Class 9 Maths Ganita Manjari Chapter 4 MCQ – Exploring Algebraic Identities

An identity states an equality that holds for every allowed value of its variables. For example, (a + b)² = a² + 2ab + b² holds for positive, negative and zero values. An ordinary equation need not hold for every value.

Expand the two squares: (n − 1)² = n² − 2n + 1 and (n + 1)² = n² + 2n + 1. Adding them gives 2n² + 2. Subtracting 2n² leaves 2, regardless of the value of n.

Each rectangle has area a × b = ab square units. Together, the two rectangles have area 2ab square units. Adding their areas to those of the two squares explains the identity (a + b)² = a² + 2ab + b².

The identity gives (a + b)² − (a² + b²) = 2ab. The given inequality means that 2ab is negative, so ab < 0. A product of two real numbers is negative only when the numbers have opposite signs.

Expanding the left side gives a² + 2ab + b². For this to equal a² + b², the extra term 2ab must be zero. Therefore, ab = 0, which means that a = 0 or b = 0, or both.

Use (a + b)² = a² + 2ab + b² with a = 3x and b = 2y. The middle term is 2 × 3x × 2y = 12xy. Thus, the expansion is 9x² + 12xy + 4y².

Use (a − b)² = a² − 2ab + b² with a = 5p and b = 3q. The middle term is −2 × 5p × 3q = −30pq. The last term remains positive because (3q)² = 9q².

Write 102 as 100 + 2 and use the square-of-a-sum identity. We get 102² = 100² + 2 × 100 × 2 + 2² = 10000 + 400 + 4 = 10404. The middle term must be included.

Apply (a − b)² = a² − 2ab + b². This gives 98² = 10000 − 400 + 4 = 9604. Subtracting only 2² from 100² would use the wrong expression, since a square of a difference includes a middle term.

First take out the common factor 2 to get 2(9x² + 12xy + 4y²). The expression inside the brackets is (3x + 2y)² because its middle term is 2 × 3x × 2y. Hence, the factorisation is 2(3x + 2y)².

Use the square-of-three-terms identity with terms 2x, −y and 3z. Their squares are 4x², y² and 9z². Twice their pairwise products gives −4xy, −6yz and 12xz. Adding these six terms produces the expression in option D.

The identity (a + b + c)² = a² + b² + c² + 2(ab + bc + ca) connects the given quantities. Substituting gives 36 = a² + b² + c² + 22. Therefore, the required sum of squares is 14.

The expression is a difference of squares: 9x² − 25 = (3x)² − 5². Apply a² − b² = (a − b)(a + b). This gives (3x − 5)(3x + 5), with the middle terms cancelling when multiplied.

Write the product as (100 + 3)(100 − 3). Using (a + b)(a − b) = a² − b², we get 100² − 3² = 10000 − 9 = 9991. This avoids multiplying the two numbers directly.

The total area of the pieces is x² + 8x + 15. To factor it, choose two numbers with sum 8 and product 15: they are 3 and 5. Thus, the area is (x + 3)(x + 5), giving the stated dimensions.

Expanding the proposed factors gives x² + 7x + 10. The middle term is correct, but the constant term is not. The chosen numbers must satisfy both conditions: sum 7 and product 12. The suitable pair is 3 and 4.

Distribute each term: (ax + b)(cx + d) = acx² + adx + bcx + bd. The two terms containing x combine to give (ad + bc)x. Therefore, the coefficient of x is ad + bc.

Split 7x as 3x + 4x. Then 6x² + 3x + 4x + 2 = 3x(2x + 1) + 2(2x + 1). Taking out the common factor gives (3x + 2)(2x + 1), whose product matches all three original terms.

We need two numbers whose sum is −1 and whose product is −20. The numbers −5 and 4 satisfy both requirements. Therefore, x² − x − 20 = (x − 5)(x + 4). Expanding gives the middle term 4x − 5x = −x.

Use (a + b)³ = a³ + 3a²b + 3ab² + b³ with a = x and b = 2y. The middle terms become 6x²y and 12xy², while (2y)³ = 8y³. All four terms are required.

Apply (a − b)³ = a³ − 3a²b + 3ab² − b³ with a = 2x and b = 3. The four terms are 8x³, −36x², 54x and −27. Their alternating signs are important in this expansion.

Each of the first three cuboids has volume a²b, giving 3a²b altogether. Each of the other three has volume ab², giving 3ab². Their combined volume is 3a²b + 3ab², which supplies the middle terms of (a + b)³.

Recognise 8x³ + 27 as (2x)³ + 3³. Use a³ + b³ = (a + b)(a² − ab + b²). Substituting a = 2x and b = 3 gives (2x + 3)(4x² − 6x + 9).

The expression is (3p)³ − (2q)³. Apply a³ − b³ = (a − b)(a² + ab + b²). The second factor is 9p² + 6pq + 4q², so the middle term in that factor has a positive sign.

Use x³ + y³ + z³ − 3xyz = (x + y + z)(x² + y² + z² − xy − xz − yz). Since the first factor is zero, the product is zero. Moving 3xyz to the other side gives the required equality.

First, a² + b² + c² = 9² − 2(26) = 29. The three-variable cubic identity then gives a³ + b³ + c³ − 3abc = (a + b + c)[a² + b² + c² − (ab + bc + ca)] = 9(29 − 26) = 27.

Factor the numerator as (x − 3)(x + 3) and the denominator as (x + 2)(x + 3). The factor x + 3 is nonzero under the stated restrictions, so it can be cancelled. The original restrictions still apply after simplification.

The original denominator is x − 2, which is zero when x = 2. Although cancelling this factor gives x + 2, cancellation is valid only when the factor is nonzero. The original expression therefore remains undefined at x = 2.

Let the length be x metres, so the breadth is x − 4 metres. Then x(x − 4) = 96, giving (x − 12)(x + 8) = 0. Reject the negative length −8. Thus, the length is 12 m and breadth is 8 m.

Factor n³ − n as n(n² − 1) = n(n − 1)(n + 1). These are three consecutive integers. At least one is even and one is divisible by 3, so their product is divisible by 6. For n = 1, the product is 0, also divisible by 6.