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Algebraic Expressions and Identities - Applying Identities

Grade 8CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

An Algebraic Identity is an equality that holds true for every value of the variable(s) present in it. Think of it like a balance scale where both sides are identical in value regardless of the weights (numbers) you substitute into the letters.

The identity (a+b)2=a2+2ab+b2(a + b)^2 = a^2 + 2ab + b^2 represents the area of a large square with side length (a+b)(a+b). If you visualize this square divided into four sections based on lengths aa and bb, you will see one square of area a2a^2, another square of area b2b^2, and two identical rectangles each with area abab.

The identity (ab)2=a22ab+b2(a - b)^2 = a^2 - 2ab + b^2 is used for the square of a binomial difference. Visually, this can be understood by starting with a square of side aa and removing two strips of width bb, then adding back the small square b2b^2 that was subtracted twice at the corner.

The identity (a+b)(ab)=a2b2(a + b)(a - b) = a^2 - b^2 is known as the difference of two squares. Geometrically, if you have a square of area a2a^2 and remove a smaller square of area b2b^2 from the corner, the remaining L-shaped area can be rearranged into a rectangle with dimensions (a+b)(a+b) and (ab)(a-b).

The identity (x+a)(x+b)=x2+(a+b)x+ab(x + a)(x + b) = x^2 + (a + b)x + ab applies when the first terms of two binomials are identical. Imagine a rectangle with width (x+a)(x + a) and height (x+b)(x + b); its total area is composed of a square x2x^2, two rectangles axax and bxbx, and a smaller rectangle abab.

Identities are powerful tools for mental math. For example, to calculate 1032103^2, you can visualize it as (100+3)2(100 + 3)^2 and apply the square of a sum identity to avoid tedious long multiplication.

When applying identities, the first step is always identifying the values of aa and bb (or xx) by comparing the given expression with the standard form of the identity. For example, in (3x5y)2(3x - 5y)^2, a=3xa = 3x and b=5yb = 5y.

📐Formulae

(a+b)2=a2+2ab+b2(a + b)^2 = a^2 + 2ab + b^2

(ab)2=a22ab+b2(a - b)^2 = a^2 - 2ab + b^2

(a+b)(ab)=a2b2(a + b)(a - b) = a^2 - b^2

(x+a)(x+b)=x2+(a+b)x+ab(x + a)(x + b) = x^2 + (a + b)x + ab

💡Examples

Problem 1:

Use a suitable identity to find the value of (2x+3y)2(2x + 3y)^2.

Solution:

  1. Identify the identity: This matches (a+b)2=a2+2ab+b2(a + b)^2 = a^2 + 2ab + b^2. \ 2. Identify aa and bb: Here, a=2xa = 2x and b=3yb = 3y. \ 3. Substitute values: (2x+3y)2=(2x)2+2(2x)(3y)+(3y)2(2x + 3y)^2 = (2x)^2 + 2(2x)(3y) + (3y)^2. \ 4. Simplify: 4x2+12xy+9y24x^2 + 12xy + 9y^2.

Explanation:

We use the Square of a Binomial Sum identity because the expression is a sum of two terms raised to the power of 2.

Problem 2:

Evaluate 105×95105 \times 95 using algebraic identities.

Solution:

  1. Rewrite the numbers relative to a common base: 105=(100+5)105 = (100 + 5) and 95=(1005)95 = (100 - 5). \ 2. Identify the identity: This matches (a+b)(ab)=a2b2(a + b)(a - b) = a^2 - b^2. \ 3. Identify aa and bb: Here, a=100a = 100 and b=5b = 5. \ 4. Substitute values: (100+5)(1005)=100252(100 + 5)(100 - 5) = 100^2 - 5^2. \ 5. Calculate: 1000025=997510000 - 25 = 9975.

Explanation:

The Difference of Two Squares identity is applied here by expressing both numbers as (base+difference)(base + difference) and (basedifference)(base - difference).