Question 1

Find the values of the following using the identity (a – b)2 = a2 – 2ab + b2.
(i) (79)2 (ii) (193)2 (iii) (299)2

Accepted Answer

Using the identity (a - b)^2 = a^2 - 2ab + b^2, we rewrite each number as (a - b) and calculate:

(i) 79^2 = (80 - 1)^2 = 80^2 - 2*80*1 + 1^2 = 6400 - 160 + 1 = 6241

(ii) 193^2 = (200 - 7)^2 = 200^2 - 2*200*7 + 7^2 = 40000 - 2800 + 49 = 37249

(iii) 299^2 = (300 - 1)^2 = 300^2 - 2*300*1 + 1^2 = 90000 - 600 + 1 = 89401 — For each number, express it as (a - b) where a is a round number close to the number, then apply the identity (a - b)^2 = a^2 - 2ab + b^2 to simplify calculations.

Question 2

Find the following squares using one of the above identities. Determine which of these identities will make these calculations easier.
(i) 1172 (ii) 782 (iii) 1982 (iv) 2142 (v) 11042 (vi) 11202

Accepted Answer

Use the identity (a + b)^2 = a^2 + 2ab + b^2 or (a - b)^2 = a^2 - 2ab + b^2 as appropriate:

(i) 117^2 = (100 + 17)^2 = 10000 + 2*100*17 + 289 = 10000 + 3400 + 289 = 13689

(ii) 78^2 = (80 - 2)^2 = 6400 - 320 + 4 = 6084

(iii) 198^2 = (200 - 2)^2 = 40000 - 800 + 4 = 39204

(iv) 214^2 = (200 + 14)^2 = 40000 + 5600 + 196 = 45796

(v) 1104^2 = (1100 + 4)^2 = 1,210,000 + 8,800 + 16 = 1,218,816

(vi) 1120^2 = (1100 + 20)^2 = 1,210,000 + 44,000 + 400 = 1,254,400 — Choose a and b such that the number is expressed as (a + b) or (a - b) with a round number to simplify squaring using the identities.

Question 3

Factor using suitable identities:
(i) 16y2 – 24y + 9
(ii) s2 + 6st + 4t2
(iii) m2 + mk + k2 + 3nk + 2mn + 9n2
(iv) p3 − 2p2 + 3p − 4
(v) 9a2 + 4b2 + c2 − 12ab + 6ac − 4bc

Accepted Answer

Factorization:

(i) 16y^2 - 24y + 9 = (4y - 3)^2 (perfect square trinomial)

(ii) s^2 + 6st + 4t^2 = (s + 2t)(s + 2t) or (s + 2t)^2

(iii) m^2 + mk + k^2 + 3nk + 2mn + 9n^2
Group terms:
(m^2 + mk + k^2) + (3nk + 2mn + 9n^2)
Try to factor as (m + an + bk)^2 or factor by grouping:
Rewrite as (m + 3n + k)(m + 3n + k) = (m + 3n + k)^2

(iv) p^3 − 2p^2 + 3p − 4
Group:
(p^3 - 2p^2) + (3p - 4) = p^2(p - 2) + 1(3p - 4)
No common factor; try factor by grouping or synthetic division:
Try p = 1:
1 - 2 + 3 - 4 = -2 ≠ 0
Try p = 2:
8 - 8 + 6 - 4 = 2 ≠ 0
Try p = 4:
64 - 32 + 12 - 4 = 40 ≠ 0
No rational roots; factorization not straightforward.

(v) 9a^2 + 4b^2 + c^2 − 12ab + 6ac − 4bc
Try to write as (3a - 2b + c)^2:
(3a)^2 = 9a^2
(-2b)^2 = 4b^2
(c)^2 = c^2
Cross terms:
2*(3a)*(-2b) = -12ab
2*(3a)*c = 6ac
2*(-2b)*c = -4bc
So, expression = (3a - 2b + c)^2 — Use perfect square trinomial identities and factor by grouping or trial to factor the expressions.

Question 4

Expand the following using the identity (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca:
(i) (p + 3q + 7r)2
(ii) (3x – 2y + 4z)2

Accepted Answer

(i) (p + 3q + 7r)^2 = p^2 + (3q)^2 + (7r)^2 + 2*p*3q + 2*3q*7r + 2*7r*p
= p^2 + 9q^2 + 49r^2 + 6pq + 42qr + 14rp

(ii) (3x - 2y + 4z)^2 = (3x)^2 + (-2y)^2 + (4z)^2 + 2*3x*(-2y) + 2*(-2y)*4z + 2*4z*3x
= 9x^2 + 4y^2 + 16z^2 - 12xy - 16yz + 24zx — Apply the identity (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca by substituting the given expressions.

Question 5

Is this an identity?
(a+b−c)2 +(a−b+c)2 +(a−b−c)2 = 2a2 + 2b2 + 2c2

Accepted Answer

Yes, it is an identity.

Proof:
Expand each square:

(a + b - c)^2 = a^2 + b^2 + c^2 + 2ab - 2ac - 2bc
(a - b + c)^2 = a^2 + b^2 + c^2 - 2ab + 2ac - 2bc
(a - b - c)^2 = a^2 + b^2 + c^2 - 2ab - 2ac + 2bc

Add all three:
= (a^2 + b^2 + c^2 + 2ab - 2ac - 2bc) + (a^2 + b^2 + c^2 - 2ab + 2ac - 2bc) + (a^2 + b^2 + c^2 - 2ab - 2ac + 2bc)
= 3a^2 + 3b^2 + 3c^2 + (2ab - 2ab - 2ab) + (-2ac + 2ac - 2ac) + (-2bc - 2bc + 2bc)
= 3a^2 + 3b^2 + 3c^2 - 2ab - 2ac - 2bc
But the cross terms cancel out differently:
Actually, summing carefully:
Sum of 2ab terms: 2ab - 2ab - 2ab = -2ab
Sum of -2ac terms: -2ac + 2ac - 2ac = -2ac
Sum of -2bc terms: -2bc - 2bc + 2bc = -2bc
But this contradicts the RHS.

Re-examining, the cross terms cancel out:
Sum of 2ab terms: 2ab - 2ab - 2ab = -2ab
Sum of -2ac terms: -2ac + 2ac - 2ac = -2ac
Sum of -2bc terms: -2bc - 2bc + 2bc = -2bc
So total cross terms sum to -2ab - 2ac - 2bc
Therefore, total sum = 3a^2 + 3b^2 + 3c^2 - 2ab - 2ac - 2bc
But RHS is 2a^2 + 2b^2 + 2c^2
Hence, the given equality is not true as stated.

Therefore, the expression is NOT an identity. — Expanding each term and adding shows the left side does not equal the right side, so it is not an identity.

Question 6

Try to evaluate the following using a suitable identity:
(i) 352 (ii) 652 (iii) 852 (iv) 1052
Do you observe any interesting pattern?

Accepted Answer

Use the identity a^2 = (a + b)(a - b) + b^2:

(i) 35^2 = (30 + 5)(30 - 5) + 5^2 = 25*35 + 25 = 875 + 25 = 900

(ii) 65^2 = (60 + 5)(60 - 5) + 5^2 = 65*55 + 25 = 3575 + 25 = 3600

(iii) 85^2 = (80 + 5)(80 - 5) + 5^2 = 85*75 + 25 = 6375 + 25 = 6400

(iv) 105^2 = (100 + 5)(100 - 5) + 5^2 = 105*95 + 25 = 9975 + 25 = 10000

Pattern observed: The numbers are expressed as (a + b) and (a - b) where b = 5, and the squares are computed quickly using the identity. — This method simplifies squaring numbers near multiples of 10 by using the difference of squares identity.

Question 7

Observe the two rows of figures below. They represent an algebraic identity. Try to identify it.
a + b + c
2c
a + b – c
a – b + c
2b
a – b – c
2a 2b 2c

Accepted Answer

The figures represent the identity:
(a + b + c)(a + b - c)(a - b + c)(a - b - c) = (a^2 - b^2 - c^2)^2 or related to the expansion of (a + b + c)^2 and its variants.

More specifically, the identity is related to the sum of squares and differences of the terms a, b, and c.

This is a hint to explore the factorization or expansion involving these terms. — The two rows show variations of sums and differences of a, b, and c, indicating an algebraic identity involving these expressions.

Question 8

Figure out the product of x + 2 and x + 3 using algebra tiles.

Accepted Answer

Using algebra tiles, the product (x + 2)(x + 3) can be visualized as:

x * x = x^2 (one x^2-tile)

x * 3 = 3x (three x-tiles)

2 * x = 2x (two x-tiles)

2 * 3 = 6 (six unit tiles)

Adding all parts: x^2 + 3x + 2x + 6 = x^2 + 5x + 6 — Algebra tiles represent each term visually, and their total area corresponds to the expanded product.

Question 9

Lay out algebra tiles for x2 + 11x + 30 in such a way that you will see its factors.

Accepted Answer

To factor x^2 + 11x + 30 using algebra tiles, arrange:

- One x^2-tile
- Eleven x-tiles split as 5x and 6x
- Thirty unit tiles arranged in a 5 by 6 rectangle

This arrangement shows the rectangle dimensions as (x + 5) and (x + 6), so:

x^2 + 11x + 30 = (x + 5)(x + 6) — The algebra tiles visually represent the factorization by forming a rectangle whose sides correspond to the factors.

Question 10

Fill in the blanks with the appropriate expressions to make the equation true.
(px + a) (qx + b) = (_____)x2 + (_____)x + _____ . Also, verify your answer using the distributive property.

Accepted Answer

Expanding (px + a)(qx + b):

= px * qx + px * b + a * qx + a * b
= p q x^2 + (p b + a q) x + a b

Therefore,
(px + a)(qx + b) = (p q) x^2 + (p b + a q) x + a b

Verification by distributive property confirms this expansion. — Multiply each term in the first bracket by each term in the second bracket and combine like terms.

Question 11

Consider any three consecutive square numbers, for example, 1, 4, and 9. Add the smallest and the largest squares and then subtract twice the middle square. What is the result?

Accepted Answer

2 — For the consecutive squares 1, 4, and 9, sum of smallest and largest is 1 + 9 = 10. Twice the middle square is 2 × 4 = 8. Subtracting gives 10 – 8 = 2. This pattern holds for any three consecutive squares.

Question 12

Which of the following algebraic expressions represents the area of a square with side length (a + b)?

Accepted Answer

a² + 2ab + b² — The area of a square with side (a + b) is (a + b)², which expands to a² + 2ab + b² by the algebraic identity for the square of a sum.

Question 13

Verify the identity (a + b)² = a² + 2ab + b² for a = –2 and b = –3 by calculating both sides.

Accepted Answer

For a = –2 and b = –3, (a + b) = –5, so (a + b)² = 25. Calculating the right side: a² = 4, b² = 9, and 2ab = 2 × (–2) × (–3) = 12. Sum is 4 + 12 + 9 = 25. Hence, the identity holds. — Given a = –2 and b = –3, (a + b) = –5, so (a + b)² = (–5)² = 25. On the right side, a² = (–2)² = 4, b² = (–3)² = 9, and 2ab = 2 × (–2) × (–3) = 12. Summing these gives 4 + 12 + 9 = 25, which matches the left side, verifying the identity for negative values.

Question 14

What is the difference between an algebraic identity and an equation?

Accepted Answer

An algebraic identity is an equation that is true for all values of the variables involved, such as (a + b)² = a² + 2ab + b². An equation is true only for specific values of the variables, for example, x² – 1 = 24 is true only for x = 5 or x = –5. — An algebraic identity holds universally for every possible value of the variables, making it always true. In contrast, an equation is conditionally true, depending on particular values of variables that satisfy it. This distinction helps in knowing when expressions can be simplified using identities.

Question 15

Given a = 10 and b = 2, compare the values of (a + b)² and a² + b². Which is greater?

Accepted Answer

(a + b)² is greater than a² + b² — For a = 10 and b = 2, (a + b)² = (12)² = 144, while a² + b² = 100 + 4 = 104. Therefore, (a + b)² > a² + b². This is because (a + b)² = a² + 2ab + b², and the term 2ab adds a positive amount when a and b are positive.

Exploring Algebraic Identities

Exploring Algebraic Identities — Study Notes

Introduction

What is an Identity?

Standard Algebraic Identities

Practice Questions — Exploring Algebraic Identities

All 8 Chapters in Mathematics