Using an identity, \(103^2\) equals
- (a)\(10900\)
- (b)\(10069\)
- (c)\(10609\)
- (d)\(10309\)
Revision notes, 17 board-style questions with the step marks shown, and a four-step route from the basics to 95+, each step ending in a short checkpoint.
Algebra unit: 20 of 80 theory marks (Introduction to Polynomials, Sequences and Progressions, Exploring Algebraic Identities, Linear Equations in Two Variables).
In the NCERT book Ganita Manjari: Part I, Chapter 4, Exploring Algebraic Identities.
An identity is an equality that is true for every value of the variables, such as \((a + b)^2 = a^2 + 2ab + b^2\). An equation such as \(2x + 1 = 7\) is true only for some values. Many identities can be seen as areas: \((a + b)^2\) is a square split into \(a^2\), \(b^2\) and two \(ab\) rectangles.
Look for a pattern: a difference of two squares (\(49x^2 - 25 = (7x - 5)(7x + 5)\)), a perfect square (\(x^2 - 10x + 25 = (x - 5)^2\)), a sum or difference of cubes, or \(x^2 + px + q\) with two numbers whose sum is \(p\) and product \(q\).
\(103^2 = (100 + 3)^2 = 10609\); \(98^2 = (100 - 2)^2 = 9604\); \(47 \times 53 = 50^2 - 3^2 = 2491\).
If \(a + b = 9\) and \(ab = 14\), find \(a^2 + b^2\).
\(a^2 + b^2 = 81 - 28 = 53\).
Factorise \(27x^3 - 8\).
\((3x)^3 - 2^3 = (3x - 2)(9x^2 + 6x + 4)\).
Evaluate \(59 \times 61\) without multiplying directly.
\((60 - 1)(60 + 1) = 3600 - 1 = 3599\).
Topics in this chapter: Mental arithmetic with identities · Factorising with identities · Square identities · Cube identities.
Work through the steps in order. Take each checkpoint closed book, about 1.5 minutes per mark; pass at 80% to move on. Two misses in a row means going back one step.
You can pick the right identity to expand a square, a product or a cube, and use one for quick arithmetic.
You can factorise with identities and find values such as a² + b² or a³ − b³ from given sums and products.
You can expand, evaluate and factorise in multi-part questions and use identities for areas and volumes.
You can chain identities, such as from x − 1/x to x⁴ + 1/x⁴, and set up identity-based models.
Original questions in the board's styles. Multiple-choice answers are checked as you go; for written answers, compare your working with the step mark scheme and record your marks. 3 of the 17 are competency-based (case studies and questions set in a real-life situation): the "Competency-based" button shows just those.
Using an identity, \(103^2\) equals
\(49x^2 - 25\) factorises as
If \(a + b = 7\) and \(ab = 10\), then \(a^2 + b^2\) is
Stretch yourself: original algebra problems at Intermediate level (ages 13 to 16), with hints and full solutions: Problem I24 · Problem I106 · Problem I113 · Problem I120. All 29 →
Want it against the clock? Take a timed 30-mark chapter test on Exploring Algebraic Identities, new questions each time, or climb the Difficulty ladder, levels 1 to 10, three questions a level (CBSE Essentials or the free trial).