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Algebraic Expressions and Identities - Application of Identities

Grade 8ICSE

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

🔑Concepts

Definition of an Identity: An algebraic identity is an equality that remains true for all possible values of the variables involved. Unlike a standard equation which is only true for specific values, an identity represents a universal rule for transforming expressions.

Identity I (Square of a Sum): The expression (a+b)2=a2+2ab+b2(a + b)^2 = a^2 + 2ab + b^2 is the expansion of a binomial square. Visually, imagine a large square with a side length of (a+b)(a + b). The total area is composed of one square with area a2a^2, another square with area b2b^2, and two identical rectangles each with area abab, filling the total space of the (a+b)(a+b) by (a+b)(a+b) square.

Identity II (Square of a Difference): The expression (ab)2=a22ab+b2(a - b)^2 = a^2 - 2ab + b^2 handles the subtraction of terms. Geometrically, this represents the area of a smaller square (with side aba - b) that remains when you start with a square of side aa and subtract the regions representing 2ab2ab while adding back the b2b^2 that was subtracted twice.

Identity III (Difference of Two Squares): The product (a+b)(ab)=a2b2(a + b)(a - b) = a^2 - b^2 is a fundamental tool for factorization. Visually, if you take a square of area a2a^2 and cut out a smaller square of area b2b^2 from the corner, the remaining L-shaped region can be sliced and rearranged into a single rectangle with dimensions (a+b)(a + b) and (ab)(a - b).

Identity IV (Product of Binomials with a Common Term): The identity (x+a)(x+b)=x2+(a+b)x+ab(x + a)(x + b) = x^2 + (a + b)x + ab is used when the first terms of two binomials are identical. The result is a trinomial where the middle term's coefficient is the sum of the non-common terms and the final constant is their product.

Numerical Applications: Identities can be used to perform complex arithmetic mentally. For example, calculating 102×98102 \times 98 can be simplified using the Difference of Squares identity by rewriting it as (100+2)(1002)(100 + 2)(100 - 2), which equals 100222100^2 - 2^2.

Simplification and Evaluation: Standard identities are often used to simplify long algebraic strings or to find the value of an expression like x2+1x2x^2 + \frac{1}{x^2} given the value of x+1xx + \frac{1}{x} by squaring both sides of the given equation.

📐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

(a+b+c)2=a2+b2+c2+2ab+2bc+2ca(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca

a2+b2=(a+b)22aba^2 + b^2 = (a + b)^2 - 2ab

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

💡Examples

Problem 1:

Using a suitable identity, evaluate 105×95105 \times 95.

Solution:

Step 1: Rewrite the numbers in terms of a common base: 105=(100+5)105 = (100 + 5) 95=(1005)95 = (100 - 5) Step 2: Identify the identity to use. This fits the form (a+b)(ab)=a2b2(a + b)(a - b) = a^2 - b^2. Step 3: Substitute a=100a = 100 and b=5b = 5: (100+5)(1005)=100252(100 + 5)(100 - 5) = 100^2 - 5^2 Step 4: Calculate the squares: 100002510000 - 25 Step 5: Subtract to find the final result: 99759975

Explanation:

This approach uses the 'Difference of Two Squares' identity to simplify multiplication by converting it into a simple subtraction of two squares.

Problem 2:

If x+1x=6x + \frac{1}{x} = 6, find the value of x2+1x2x^2 + \frac{1}{x^2}.

Solution:

Step 1: Start with the given equation x+1x=6x + \frac{1}{x} = 6. Step 2: Square both sides of the equation: (x+1x)2=62(x + \frac{1}{x})^2 = 6^2 Step 3: Apply the identity (a+b)2=a2+2ab+b2(a + b)^2 = a^2 + 2ab + b^2 where a=xa = x and b=1xb = \frac{1}{x}: x2+2(x)(1x)+(1x)2=36x^2 + 2(x)(\frac{1}{x}) + (\frac{1}{x})^2 = 36 Step 4: Simplify the middle term 2(x)(1x)2(x)(\frac{1}{x}) which becomes 22: x2+2+1x2=36x^2 + 2 + \frac{1}{x^2} = 36 Step 5: Subtract 2 from both sides to isolate x2+1x2x^2 + \frac{1}{x^2}: x2+1x2=362=34x^2 + \frac{1}{x^2} = 36 - 2 = 34

Explanation:

This problem uses the Square of a Sum identity to find the value of a derived expression without ever needing to solve for the value of xx itself.