Tips & Tricks for Algebra

Algebra is key to aptitude tests, helping solve problems quickly and efficiently. By mastering simple tricks like simplifying equations, spotting patterns, and using shortcuts, you can save time and boost accuracy. This article covers practical algebra techniques to make solving aptitude questions easier and faster.

Important Algebraic Formulas

Here is a list of essential algebraic formulas often used in aptitude tests:

Basic Algebraic Identities

• (a + b)² = a² + 2ab + b²
• (a − b)² = a² − 2ab + b²
• a² − b² = (a + b) (a − b)
• (x + a) (x + b) = x² + ax + bx + ab
• (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
• (a + b - c)² = a² + b² + c² + 2ab - 2bc - 2ca
• (a - b - c)² = a² + b² + c² - 2ab + 2bc - 2ca
• 2(a² + b²) = (a + b)² + (a - b)²
• (a + b)² - (a − b)² = 4ab

Factorization Formulas

• x³ + y³ = (x + y) (x² − xy + y²)
• x³ − y³ = (x − y) (x² + xy + y²)

Cubic Identities

• (a + b)³ = a³ + b³ + 3a²b + 3ab²
• (a − b)³ = a³ − b³ − 3a²b + 3ab²

Quadratic Equation Formula

For ax² + bx + c = 0

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

For Arithematic Progression (AP)

• n-th term of an AP: a_n = a + (n − 1)d
• Sum of first n terms: S_n = n/2[2a + (n − 1)d]

Geometric Progression (GP)

• n-th term of a GP: a_n = arⁿ⁻¹
• The sum of the first n terms, S_n = a (1 – rⁿ) / (1 – r)
• Sum of infinite terms when r < 1, S = a / (1 – r)

Logarithmic Formulas

• log_a (xy) = log_a x + log_a y
• log_a (x/y) = log_a x – log_a y
• log_a x^m = m log_a x
• log_a a = 1
• log_a 1 = 0

Read More:

• Basics of Algebra.
• Algebraic Formula.

Tricks to Solve Algebraic Questions

Below are some of the tricks and shortcut formulas to quickly solve the Algebraic equations.

Trick 1: Simplifying Powers of x ± 1/x: A Step-by-Step Approach

Given: x + 1/x = k, find x² + 1/x².

Step 1: Square both sides

• (x + 1/x)² = k²

Expand

• x² + 1/x² + 2 = k²

Rearrange

• x² + 1/x² = k²- 2

Note: similarly for x - 1/x = k the value will be K² + 2

Trick 2: Trick for Simplifying Higher Powers of x + 1/x and x − 1/x

The higher powers of x + 1/x² can be solved quickly by applying the formulas below:

• x² + 1/x^{2 =} a² - 2
• x³ + 1/x^{3 =} a³ - 3a
• x⁴ + 1/x^{4 =} a⁴ - 4a² + 2
• x⁵ + 1/x^{5 =} a⁵ - 5a³ + 5a
• x⁶ + 1/x^{6 =} a⁶ - 6a⁴ + 9a² - 2

Similarly for x - 1/x = a

• x² + 1/x^{2 =} a² + 2
• x³ + 1/x^{3 =} a³ + 3a
• x⁴ + 1/x^{4 =} a⁴ + 4a² + 2
• x⁵ + 1/x^{5 =} a⁵ + 5a³ + 5a
• x⁶ + 1/x^{6 =} a⁶ + 6a⁴ + 9a² + 2

Trick 3: Simplifying Powers of x + 1/x = 2

• If x + 1/x = 2 , then for any positive integer n, the value of any equation x^m + 1/xⁿ will also be equal to 2.
• If x + 1/x = 2 , then for any positive integer n, the value of any equation xⁿ + 1/xⁿ will also be equal to 2.
• If x + 1/x = 2 , then for any positive integer n, the value of any equation xⁿ - 1/xⁿ will also be equal to 0.
• If x + 1/x = 2 , then for any positive integer n, the value of any equation x^m - 1/xⁿ be equal to 0.

Trick 4: Quick Formulas for x − 1/x​ and x + 1/x​ Relations

This trick helps you find the value of x − 1/x or x + 1/x​ using a simple formula when the other is given.

The value of x - 1/x can be easily be found by applying the formula √m² - 4

Similarly for the second equation

The value of x + 1/x can be easily be found by applying the formula √n² + 4

Algebra Tips & Tricks - Examples

Example 1: If x + 1/x = 18 then find the value of x - 1/x..

Solution:

Applying the formula: √m² - 4

= √18² - 4

= √324 - 4

= ±√320 or ±8√5

Example 2: If x - 1/x = 11 then find the value of x + 1/x.

Solution:

Applying the formula: √n² + 4

= √11² + 4

= √121 + 4

= ±√125 or ±5√5

Practice Algebra:

• Algebra Quiz
• Algebra Practice Questions