Exploring Algebraic Identities | Exercise 4.2

Question 2

Find the values of the following using the identity (a−b)2=a2−2ab+b2(a - b)^2 = a^2 - 2ab + b^2.

(i) (79)2(79)^2

(ii) (193)2(193)^2

(iii) (299)2(299)^2

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Solution
Understand the Question
  • To calculate the square of a number near a multiple of 1010 or 100100 without direct long multiplication, express the number as a difference: (a−b)(a - b).
  • Apply the algebraic identity: (a−b)2=a2−2ab+b2(a - b)^2 = a^2 - 2ab + b^2
  • Substitute the values of aa and bb and simplify the arithmetic.

(i) Evaluate (79)2(79)^2 using the identity (a−b)2=a2−2ab+b2(a - b)^2 = a^2 - 2ab + b^2.

Step 1 · Evaluate (79)2(79)^2

Express 7979 as (80−1)(80 - 1).

Here, a=80a = 80 and b=1b = 1.

(79)2=(80−1)2=(80)2−2×80×1+(1)2=6400−160+1=6240+1=6241\begin{aligned} (79)^2 &= (80 - 1)^2 \\ &= (80)^2 - 2 \times 80 \times 1 + (1)^2 \\ &= 6400 - 160 + 1 \\ &= 6240 + 1 \\ &= 6241 \end{aligned}
Answer

(i) 62416241

(ii) Evaluate (193)2(193)^2 using the identity (a−b)2=a2−2ab+b2(a - b)^2 = a^2 - 2ab + b^2.

Step 1 · Evaluate (193)2(193)^2

Express 193193 as (200−7)(200 - 7).

Here, a=200a = 200 and b=7b = 7.

(193)2=(200−7)2=(200)2−2×200×7+(7)2=40000−2800+49=37200+49=37249\begin{aligned} (193)^2 &= (200 - 7)^2 \\ &= (200)^2 - 2 \times 200 \times 7 + (7)^2 \\ &= 40000 - 2800 + 49 \\ &= 37200 + 49 \\ &= 37249 \end{aligned}
Answer

(ii) 3724937249

(iii) Evaluate (299)2(299)^2 using the identity (a−b)2=a2−2ab+b2(a - b)^2 = a^2 - 2ab + b^2.

Step 1 · Evaluate (299)2(299)^2

Express 299299 as (300−1)(300 - 1).

Here, a=300a = 300 and b=1b = 1.

(299)2=(300−1)2=(300)2−2×300×1+(1)2=90000−600+1=89400+1=89401\begin{aligned} (299)^2 &= (300 - 1)^2 \\ &= (300)^2 - 2 \times 300 \times 1 + (1)^2 \\ &= 90000 - 600 + 1 \\ &= 89400 + 1 \\ &= 89401 \end{aligned}
Answer

(iii) 8940189401

Common Mistakes
  • Sign Error in the Last Term: Writing (a−b)2=a2−2ab−b2(a - b)^2 = a^2 - 2ab - b^2 instead of a2−2ab+b2a^2 - 2ab + b^2. The last term +b2+b^2 is always positive.
  • Using the Addition Identity: Splitting 7979 as (70+9)(70 + 9) uses (a+b)2(a + b)^2, but the question specifically asks to use the (a−b)2(a - b)^2 identity.

More questions in Exercise 4.2

Q1

Factor completely:

(i) 9x2+24xy+16y29x^2 + 24xy + 16y^2

(ii) 4s2+20st+25t24s^2 + 20st + 25t^2

(iii) 49x2+28xy+4y249x^2 + 28xy + 4y^2

(iv) 64p2+323pq+49q264p^2 + \dfrac{32}{3}pq + \dfrac{4}{9}q^2

*(v) 3a2+4ab+43b23a^2 + 4ab + \dfrac{4}{3}b^2

*(vi) 95s2+6sv+5v2\dfrac{9}{5}s^2 + 6sv + 5v^2

(Hint: 2 was taken out as a common factor in Example 7. Is it possible to do something similar in Exercises (v) and (vi) above?)

Q2

Find the values of the following using the identity (a−b)2=a2−2ab+b2(a - b)^2 = a^2 - 2ab + b^2.

(i) (79)2(79)^2

(ii) (193)2(193)^2

(iii) (299)2(299)^2

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