SSC CGL Maths · Algebraic Identities and Simplification

Reciprocal Sums x ± 1/x

From x + 1/x or x − 1/x to their squares and fourth powers, switching between the two signs, and getting x + 1/x out of an equation or a surd.

All Algebraic Identities and Simplification notes

Why this matters

35 questions in the bank, and seven of them come from the 2025 shifts. Every one rests on one fact: x times 1/x is 1, so when you square x ± 1/x the cross term is just 2. Learn the ladder and the sign rule and most of these take a single line.

Concept 1 of 3: Squaring x + 1/x and x − 1/x

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

the middle term is

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

: the x cancels. So each rung up the ladder costs only "square it, then take away 2".

Definition

$x^2 + \frac{1}{x^2} = \left(x + \frac{1}{x}\right)^2 - 2 = \left(x - \frac{1}{x}\right)^2 + 2$ .
Climb again: $x^4 + \frac{1}{x^4} = \left(x^2 + \frac{1}{x^2}\right)^2 - 2$ . Every rung is "square, minus 2".
A number in front: for $4y + \frac{4}{y} = m$ , divide by 4 first to get $y + \frac{1}{y}$ .
Two terms that still multiply to 1: in $3t + \frac{1}{3t}$ , the product is 1, so the same rule works with $3t$ in place of x: $9t^2 + \frac{1}{9t^2} = \left(3t + \frac{1}{3t}\right)^2 - 2$ .
If the terms do not multiply to 1, as in $x + \frac{1}{16x}$ , multiply through first: $4x + \frac{1}{4x}$ has product 1.
Square roots: $\sqrt{x} + \frac{1}{\sqrt{x}} = k$ gives $x + \frac{1}{x} = k^2 - 2$ .

The squaring ladder

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

Worked example

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

, find

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

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2025 · Tier I, 15 Sep 2025, Shift 3 · Q74Easy

Example 1 · Algebraic Identities and Simplification · Reciprocal Sums x ± 1/x

Given,

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

, find the value of

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

Plus 2 or minus 2

From

x + \frac{1}{x}

, subtract 2; from

x - \frac{1}{x}

, add 2.

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

, so

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

is that square plus 2.

A number in front left inside

From

5y + \frac{5}{y} = 12

y + \frac{1}{y}

\frac{12}{5}

, not 12. Divide by the number in front first, then square.

Drill 14 more on squaring x + 1/x and x − 1/x

Concept 2 of 3: Switching between x + 1/x and x − 1/x

The squares of

x + \frac{1}{x}

and

x - \frac{1}{x}

differ by exactly 4. So either one gives the other, up to a sign that you choose from what you know about x.

Definition

$\left(x - \frac{1}{x}\right)^2 = \left(x + \frac{1}{x}\right)^2 - 4$ and $\left(x + \frac{1}{x}\right)^2 = \left(x - \frac{1}{x}\right)^2 + 4$ .
$x^2 - \frac{1}{x^2} = \left(x + \frac{1}{x}\right)\left(x - \frac{1}{x}\right)$ , and $x^4 - \frac{1}{x^4} = \left(x^2 + \frac{1}{x^2}\right)\left(x^2 - \frac{1}{x^2}\right)$ .
Choosing the sign: if $x > 1$ then $x > \frac{1}{x}$ , so $x - \frac{1}{x}$ is positive. If $0 < x < 1$ then $\frac{1}{x}$ is the bigger one, so $x - \frac{1}{x}$ and $x^2 - \frac{1}{x^2}$ are negative. $x + \frac{1}{x}$ is positive whenever x is.
If the question gives no condition and the options show only one sign, take that one.

Switching the sign

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

Worked example

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

and

0 < x < 1

, find

x^2 - \frac{1}{x^2}

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2023 · Tier I, 17 Jul 2023, Shift 1 · Q72Moderate

Example 2 · Algebraic Identities and Simplification · Reciprocal Sums x ± 1/x

(x^2 + \frac{1}{x^2}) = 7

, and

0 < x < 1

, find the value of

x^2 - \frac{1}{x^2}

Losing the sign when x is between 0 and 1

For

0 < x < 1

\frac{1}{x} > x

, so

x - \frac{1}{x}

and

x^2 - \frac{1}{x^2}

are negative. The positive root is printed as a wrong option.

Minus 4, not minus 2

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

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

. The 2 belongs to

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

; the difference between the two squares is 4.

Drill 8 more on switching between x + 1/x and x − 1/x

Concept 3 of 3: Getting x + 1/x from an equation or a surd

An equation like

x^2 - kx + 1 = 0

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

in disguise: divide by x. A surd like

3 + 2\sqrt{2}

has a reciprocal that is its conjugate. Either way you land back on the ladder.

Definition

Divide by x: $x^2 - kx + 1 = 0$ gives $x + \frac{1}{x} = k$ , and $x^2 - kx - 1 = 0$ gives $x - \frac{1}{x} = k$ . This needs the $x^2$ term and the constant to be equal (or opposite).
Multiply by $x^2$ to go back: from $x^2 + \frac{1}{x^2} = m$ , $x^4 + 1 = mx^2$ .
A surd and its reciprocal: when $(a + \sqrt{b})(a - \sqrt{b}) = a^2 - b = 1$ , the reciprocal of $a + \sqrt{b}$ is $a - \sqrt{b}$ . So $\frac{1}{3 + 2\sqrt{2}} = 3 - 2\sqrt{2}$ , since $9 - 8 = 1$ .
$x + \frac{1}{x} = 2$ means $x = 1$ , because $(x - 1)^2 = 0$ ; and $-2$ means $x = -1$ . Then every power is 1 (or ±1), whatever the exponent.
Inside a fraction: divide the top and the bottom by x. $\frac{3x^2 - 5x + 3}{x^2 + x + 1}$ becomes $\frac{3\left(x + \frac{1}{x}\right) - 5}{\left(x + \frac{1}{x}\right) + 1}$ .

From a quadratic to x ± 1/x

x^2 - kx + 1 = 0 \Rightarrow x + \frac{1}{x} = k \qquad x^2 - kx - 1 = 0 \Rightarrow x - \frac{1}{x} = k

Worked example

x^2 - 6x + 1 = 0

, find

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

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2025 · Tier I, 20 Sep 2025, Shift 1 · Q71Easy

Example 3 · Algebraic Identities and Simplification · Reciprocal Sums x ± 1/x

x = 2 + \sqrt{3}

, find

x - \frac{1}{x}

Dividing an equation whose ends do not match

x^2 - 6x + 4 = 0

divided by x gives

x + \frac{4}{x} = 6

, not

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

. The trick needs the

x^2

coefficient and the constant to be equal.

A conjugate is the reciprocal only when the product is 1

\frac{1}{2 + \sqrt{5}}

is not

2 - \sqrt{5}

: the product is

4 - 5 = -1

, so the reciprocal is

\sqrt{5} - 2

. Work out

a^2 - b

first.

Drill 10 more on getting x + 1/x from an equation or a surd

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Squaring x + 1/x and x − 1/x
The squaring ladder
$x^2 + \frac{1}{x^2} = \left(x + \frac{1}{x}\right)^2 - 2 \qquad x^2 + \frac{1}{x^2} = \left(x - \frac{1}{x}\right)^2 + 2 \qquad x^4 + \frac{1}{x^4} = \left(x^2 + \frac{1}{x^2}\right)^2 - 2$
Switching between x + 1/x and x − 1/x
Switching the sign
$\left(x - \frac{1}{x}\right)^2 = \left(x + \frac{1}{x}\right)^2 - 4 \qquad x^2 - \frac{1}{x^2} = \left(x + \frac{1}{x}\right)\left(x - \frac{1}{x}\right) \qquad x^4 - \frac{1}{x^4} = \left(x^2 + \frac{1}{x^2}\right)\left(x^2 - \frac{1}{x^2}\right)$
Getting x + 1/x from an equation or a surd
From a quadratic to x ± 1/x
$x^2 - kx + 1 = 0 \Rightarrow x + \frac{1}{x} = k \qquad x^2 - kx - 1 = 0 \Rightarrow x - \frac{1}{x} = k$

Watch out for (6)

Plus 2 or minus 2
→ Squaring x + 1/x and x − 1/x
A number in front left inside
→ Squaring x + 1/x and x − 1/x
Losing the sign when x is between 0 and 1
→ Switching between x + 1/x and x − 1/x
Minus 4, not minus 2
→ Switching between x + 1/x and x − 1/x
Dividing an equation whose ends do not match
→ Getting x + 1/x from an equation or a surd
A conjugate is the reciprocal only when the product is 1
→ Getting x + 1/x from an equation or a surd

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