Reciprocal Forms (x + 1/x) and Surd Values

Why $x + \frac{1}{x}$ Works So Neatly

Roughly one algebra question in three in SSC Tier 1 starts with a line like "if $x + \frac{1}{x} = 6$", or gives $x$ as a surd such as $2 + \sqrt{3}$. They all run on one fact: a number times its reciprocal is 1. So when you square or cube $x + \frac{1}{x}$, the cross terms turn into plain numbers.

Squaring.

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

So if $x + \frac{1}{x} = k$, then

$$x^2 + \frac{1}{x^2} = k^2 - 2$$

Cubing. Use $(a + b)^3 = a^3 + b^3 + 3ab(a + b)$ with $ab = x \cdot \frac{1}{x} = 1$:

$$\left(x + \frac{1}{x}\right)^3 = x^3 + \frac{1}{x^3} + 3\left(x + \frac{1}{x}\right) \;\Rightarrow\; x^3 + \frac{1}{x^3} = k^3 - 3k$$

With $x + \frac{1}{x} = 6$: $x^2 + \frac{1}{x^2} = 36 - 2 = 34$ and $x^3 + \frac{1}{x^3} = 216 - 18 = 198$.

A second road to the cube. From $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$,

$$x^3 + \frac{1}{x^3} = \left(x + \frac{1}{x}\right)\left(x^2 + \frac{1}{x^2} - 1\right) = k(k^2 - 3)$$

This is the quicker one when $x^2 + \frac{1}{x^2}$ is already known: 6 × 33 = 198 again.

Going backwards. If $x^2 + \frac{1}{x^2} = 79$, add 2 and take the square root: $\left(x + \frac{1}{x}\right)^2 = 81$, so $x + \frac{1}{x} = 9$ for positive $x$ (and $-9$ for negative $x$). Then $x^3 + \frac{1}{x^3} = 729 - 27 = 702$.

Given the cube instead, try small whole numbers in $k^3 - 3k$. For $x^3 + \frac{1}{x^3} = 52$, $k = 4$ fits because 64 − 12 = 52, and it is the only real value that does.

Where $k$ can sit. For positive $x$, $x + \frac{1}{x}$ is at least 2, and it equals 2 only when $x = 1$ (the Linear Equations and Maximum or Minimum Values section shows why). For negative $x$ it is at most $-2$. So $x + \frac{1}{x} = 1$ or $\sqrt{3}$ cannot happen for a real $x$. SSC still asks such values, and they have their own rules later in this section.

The plus family always subtracts: $k^2 - 2$ and $k^3 - 3k$. Two wrong options you will often meet are $k^2 + 2$ and $k^3 - 3$, where the 3 forgot to multiply $k$. For $k = 6$ those are 38 and 213, placed right beside the correct 34 and 198.

Higher Powers: Fourth, Fifth and Sixth

Every higher power is built from the ones you already have. Squaring doubles the power, and multiplying two values adds the powers.

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

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

$$x^6 + \frac{1}{x^6} = \left(x^3 + \frac{1}{x^3}\right)^2 - 2$$

The fifth-power line needs a word. Multiplying out the two brackets gives $x^5 + \frac{1}{x^5}$ and also $x + \frac{1}{x}$ (from $x^2 \cdot \frac{1}{x^3}$ and $\frac{1}{x^2} \cdot x^3$), so $k$ is taken off at the end. Any two powers work the same way: $x^7 + \frac{1}{x^7} = \left(x^3 + \frac{1}{x^3}\right)\left(x^4 + \frac{1}{x^4}\right) - \left(x + \frac{1}{x}\right)$.

The ladder. Multiplying by $x + \frac{1}{x}$ climbs one power at a time:

$$x^{n+1} + \frac{1}{x^{n+1}} = k\left(x^n + \frac{1}{x^n}\right) - \left(x^{n-1} + \frac{1}{x^{n-1}}\right)$$

Each new value is $k$ times the last one, minus the one before it. Start the ladder with 2 (that is $x^0 + \frac{1}{x^0}$) and then $k$. For $k = 3$: 2, 3, 7, 18, 47, 123, 322.

$x + \frac{1}{x}$	$x^2 + \frac{1}{x^2}$	$x^3 + \frac{1}{x^3}$	$x^4 + \frac{1}{x^4}$	$x^5 + \frac{1}{x^5}$	$x^6 + \frac{1}{x^6}$
3	7	18	47	123	322
4	14	52	194	724	2702
5	23	110	527	2525	12098

Check the middle row with the formulas: $14^2 - 2 = 194$, $14 \times 52 - 4 = 724$ and $52^2 - 2 = 2702$.

Coming down. Given a high power, add 2 and take the square root, as many times as needed. If $x^4 + \frac{1}{x^4} = 194$, then $x^2 + \frac{1}{x^2} = \sqrt{196} = 14$, and $x + \frac{1}{x} = \sqrt{16} = 4$ for positive $x$. For $x^8 + \frac{1}{x^8} = 2207$ the chain is 2209 = $47^2$, then 49 = $7^2$, then 9 = $3^2$, so $x + \frac{1}{x} = 3$ for positive $x$. The root at each middle step is positive, because $x^2 + \frac{1}{x^2}$ and $x^4 + \frac{1}{x^4}$ are never negative for real $x$. Only the last step, down to $x + \frac{1}{x}$, needs the sign of $x$.

Note that $x^4 + \frac{1}{x^4}$ is not $k^4 - 2$. Go through the square: $(k^2 - 2)^2 - 2$. For $k = 3$ that is 47, while $3^4 - 2 = 79$ is the option that shortcut leads to. For the sixth power, squaring the cube value is faster than climbing the ladder.

The Minus Family: $x - \frac{1}{x}$

Now the two terms are subtracted. The product is still 1, but the minus sign turns every "take away" into "add".

$$\left(x - \frac{1}{x}\right)^2 = x^2 - 2 + \frac{1}{x^2} \;\Rightarrow\; x^2 + \frac{1}{x^2} = m^2 + 2$$

For the cube, $(a - b)^3 = a^3 - b^3 - 3ab(a - b)$ with $ab = 1$ gives

$$x^3 - \frac{1}{x^3} = m^3 + 3m = m(m^2 + 3)$$

where $m = x - \frac{1}{x}$. With $m = 3$: $x^2 + \frac{1}{x^2} = 9 + 2 = 11$, $x^3 - \frac{1}{x^3} = 27 + 9 = 36$, and $x^4 + \frac{1}{x^4} = 11^2 - 2 = 119$.

The even powers with a plus sign ($x^2 + \frac{1}{x^2}$, $x^4 + \frac{1}{x^4}$, $x^6 + \frac{1}{x^6}$) can come from either family. Once you have $x^2 + \frac{1}{x^2}$, carry on exactly as in the plus family. For the sixth power you can also square the cube: $x^6 + \frac{1}{x^6} = \left(x^3 - \frac{1}{x^3}\right)^2 + 2 = 1296 + 2 = 1298$.

[Tier 2] Odd powers. The multiplying trick still works:

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

With $m = 3$ that is 11 × 36 − 3 = 393.

Spotting it in an equation. Dividing $x^2 - mx - 1 = 0$ by $x$ gives $x - m - \frac{1}{x} = 0$, so $x - \frac{1}{x} = m$. Look at the last number: $-1$ means the minus family, $+1$ means the plus family. For $x^2 - 4x - 1 = 0$, $x - \frac{1}{x} = 4$ and $x^2 + \frac{1}{x^2} = 18$.

Given	Square	Cube
$x + \frac{1}{x} = k$	$x^2 + \frac{1}{x^2} = k^2 - 2$	$x^3 + \frac{1}{x^3} = k^3 - 3k$
$x - \frac{1}{x} = m$	$x^2 + \frac{1}{x^2} = m^2 + 2$	$x^3 - \frac{1}{x^3} = m^3 + 3m$

[SSC Tip] Plus family subtracts, minus family adds. A very common slip is $x^3 - \frac{1}{x^3} = m^3 - 3m$. For $m = 3$ it gives 18 instead of 36, and 18 will be among the options.

Switching Between $x + \frac{1}{x}$ and $x - \frac{1}{x}$

Plenty of questions give one family and ask for the other: "$x + \frac{1}{x} = 5$, find $x^2 - \frac{1}{x^2}$". The bridge between them is

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

because the two squares differ only in the $+2$ and the $-2$. So

$$x - \frac{1}{x} = \pm\sqrt{k^2 - 4} \qquad x + \frac{1}{x} = \pm\sqrt{m^2 + 4}$$

The $\pm$ is genuine. Which sign is right depends on where $x$ lies, and that is the next part of this section.

Differences of even powers are products.

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

$$x^4 - \frac{1}{x^4} = \left(x^2 + \frac{1}{x^2}\right)\left(x^2 - \frac{1}{x^2}\right) \qquad x^6 - \frac{1}{x^6} = \left(x^3 + \frac{1}{x^3}\right)\left(x^3 - \frac{1}{x^3}\right)$$

A full run with $x + \frac{1}{x} = 5$ and $x > 1$.

• $x - \frac{1}{x} = \sqrt{25 - 4} = \sqrt{21}$
• $x^2 - \frac{1}{x^2} = 5\sqrt{21}$
• $x^3 - \frac{1}{x^3} = m^3 + 3m = 21\sqrt{21} + 3\sqrt{21} = 24\sqrt{21}$
• $x^4 - \frac{1}{x^4} = 23 \times 5\sqrt{21} = 115\sqrt{21}$

If instead $0 < x < 1$, every line above changes sign: $-\sqrt{21}$, $-5\sqrt{21}$, $-24\sqrt{21}$ and $-115\sqrt{21}$. The plus-family values (23, 110, 527) stay as they are.

The other direction. $x - \frac{1}{x} = 3$ with $x > 0$ gives $x + \frac{1}{x} = \sqrt{9 + 4} = \sqrt{13}$. Then $x^3 + \frac{1}{x^3} = 13\sqrt{13} - 3\sqrt{13} = 10\sqrt{13}$, and $x^2 - \frac{1}{x^2} = 3\sqrt{13}$.

[Tier 2] A mixed equation. "$\left(x + \frac{1}{x}\right)^2 - 3\left(x - \frac{1}{x}\right) = 14$ and $x > 1$; find $x - \frac{1}{x}$." Write the first bracket as $m^2 + 4$. Then $m^2 - 3m - 10 = 0$, so $(m - 5)(m + 2) = 0$ and $m = 5$ or $m = -2$. Since $x > 1$, the difference is positive: $x - \frac{1}{x} = 5$.

[SSC Tip] Going from plus to minus, the 4 is taken away: $\sqrt{k^2 - 4}$. Going from minus to plus, it is added: $\sqrt{m^2 + 4}$. Swapping them for $k = 5$ gives $\sqrt{29}$, which is a regular wrong option.

Which Sign? $x > 1$, $0 < x < 1$ and Negative $x$

Take $x + \frac{1}{x} = 5$ again. It comes from $x^2 - 5x + 1 = 0$, which has two roots:

$$x = \frac{5 + \sqrt{21}}{2} \approx 4.79 \qquad x = \frac{5 - \sqrt{21}}{2} \approx 0.21$$

The roots multiply to 1, so each is the reciprocal of the other: one is above 1 and the other is below 1. Both give the same $x + \frac{1}{x}$, but $x - \frac{1}{x}$ is $\sqrt{21}$ for the first and $-\sqrt{21}$ for the second. That is why a well-set question says which $x$ it means.

The whole rule is "which is bigger, $x$ or $\frac{1}{x}$?"

Where $x$ lies	$x + \frac{1}{x}$	$x - \frac{1}{x}$	$x^2 - \frac{1}{x^2}$
$x > 1$	more than 2	positive	positive
$0 < x < 1$	more than 2	negative	negative
$-1 < x < 0$	less than $-2$	positive	negative
$x < -1$	less than $-2$	negative	positive

Check the rows with $x = 2$, $\frac{1}{2}$, $-\frac{1}{2}$ and $-2$: $x - \frac{1}{x}$ comes out as 1.5, −1.5, 1.5 and −1.5. The last column is positive exactly when $x$ is further from zero than 1, because then $x^2$ is bigger than $\frac{1}{x^2}$. Odd differences such as $x^3 - \frac{1}{x^3}$ always have the same sign as $x - \frac{1}{x}$.

Negative $k$. "$x + \frac{1}{x} = -6$ and $x < -1$; find $x^2 - \frac{1}{x^2}$." From the table, $x - \frac{1}{x}$ is negative, so $x - \frac{1}{x} = -\sqrt{36 - 4} = -4\sqrt{2}$. Then $x^2 - \frac{1}{x^2} = (-6)(-4\sqrt{2}) = 24\sqrt{2}$. Two minus signs make a plus, and the table agrees that the answer is positive.

From the minus side. $x - \frac{1}{x} = 3$ has roots of about 3.30 and −0.30. So $x + \frac{1}{x} = \sqrt{13}$ when $x > 0$, and $-\sqrt{13}$ when $x < 0$.

If the question names no range. Look at the options. Usually only one of $\sqrt{21}$ and $-\sqrt{21}$ is there, and that is the one to mark. If an option reads $\pm\sqrt{21}$, that is what the setter wants.

[SSC Tip] The sign only matters when your working takes a square root, as in $\sqrt{k^2 - 4}$. Values such as $x^2 + \frac{1}{x^2}$, $x^4 + \frac{1}{x^4}$ and $x^3 + \frac{1}{x^3}$ worked out from a given $k$ never need it, so don't spend time on the range for those.

Square Roots: $\sqrt{x} + \frac{1}{\sqrt{x}}$

For positive $x$, $\sqrt{x}$ and $\frac{1}{\sqrt{x}}$ are another reciprocal pair, so the squaring trick works one level down:

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

So if $x + \frac{1}{x} = k$,

$$\sqrt{x} + \frac{1}{\sqrt{x}} = \sqrt{k + 2} \qquad \sqrt{x} - \frac{1}{\sqrt{x}} = \pm\sqrt{k - 2}$$

The first is always positive. The second follows the sign rule: positive for $x > 1$, negative for $0 < x < 1$.

With $x + \frac{1}{x} = 14$: $\sqrt{x} + \frac{1}{\sqrt{x}} = \sqrt{16} = 4$, and for $x > 1$, $\sqrt{x} - \frac{1}{\sqrt{x}} = \sqrt{12} = 2\sqrt{3}$.

Going up instead. If $\sqrt{x} + \frac{1}{\sqrt{x}} = 3$, then $x + \frac{1}{x} = 9 - 2 = 7$ and $x^2 + \frac{1}{x^2} = 49 - 2 = 47$. Think of $\sqrt{x}$ as the letter: $x$ is its square and $x^2$ its fourth power, so you are on the ladder for $k = 3$ (2, 3, 7, 18, 47).

Index notation. $x^{\frac{1}{2}} + x^{-\frac{1}{2}}$ is the same as $\sqrt{x} + \frac{1}{\sqrt{x}}$, and $x^{\frac{3}{2}} + x^{-\frac{3}{2}}$ is its cube version:

$$x^{\frac{3}{2}} + x^{-\frac{3}{2}} = \left(\sqrt{x} + \frac{1}{\sqrt{x}}\right)^3 - 3\left(\sqrt{x} + \frac{1}{\sqrt{x}}\right)$$

With $x + \frac{1}{x} = 14$ this is $4^3 - 3 \times 4 = 52$.

Two levels down. $x + \frac{1}{x} = 47$ gives $\sqrt{x} + \frac{1}{\sqrt{x}} = \sqrt{49} = 7$, and then $x^{\frac{1}{4}} + x^{-\frac{1}{4}} = \sqrt{7 + 2} = 3$.

When $x$ itself is a surd like $7 + 4\sqrt{3}$ and $\sqrt{x}$ is asked, it is quicker to find $\sqrt{x}$ directly. That comes up with the surd values near the end of this section.

Going down a level adds 2 under the root, $\sqrt{k + 2}$. Going up a level takes 2 away, $k^2 - 2$. Mixing them up turns the right answer 4 into $\sqrt{12}$ for $x + \frac{1}{x} = 14$, and $\sqrt{12}$ (or $2\sqrt{3}$) is usually one of the choices.

Scaled Forms: $ax + \frac{b}{x}$

Questions like $5t + \frac{1}{5t} = 4$ or $x + \frac{1}{4x} = 3$ look new, but the idea is the same. What matters is the product of the two terms, because that product is what turns into a plain number when you square or cube.

$$\left(ax + \frac{b}{x}\right)^2 = a^2x^2 + \frac{b^2}{x^2} + 2ab$$

$$\left(ax + \frac{b}{x}\right)^3 = a^3x^3 + \frac{b^3}{x^3} + 3ab\left(ax + \frac{b}{x}\right)$$

Case 1: the product is 1. In $5t + \frac{1}{5t}$ the two terms multiply to 1. Treat $5t$ as one letter, say $y$, so $y + \frac{1}{y} = 4$. Then

$$25t^2 + \frac{1}{25t^2} = 16 - 2 = 14 \qquad 125t^3 + \frac{1}{125t^3} = 64 - 12 = 52$$

Case 2: the product is a fraction. In $x + \frac{1}{4x} = 3$ the product is $\frac{1}{4}$. Squaring gives $x^2 + \frac{1}{16x^2} + 2 \times \frac{1}{4} = 9$, so $x^2 + \frac{1}{16x^2} = \frac{17}{2}$. Cubing gives $x^3 + \frac{1}{64x^3} = 27 - 3 \times \frac{1}{4} \times 3 = \frac{99}{4}$.

You can also turn case 2 into case 1 by multiplying through by 2: $2x + \frac{1}{2x} = 6$, so $4x^2 + \frac{1}{4x^2} = 34$, and dividing by 4 gives $\frac{17}{2}$ again. Choose whichever way matches what the question asks for.

Case 3: two different numbers. $4x + \frac{1}{3x} = 3$ has the product $\frac{4}{3}$:

$$16x^2 + \frac{1}{9x^2} = 9 - \frac{8}{3} = \frac{19}{3} \qquad 64x^3 + \frac{1}{27x^3} = 27 - 3 \times \frac{4}{3} \times 3 = 15$$

With a minus sign. $3x - \frac{2}{x} = 5$ has the product 6. As in the minus family, the square and the cube both add:

$$9x^2 + \frac{4}{x^2} = 25 + 12 = 37 \qquad 27x^3 - \frac{8}{x^3} = 125 + 3 \times 6 \times 5 = 215$$

(Here $x = 2$ fits the equation, and $36 + 1 = 37$ and $216 - 1 = 215$ confirm both.)

A common factor in disguise. $4x + \frac{4}{x} = 18$ is just $x + \frac{1}{x} = \frac{9}{2}$. Divide first, then use the usual formulas.

[SSC Tip] The cross term is twice the product of the two terms, which is 2 only when the product is 1. In $x + \frac{1}{4x} = 3$, taking off 2 instead of $\frac{1}{2}$ gives 7 instead of $\frac{17}{2}$, and 7 will be an option.

Equations in Disguise and Long Expressions

The same $x + \frac{1}{x}$ can be hidden inside a quadratic, a fraction or a long ratio. In every case the move is to divide by a power of $x$.

Quadratics whose end numbers match.

$$x^2 - 7x + 1 = 0 \;\Rightarrow\; x - 7 + \frac{1}{x} = 0 \;\Rightarrow\; x + \frac{1}{x} = 7$$

So $x^2 + \frac{1}{x^2} = 47$ and $x^3 + \frac{1}{x^3} = 343 - 21 = 322$. The same shape hides in $x^2 + 1 = 7x$ and in $x = 7 - \frac{1}{x}$. When the end numbers match but are not 1, divide them out too: $3x^2 - 10x + 3 = 0$ gives $x + \frac{1}{x} = \frac{10}{3}$. If the last term is $-1$, as in $x^2 - 3x - 1 = 0$, you get the minus family, $x - \frac{1}{x} = 3$.

Putting the result back. "If $x^2 - 7x + 1 = 0$, find $x^4 - 47x^2 + 5$." Multiply $x^2 + \frac{1}{x^2} = 47$ by $x^2$: $x^4 + 1 = 47x^2$, so $x^4 - 47x^2 = -1$ and the answer is 4.

A fraction with $x$ alone on top. Flip both sides first. $\frac{x}{x^2 + 1} = \frac{1}{8}$ becomes $\frac{x^2 + 1}{x} = 8$, that is $x + \frac{1}{x} = 8$. With a middle term, $\frac{x}{x^2 - 4x + 1} = \frac{1}{2}$ becomes $x - 4 + \frac{1}{x} = 2$, so $x + \frac{1}{x} = 6$.

Divide by the middle power. For a degree-4 expression divide the top and the bottom by $x^2$; for degree 6, by $x^3$; for degree 2, by $x$.

$$\frac{x^4 + 1}{x^2} = x^2 + \frac{1}{x^2} \qquad \frac{x^6 + 1}{x^3} = x^3 + \frac{1}{x^3} \qquad \frac{x^2}{x^4 + 1} = \frac{1}{x^2 + \frac{1}{x^2}}$$

With $x + \frac{1}{x} = 4$, so that $x^2 + \frac{1}{x^2} = 14$ and $x^3 + \frac{1}{x^3} = 52$:

$$\frac{x^4 + 3x^2 + 1}{x^4 + 1} = \frac{x^2 + \frac{1}{x^2} + 3}{x^2 + \frac{1}{x^2}} = \frac{17}{14} \qquad \frac{2x^2 - 3x + 2}{3x^2 + 5x + 3} = \frac{2 \times 4 - 3}{3 \times 4 + 5} = \frac{5}{17}$$

Longer tops split into reciprocal pairs:

$$\frac{x^6 + x^4 + x^2 + 1}{x^3} = \left(x^3 + \frac{1}{x^3}\right) + \left(x + \frac{1}{x}\right) = 52 + 4 = 56$$

$$\frac{x^4 + x^3 + x^2 + x + 1}{x^2} = \left(x^2 + \frac{1}{x^2}\right) + \left(x + \frac{1}{x}\right) + 1 = 19$$

When the trick works. The numbers in front of the powers must read the same from both ends: 1, 3, 1 or 2, −3, 2 or 1, 1, 1, 1, 1. If they don't, as in $x^4 + 2x^2 + 3$, the expression does not come down to $k$ alone.

Special Values: 2, −2, 1, −1, $\sqrt{3}$ and $\sqrt{2}$

Some values of $x + \frac{1}{x}$ make $x$, or a power of $x$, very simple. SSC sets questions on all six.

$k = 2$ and $k = -2$. $x + \frac{1}{x} = 2$ means $x^2 - 2x + 1 = 0$, that is $(x - 1)^2 = 0$, so $x = 1$. Every $x^n + \frac{1}{x^n}$ is then 2 and every $x^n - \frac{1}{x^n}$ is 0. In the same way $x + \frac{1}{x} = -2$ gives $x = -1$, and $x^n + \frac{1}{x^n}$ is 2 for even $n$ and $-2$ for odd $n$.

$k = 1$. $x + \frac{1}{x} = 1$ means $x^2 - x + 1 = 0$. Multiply by $x + 1$: $(x + 1)(x^2 - x + 1) = x^3 + 1 = 0$, so

$$x^3 = -1 \qquad x^6 = 1$$

The formula agrees: $x^3 + \frac{1}{x^3} = 1 - 3 = -2$, which is $(-1) + (-1)$.

$k = -1$. $x^2 + x + 1 = 0$. Multiply by $x - 1$: $x^3 - 1 = 0$, so $x^3 = 1$.

$k = \sqrt{3}$. $x^3 + \frac{1}{x^3} = 3\sqrt{3} - 3\sqrt{3} = 0$. Multiply by $x^3$: $x^6 + 1 = 0$, so

$$x^6 = -1 \qquad x^{12} = 1$$

$k = \sqrt{2}$. $x^2 + \frac{1}{x^2} = 2 - 2 = 0$. Multiply by $x^2$: $x^4 + 1 = 0$, so

$$x^4 = -1 \qquad x^8 = 1$$

The negative values $-\sqrt{3}$ and $-\sqrt{2}$ lead to the same $x^6 = -1$ and $x^4 = -1$, because the working uses only $k^2$. So a question that says $\left(x + \frac{1}{x}\right)^2 = 3$ is the $\sqrt{3}$ case whichever sign is meant.

$x + \frac{1}{x}$	Key fact	Powers of $x$ repeat every
$2$	$x = 1$	1
$-2$	$x = -1$	2
$1$	$x^3 = -1$	6
$-1$	$x^3 = 1$	3
$\pm\sqrt{3}$	$x^6 = -1$	12
$\pm\sqrt{2}$	$x^4 = -1$	8

These $x$ are not real numbers. For any $k$ strictly between $-2$ and 2, no real $x$ fits. SSC sets these anyway, and the algebra works the same way. Treat $x$ as a symbol that obeys its key fact, such as $x^3 = -1$, and the values SSC asks for come out as ordinary numbers.

High Powers and the $x^2 + 1 = x$ Type

Once you know $x^3 = -1$ or $x^4 = -1$, a power like $x^{100}$ is quick: split off as many copies of the key power as you can.

$k = 1$, so $x^3 = -1$.

• $x^{30} = (x^3)^{10} = 1$ and $x^{21} = (x^3)^7 = -1$.
• $x^{100} = (x^3)^{33} \cdot x = -x$. So $x^{100} + \frac{1}{x^{100}} = -\left(x + \frac{1}{x}\right) = -1$.

For $x^n + \frac{1}{x^n}$ it is often quicker to write out one full cycle. For $k = 1$ the ladder gives

$n$	1	2	3	4	5	6
$x^n + \frac{1}{x^n}$	$1$	$-1$	$-2$	$-1$	$1$	$2$

and then the same six values repeat. For $n = 100$: 100 = 6 × 16 + 4, so the value is the one under 4, which is $-1$, as before. For $k = -1$ the cycle is shorter: $-1$, $-1$, 2, and repeat.

$k = \sqrt{2}$, so $x^4 = -1$. $x^{50} = (x^4)^{12} \cdot x^2 = x^2$, so $x^{50} + \frac{1}{x^{50}} = x^2 + \frac{1}{x^2} = 0$.

The $x^2 + 1 = x$ type. Dividing by $x$ gives $x + \frac{1}{x} = 1$, so this is the $k = 1$ case in disguise, with $x^3 = -1$. In the same way $x^2 + x + 1 = 0$ is the $k = -1$ case, with $x^3 = 1$. "If $x^2 + 1 = x$, find $x^{18} + x^{12} + x^6 + 1$": each power is a multiple of 6, so each term is 1, and the answer is 4.

Two facts make many of these one-liners. With $x^2 - x + 1 = 0$, any three neighbouring powers with signs $+, -, +$ cancel: $x^{n+2} - x^{n+1} + x^n = x^n(x^2 - x + 1) = 0$. With $x^2 + x + 1 = 0$, any three neighbouring powers added together cancel.

When an $x$ is left over. Not every expression comes out as a number. With $x^2 + 1 = x$, the powers in $x^{14} + x^8 + 1$ reduce to $x^2 + x^2 + 1 = 2x^2 + 1$, and since $x^2 = x - 1$ this is $2x - 1$. That depends on which root $x$ is, so the honest answer is $2x - 1$. If your working leaves an $x$ and every option is a plain number, recheck the reduction and the question before you mark anything.

Bigger polynomials reduced with a given equation, like $x^3 - 6x^2 + \cdots$, are in the Polynomials and Quadratic Equations section.

Reduce the power first, then add. $x^{2025}$ with $x^3 = 1$ is just 1, because 2025 is a multiple of 3; writing out the cycle for that would waste a minute.

Surd Values: When $\frac{1}{x}$ Is the Conjugate

Many questions give $x$ as a surd, like $x = 2 + \sqrt{3}$, and ask for $x^2 + \frac{1}{x^2}$ or something like it. Don't substitute and expand. Find $\frac{1}{x}$ first.

The product test. For $x = p + \sqrt{q}$, multiply by the conjugate: $(p + \sqrt{q})(p - \sqrt{q}) = p^2 - q$. If that is 1, the conjugate is $\frac{1}{x}$.

• $x = 2 + \sqrt{3}$: 4 − 3 = 1, so $\frac{1}{x} = 2 - \sqrt{3}$ and $x + \frac{1}{x} = 4$.
• The $k = 4$ ladder does the rest: $x^2 + \frac{1}{x^2} = 14$, $x^3 + \frac{1}{x^3} = 52$, $x^4 + \frac{1}{x^4} = 194$, and $x - \frac{1}{x} = 2\sqrt{3}$ (positive, as $x > 1$).

Pairs whose product is 1, worth spotting at sight:

$x$	$\frac{1}{x}$	$x + \frac{1}{x}$	$x - \frac{1}{x}$
$2 + \sqrt{3}$	$2 - \sqrt{3}$	$4$	$2\sqrt{3}$
$3 + 2\sqrt{2}$	$3 - 2\sqrt{2}$	$6$	$4\sqrt{2}$
$4 + \sqrt{15}$	$4 - \sqrt{15}$	$8$	$2\sqrt{15}$
$\sqrt{2} + 1$	$\sqrt{2} - 1$	$2\sqrt{2}$	$2$
$\sqrt{3} + \sqrt{2}$	$\sqrt{3} - \sqrt{2}$	$2\sqrt{3}$	$2\sqrt{2}$
$\sqrt{n + 1} + \sqrt{n}$	$\sqrt{n + 1} - \sqrt{n}$	$2\sqrt{n + 1}$	$2\sqrt{n}$

If the product is some other number $d$, then $\frac{1}{x} = \frac{p - \sqrt{q}}{d}$.

The minus family from a surd. $x = \sqrt{2} + 1$ gives $x - \frac{1}{x} = 2$, so $x^2 + \frac{1}{x^2} = 6$ and $x^3 - \frac{1}{x^3} = 8 + 6 = 14$.

When $\sqrt{x}$ is asked. If $x$ looks like $a + 2\sqrt{b}$, it is often a perfect square: find two numbers that add to $a$ and multiply to $b$. So $7 + 4\sqrt{3} = 7 + 2\sqrt{12} = (2 + \sqrt{3})^2$, $5 + 2\sqrt{6} = (\sqrt{3} + \sqrt{2})^2$ and $9 + 4\sqrt{5} = (\sqrt{5} + 2)^2$. For $x = 7 + 4\sqrt{3}$, the square root is the $2 + \sqrt{3}$ from above:

$$\sqrt{x} = 2 + \sqrt{3} \qquad \frac{1}{\sqrt{x}} = 2 - \sqrt{3} \qquad \sqrt{x} + \frac{1}{\sqrt{x}} = 4 \qquad \sqrt{x} - \frac{1}{\sqrt{x}} = 2\sqrt{3}$$

The rule from the square roots part agrees: here $x + \frac{1}{x} = 14$ (since 49 − 48 = 1), and $\sqrt{14 + 2} = 4$.

Big powers. For $a = \sqrt{2} + 1$, $a^2 + a^{-2} = 6$. Treat $a^2$ as the letter: $a^6 + a^{-6} = 6^3 - 3 \times 6 = 198$, and $a^{12} + a^{-12} = 198^2 - 2 = 39202$. Climb with squares and cubes; never expand $(\sqrt{2} + 1)^{12}$.

Conjugate Pairs, Fraction Forms and Squaring Out

Two conjugates, $x$ and $y$. For $x = \sqrt{a} + \sqrt{b}$ and $y = \sqrt{a} - \sqrt{b}$:

$$x + y = 2\sqrt{a} \qquad x - y = 2\sqrt{b} \qquad xy = a - b$$

Everything else follows from these through the two-variable identities of the Square and Cube Identities section:

$$x^2 + y^2 = (x + y)^2 - 2xy = 2(a + b) \qquad \frac{x}{y} + \frac{y}{x} = \frac{x^2 + y^2}{xy} \qquad x^3 + y^3 = (x + y)^3 - 3xy(x + y)$$

With $x = \sqrt{7} + \sqrt{3}$ and $y = \sqrt{7} - \sqrt{3}$: $x + y = 2\sqrt{7}$, $xy = 4$, $x^2 + y^2 = 20$, $\frac{x}{y} + \frac{y}{x} = \frac{20}{4} = 5$, and $x^3 + y^3 = 56\sqrt{7} - 24\sqrt{7} = 32\sqrt{7}$.

Fraction forms. "$x = \frac{\sqrt{3} + 1}{\sqrt{3} - 1}$ and $y = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$." Rationalise $x$:

$$x = \frac{(\sqrt{3} + 1)^2}{3 - 1} = \frac{4 + 2\sqrt{3}}{2} = 2 + \sqrt{3}$$

and $y$ is simply $\frac{1}{x} = 2 - \sqrt{3}$. Now $x + y = 4$ and $xy = 1$, so you are back on the $k = 4$ ladder: $x^2 + y^2 = 14$, $x^2 + xy + y^2 = 15$, $x^3 + y^3 = 52$. In general, for $x = \frac{\sqrt{a} + \sqrt{b}}{\sqrt{a} - \sqrt{b}}$ and $y = \frac{1}{x}$,

$$x + y = \frac{2(a + b)}{a - b} \qquad xy = 1$$

With $a = 3$ and $b = 1$ that gives 4, as above.

Squaring out. "$x = \sqrt{5} + 1$; find $x^2 - 2\sqrt{5}$." Square: $x^2 = 5 + 2\sqrt{5} + 1 = 6 + 2\sqrt{5}$, so $x^2 - 2\sqrt{5} = 6$. The question is built so that the surd part cancels; if yours doesn't cancel, look for a slip in the middle term $2pq$ of $(p + q)^2$.

Getting rid of the surd first. Put the surd alone and square: $x - 1 = \sqrt{5}$ gives $x^2 - 2x + 1 = 5$, that is $x^2 - 2x = 4$. Any expression built on $x^2 - 2x$ is now immediate: $x^2 - 2x + 9 = 13$.

In a fraction form, rationalise only $x$. Its partner $y$ is almost always $\frac{1}{x}$, so write it down without any working, and then use $x + y$ and $xy = 1$ as if they were $k$ and 1.