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Exploring Algebraic Identities - Use algebra tiles and area models to factor quadratic expressions conceptually

Grade 9CBSE

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

🔑Concepts

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Algebra tiles are a visual way to represent algebraic expressions. A large square represents x2x^2 (dimensions x×xx \times x), a rectangle represents xx (dimensions x×1x \times 1), and a small square represents 11 (dimensions 1×11 \times 1).

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Factoring a quadratic expression of the form x2+bx+cx^2 + bx + c using an area model involves arranging these tiles into a single large rectangle.

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The total area of the tiles is equal to the quadratic expression. The length and width of the resulting rectangle represent the factors of the expression.

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To form a rectangle for x2+bx+cx^2 + bx + c, we must find two numbers pp and qq such that p+q=bp + q = b and pq=cpq = c. Geometrically, this means splitting the bb rectangles into two groups to fill the corners around the x2x^2 tile and the cc unit tiles.

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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 the algebraic basis for this model, where the area of the rectangle is (x+a)×(x+b)(x+a) \times (x+b).

📐Formulae

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

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

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

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

💡Examples

Problem 1:

Factor the expression x2+5x+6x^2 + 5x + 6 using the area model concept.

Solution:

x2+5x+6=(x+2)(x+3)x^2 + 5x + 6 = (x+2)(x+3)

Explanation:

To factor x2+5x+6x^2 + 5x + 6, we use one x2x^2 tile, five xx tiles, and six 11 tiles. We arrange the x2x^2 tile at the top left. To form a rectangle, we split the five xx tiles into a group of 22 and 33. We place 22 xx tiles along one side and 33 xx tiles along the adjacent side. The six 11 tiles perfectly fill the remaining 2×32 \times 3 space. The dimensions of this rectangle are (x+2)(x+2) and (x+3)(x+3), which are the factors.

Problem 2:

Use the algebraic identity for (x+a)(x+b)(x+a)(x+b) to factor x2+7x+12x^2 + 7x + 12.

Solution:

x2+7x+12=(x+3)(x+4)x^2 + 7x + 12 = (x+3)(x+4)

Explanation:

Compare x2+7x+12x^2 + 7x + 12 with x2+(a+b)x+abx^2 + (a+b)x + ab. We need two numbers aa and bb such that a+b=7a+b = 7 and ab=12ab = 12. The pairs of factors for 1212 are (1,12)(1,12), (2,6)(2,6), and (3,4)(3,4). Only the pair (3,4)(3,4) sums to 77. Thus, a=3a=3 and b=4b=4. Substituting into the identity, we get (x+3)(x+4)(x+3)(x+4).

Problem 3:

Calculate the area of a square field if its side length is given by (x+5)(x+5) units, and express it as a quadratic trinomial.

Solution:

Area=x2+10x+25Area = x^2 + 10x + 25

Explanation:

The area of a square is (side)2(side)^2. Using the identity (a+b)2=a2+2ab+b2(a+b)^2 = a^2 + 2ab + b^2, where a=xa=x and b=5b=5: (x+5)2=x2+2(x)(5)+52=x2+10x+25(x+5)^2 = x^2 + 2(x)(5) + 5^2 = x^2 + 10x + 25