ^{5}}

equals:

BasicTry it yourself first

For

x > 0

the minimum value of

2 x + \frac{18}{x}

is:

BasicTry it yourself first

a^{2} + b^{2} = 9 ab

, then

\frac{a}{b} + \frac{b}{a}

equals:

BasicTry it yourself first

x + \frac{1}{y} = 1

and

y + \frac{1}{z} = 1

, then

x yz

equals:

StandardTry it yourself first

x + \frac{1}{x} = 3

, then

x^{300} + \frac{1}{x ^{300}}

equals:

StandardTry it yourself first

x + \frac{1}{x} = 1

, then

x^{61} + x^{58}

equals:

StandardTry it yourself first

x^{2} - x + 1 = 0

, then

x^{6} + x^{5} + x^{2}

equals:

StandardTry it yourself first

For

x > 0

the minimum value of

\frac{( x + 6 ) ( x + 6 )}{x}

is:

StandardTry it yourself first

a : b = 4 : 9

, then

\frac{a}{b} + \frac{b}{a}

equals:

StandardTry it yourself first

For real

x (x \neq = 0)

which of these is a possible value of

x + \frac{1}{x}

StandardTry it yourself first

a^{2} + b^{2} = 12 ab

, then

\frac{( a + b ) ^{2}}{ab}

equals:

StandardTry it yourself first

x + \frac{1}{y} = 1

and

y + \frac{1}{z} = 1

, then

z + \frac{1}{x}

equals:

Mixed practice — set 2

Standard to advanced

Mixed questions from the hard end of the exam.

AdvancedTry it yourself first

x + \frac{1}{x} = 2

, then

x^{24}

equals:

AdvancedTry it yourself first

x + \frac{1}{x} = 2

, then

x^{1001} + \frac{1}{x ^{1001}}

equals:

AdvancedTry it yourself first

x + \frac{1}{x} = 0

, then

x^{2026} + \frac{1}{x ^{2026}}

equals:

AdvancedTry it yourself first

x^{2} + x + 1 = 0

, then

x^{100} + x^{101} + x^{102}

equals:

AdvancedTry it yourself first

x + \frac{1}{x} = - 2

, then

x^{21} + \frac{1}{x ^{22}} + 4 x^{5}

equals:

AdvancedTry it yourself first

For

x > 0

the minimum value of

5 x + \frac{45}{x}

is:

AdvancedTry it yourself first

For

x > 0

the maximum value of

7 - 2 x - \frac{8}{x}

is:

AdvancedTry it yourself first

a^{2} + b^{2} = 12 ab

, then

\frac{a ^{2}}{b ^{2}} + \frac{b ^{2}}{a ^{2}}

equals:

AdvancedTry it yourself first

a^{2} + b^{2} = 6 ab

and

a > b > 0

, then

\frac{a + b}{a - b}

equals:

AdvancedTry it yourself first

a + b = 17

and

ab = 30

, then

\frac{a}{b} + \frac{b}{a}

equals:

AdvancedTry it yourself first

x^{2} + x + 1 = 0

, then

x^{2024} + 2 x^{2025} + x^{2026}

equals:

AdvancedTry it yourself first

x + \frac{1}{x} = - 1

, then

x^{301} + x^{302}

equals:

AdvancedTry it yourself first

x + \frac{1}{x} = 3

, then

x^{7} + \frac{1}{x ^{7}}

equals:

AdvancedTry it yourself first

x + \frac{1}{x} = - 1

, then

x^{99}

equals:

AdvancedTry it yourself first

For

x > 0

, at which

x

3 x + \frac{27}{x}

smallest?

AdvancedTry it yourself first

x + \frac{1}{x} = 3

, then

x^{777} + \frac{1}{x ^{777}}

equals:

StandardTry it yourself first

x + \frac{1}{x} = 2

, then

3 x^{11} - \frac{1}{x ^{6}}

equals:

StandardTry it yourself first

a^{2} + b^{2} = 22 ab

, then

\frac{( a - b ) ^{2}}{ab}

equals:

StandardTry it yourself first

For

x > 0

the minimum value of

6 x + \frac{24}{x}

is:

AdvancedTry it yourself first

For real

x (x \neq = 0)

which of these is a possible value of

x + \frac{1}{x}

Revision: cards and traps

Flip and recall

Read the front, think, then flip.