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Algebraic Expressions and Identities - Standard Identities: (a+b)², (a-b)², a²-b², (x+a)(x+b)

Grade 8ICSE

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

🔑Concepts

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An algebraic identity is an equality that holds true regardless of the values assigned to its variables. Unlike a standard equation which is true only for specific values of xx, an identity like (a+b)2=a2+2ab+b2(a+b)^2 = a^2 + 2ab + b^2 is a universal rule used to simplify expressions and perform mental calculations quickly.

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The identity (a+b)2=a2+2ab+b2(a+b)^2 = a^2 + 2ab + b^2 can be visualized as the total area of a large square with side length (a+b)(a+b). If you divide this large square into four sections, you get one square of area a2a^2, another square of area b2b^2, and two identical rectangles each having an area of abab. Adding these four areas together gives the complete formula.

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The identity (a−b)2=a2−2ab+b2(a-b)^2 = a^2 - 2ab + b^2 represents the area of a smaller square with side length (a−b)(a-b). Visually, if you start with a large square of area a2a^2 and remove two rectangles of area abab, you have removed the corner square b2b^2 twice. To correct this, we add b2b^2 back once, resulting in the final expression.

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The identity a2−b2=(a+b)(a−b)a^2 - b^2 = (a+b)(a-b) is known as the Difference of Two Squares. Imagine a large square of side aa with a smaller square of side bb cut out from its corner. The remaining L-shaped region can be sliced and rearranged into a single rectangle with dimensions (a+b)(a+b) and (a−b)(a-b), proving that their areas are identical.

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The identity (x+a)(x+b)=x2+(a+b)x+ab(x+a)(x+b) = x^2 + (a+b)x + ab is used when multiplying two binomials that share a common first term xx. Visually, this is a rectangle with length (x+a)(x+a) and width (x+b)(x+b). The area is composed of a square x2x^2, two rectangles with areas axax and bxbx (which combine to (a+b)x(a+b)x), and a small rectangle with area abab.

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Identities are powerful tools for mental arithmetic. For example, to find 1052105^2, we can treat it as (100+5)2(100+5)^2 and apply the first identity. Similarly, products like 98×10298 \times 102 can be solved as (100−2)(100+2)(100-2)(100+2) using the third identity, making complex multiplication much simpler.

📐Formulae

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

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

(a+b)(a−b)=a2−b2(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

(a+b)2+(a−b)2=2(a2+b2)(a+b)^2 + (a-b)^2 = 2(a^2 + b^2)

(a+b)2−(a−b)2=4ab(a+b)^2 - (a-b)^2 = 4ab

💡Examples

Problem 1:

Expand the expression (3x+5y)2(3x + 5y)^2 using a standard identity.

Solution:

Step 1: Identify the appropriate identity. Since this is in the form (a+b)2(a+b)^2, we use (a+b)2=a2+2ab+b2(a+b)^2 = a^2 + 2ab + b^2. Step 2: Substitute a=3xa = 3x and b=5yb = 5y into the formula. (3x+5y)2=(3x)2+2(3x)(5y)+(5y)2(3x + 5y)^2 = (3x)^2 + 2(3x)(5y) + (5y)^2 Step 3: Simplify each term. (3x)2=9x2(3x)^2 = 9x^2 2(3x)(5y)=30xy2(3x)(5y) = 30xy (5y)2=25y2(5y)^2 = 25y^2 Step 4: Combine the terms. Result: 9x2+30xy+25y29x^2 + 30xy + 25y^2

Explanation:

We applied the Square of a Binomial Sum identity. It is crucial to square both the coefficient and the variable in terms like (3x)2(3x)^2.

Problem 2:

Evaluate 103×97103 \times 97 using the identity a2−b2=(a+b)(a−b)a^2 - b^2 = (a+b)(a-b).

Solution:

Step 1: Express the numbers as (a+b)(a+b) and (a−b)(a-b) relative to a common base. 103=(100+3)103 = (100 + 3) and 97=(100−3)97 = (100 - 3). Step 2: Apply the identity (a+b)(a−b)=a2−b2(a+b)(a-b) = a^2 - b^2 where a=100a = 100 and b=3b = 3. 103×97=(100+3)(100−3)=1002−32103 \times 97 = (100 + 3)(100 - 3) = 100^2 - 3^2 Step 3: Calculate the squares. 1002=10000100^2 = 10000 32=93^2 = 9 Step 4: Subtract the values. 10000−9=999110000 - 9 = 9991 Result: 99919991

Explanation:

This identity is extremely efficient for multiplying numbers that are equidistant from a round number like 100.