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Exploring Algebraic Identities - Visualising Identities

Grade 9CBSE

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

🔑Concepts

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An algebraic identity is an algebraic equation that is true for all values of the variables occurring in it.

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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 area of a square with side length (a+b)(a+b). This square can be divided into a square of area a2a^2, a square of area b2b^2, and two rectangles each of area abab.

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The identity (a−b)2=a2−2ab+b2(a-b)^2 = a^2 - 2ab + b^2 is visualized by taking a square of side aa and removing two rectangles of dimensions a×ba \times b, then adding back the square b2b^2 that was subtracted twice.

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The identity a2−b2=(a−b)(a+b)a^2 - b^2 = (a-b)(a+b) can be seen as the difference between the areas of two squares, which can be rearranged to form a rectangle with sides (a−b)(a-b) and (a+b)(a+b).

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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 represents the area of a rectangle with length (x+a)(x+a) and breadth (x+b)(x+b), composed of a square of area x2x^2, two rectangles of areas axax and bxbx, and a small rectangle of area abab.

📐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

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

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

💡Examples

Problem 1:

Evaluate 1032103^2 using algebraic identities.

Solution:

1032=(100+3)2103^2 = (100 + 3)^2 Using (a+b)2=a2+2ab+b2(a+b)^2 = a^2 + 2ab + b^2 where a=100a = 100 and b=3b = 3: 1032=1002+2(100)(3)+32103^2 = 100^2 + 2(100)(3) + 3^2 1032=10000+600+9103^2 = 10000 + 600 + 9 10000600+910609\begin{array}{r} 10000 \\ 600 \\ + 9 \\ \hline 10609 \end{array}

Explanation:

We break 103103 into (100+3)(100 + 3) to make the calculation simpler using the square of a sum identity.

Problem 2:

Expand (2x+5y)2(2x + 5y)^2 using the appropriate identity.

Solution:

Using the identity (a+b)2=a2+2ab+b2(a+b)^2 = a^2 + 2ab + b^2: Let a=2xa = 2x and b=5yb = 5y. (2x+5y)2=(2x)2+2(2x)(5y)+(5y)2(2x + 5y)^2 = (2x)^2 + 2(2x)(5y) + (5y)^2 (2x+5y)2=4x2+20xy+25y2(2x + 5y)^2 = 4x^2 + 20xy + 25y^2

Explanation:

Substitute the terms 2x2x and 5y5y into the identity for the square of a binomial.

Problem 3:

Evaluate 98×10298 \times 102 using identities.

Solution:

98×102=(100−2)(100+2)98 \times 102 = (100 - 2)(100 + 2) Using the identity (a−b)(a+b)=a2−b2(a-b)(a+b) = a^2 - b^2 where a=100a = 100 and b=2b = 2: 98×102=1002−2298 \times 102 = 100^2 - 2^2 98×102=10000−498 \times 102 = 10000 - 4 10000−49996\begin{array}{r} 10000 \\ - 4 \\ \hline 9996 \end{array}

Explanation:

The product of two numbers equidistant from a central number (100100) can be calculated using the difference of squares identity.