Chapter 14 - Algebraic Formulae - Expansion of Squares
Practice set 50
1. Expand.
(i) (5a + 6b)²
Solution:
Here,
a = 5a and b = 6b
We know that,
(a + b)² = a² + 2ab + b²
∴ (5a + 6b)² = (5a)² + 2 × 5a × 6b + (6b)²
∴ (5a + 6b)² = 25a² + 60ab + 36b²
Ans: (5a + 6b)² = 25a² + 60ab + 36b²
(ii) \((\large \frac {a}{2} – \large \frac {b}{3})\)²
Solution:
Here,
a = \(\large \frac {a}{2}\) and b = \(\large \frac {b}{3}\)
We know that,
(a + b)² = a² + 2ab + b²
∴ \((\large \frac {a}{2} – \large \frac {b}{3}\))² = (\(\large \frac {a}{2}\))² + 2 × \(\large \frac {a}{2}\) × \(\large \frac {b}{3}\) + (\(\large \frac {b}{3}\))²
∴ \((\large \frac {a}{2} – \large \frac {b}{3}\))² = \(\large \frac {a²}{4}\) + \(\large \frac {ab}{3}\) + \(\large \frac {b²}{9}\)
Ans: \((\large \frac {a}{2} – \large \frac {b}{3}\))² = \(\large \frac {a²}{4}\) + \(\large \frac {ab}{3}\) + \(\large \frac {b²}{9}\)
(iii) (2p – 3q)²
Solution:
Here,
a = 2p and b = 3q
We know that,
(a – b)² = a² – 2ab + b²
∴ (2p – 3q)² = (2p)² – 2 × 2p × 3q + (3q)²
∴ (2p – 3q)² = 4p² – 12pq + 9q²
Ans: (2p – 3q)² = 4p² – 12pq + 9q²
(iv) \((x – \large \frac {2}{x}\))²
Solution:
Here,
a = x and b = \(\large \frac {2}{x}\)
We know that,
(a – b)² = a² – 2ab + b²
∴ \((x – \large \frac {2}{x}\))² = (x)² – 2 × x × \(\large \frac {2}{x}\) + (\(\large \frac {2}{x}\))²
∴ \((x – \large \frac {2}{x}\))² = x² – 4 + \(\large \frac {4}{x²}\)
Ans: \((x – \large \frac {2}{x}\))² = x² – 4 + \(\large \frac {4}{x²}\)
(v) (ax + by)²
Solution:
Here,
a = ax and b = by
We know that,
(a + b)² = a² + 2ab + b²
∴ (ax + by)² = (ax)² + 2 × ax × by + (by)²
∴ (ax + by)² = a²x² + 2axby + b²y²
Ans: (ax + by)² = a²x² + 2axby + b²y²
(vi) (7m – 4)²
Solution:
Here,
a = 7m and b = 4
We know that,
(a – b)² = a² – 2ab + b²
∴ (7m – 4)² = (7m)² – 2 × 7m × 4 + (4)²
∴ (7m – 4)² = 49m² – 56m + 16
Ans: (7m – 4)² = 49m² – 56m + 16
(vii) \((x – \large \frac {1}{2}\))²
Solution:
Here,
a = x and b = \(\large \frac {1}{2}\)
We know that,
(a – b)² = a² – 2ab + b²
∴ \((x – \large \frac {1}{2}\))² = (x)² – 2 × x × \(\large \frac {1}{2}\) + (\(\large \frac {1}{2}\))²
∴ \((x – \large \frac {1}{2}\))² = x² – x + \(\large \frac {1}{4}\)
Ans: \((x – \large \frac {1}{2}\))² = x² – x + \(\large \frac {1}{4}\)
(viii) \((a – \large \frac {1}{a}\))²
Solution:
Here,
a = a and b = \(\large \frac {1}{a}\)
We know that,
(a – b)² = a² – 2ab + b²
∴ \((a – \large \frac {1}{a}\))² = (a)² – 2 × a × \(\large \frac {1}{a}\) + (\(\large \frac {1}{a}\))²
∴ \((a – \large \frac {1}{a}\))² = a² – 2 + \(\large \frac {1}{a²}\)
Ans: \((a – \large \frac {1}{a}\))² = a² – 2 + \(\large \frac {1}{a²}\)
2. Which of the options given below is the square of the binomial \(8 – \large \frac {1}{x}\) ?
(i) 64 – \(\large \frac {1}{x²}\)
(ii) 64 + \(\large \frac {1}{x²}\)
(iii) 64 – \(\large \frac {16}{x}\) + \(\large \frac {1}{x²}\)
(iv) 64 + \(\large \frac {16}{x}\) + \(\large \frac {1}{x²}\)
Solution:
In \((8 – \large \frac {1}{x}\))²,
a = 8 and b = \(large \frac {1}{x}\)
We know that,
(a – b)² = a² – 2ab + b²
∴ \((8 – \large \frac {1}{x}\))² = (8)² – 2 × 8 × \(large \frac {1}{x}\) + \((large \frac {1}{x}\))²
∴ \((8 – \large \frac {1}{x}\))² = 64 – \(\large \frac {16}{x}\) + \(\large \frac {1}{x²}\)
OPTION (C) : 64 – \(\large \frac {16}{x}\) + \(\large \frac {1}{x²}\)
3. Of which of the binomials given below is m²n² + 14mnpq + 49p²q² the expansion?
(i) (m + n) (p + q)
(ii) (mn – pq)
(iii) (7mn + pq)
(iv) (mn + 7pq)
Solution:
We know that,
(a + b)² = a² + 2ab + b²
∴ m²n² + 14mnpq + 49p²q² = (mn)² + 2 × mn × 7pq + (7pq)²
∴ m²n² + 14mnpq + 49p²q² = (mn + 7pq)²
OPTION (D) : (mn + 7pq)
4. Use an expansion formula to find the values.
(i) (997)²
Solution:
We can write 997 as (1000 – 3)
∴ 997² = (1000 – 3)²
Here a = 1000 and b = 3
We know that,
(a – b)² = a² – 2ab + b²
∴ (1000 – 3)² = 1000² – 2 × 1000 × 3 + 3²
∴ (1000 – 3)² = 1000000 – 6000 + 9
∴ (1000 – 3)² = 1000000 – 5991
∴ (1000 – 3)² = 994009
Ans: (997)² = 994009
(ii) (102)²
Solution:
We can write 102 as (100 + 2)
∴ (102)² = (100 + 2)²
Here a = 100 and b = 2
We know that,
(a + b)² = a² + 2ab + b²
∴ (100 + 2)² = 100² + 2 × 100 × 2 + 2²
∴ (100 + 2)² = 10000 + 400 + 4
∴ (100 + 2)² = 1000000 + 404
∴ (100 + 2)² = 10404
Ans: (102)² = 10404
(iii) (97)²
Solution:
We can write 997 as (100 – 3)
∴ 97² = (100 – 3)²
Here a = 100 and b = 3
We know that,
(a – b)² = a² – 2ab + b²
∴ (100 – 3)² = 100² – 2 × 100 × 3 + 3²
∴ (100 – 3)² = 10000 – 600 + 9
∴ (100 – 3)² = 10000 – 591
∴ (100 – 3)² = 9409
Ans: (97)² = 9409
(iv) (1005)²
Solution:
We can write 997 as (1000 + 5)
∴ 997² = (1000 + 5)²
Here a = 1000 and b = 5
We know that,
(a + b)² = a² + 2ab + b²
∴ (1000 + 5)² = 1000² + 2 × 1000 × 5 + 5²
∴ (1000 + 5)² = 1000000 + 10000 + 25
∴ (1000 + 5)² = 1000000 + 10025
∴ (1000 + 5)² = 1010025
Ans: (1005)² = 1010025
Practice set 51
1. Use the formula to multiply the following.
(i) (x + y) (x – y)
Solution:
Here, a = x and b = y
We know that,
(a + b)(a – b) = a² – b²
∴ (x + y)(x – y) = x² – y²
Ans: (x + y)(x – y) = x² – y²
(ii) (3x – 5) (3x + 5)
Solution:
Here, a = 3x and b = 5
We know that,
(a + b)(a – b) = a² – b²
∴ (3x – 5) (3x + 5) = (3x)² – (5)²
∴ (3x – 5) (3x + 5) = 9x² – 25
Ans: (3x – 5) (3x + 5) = 9x² – 25
(iii) (a + 6) (a – 6)
Solution:
Here, a = a and b = 6
We know that,
(a + b)(a – b) = a² – b²
∴ (a + 6) (a – 6) = (a)² – (6)²
∴ (a + 6) (a – 6) = a² – 36
Ans: (a + 6) (a – 6) = a² – 36
(iv) \((\large \frac {x}{5}\) + 6\()\) \((\large \frac {x}{5}\) – 6\()\)
Solution:
Here, a = \(\large \frac {x}{5}\) and b = 6
We know that,
(a + b)(a – b) = a² – b²
∴ \((\large \frac {x}{5}\) + 6\()\) \((\large \frac {x}{5}\) – 6\()\) = (\(\large \frac {x}{5}\))² – (6)²
∴ \((\large \frac {x}{5}\) + 6\()\) \((\large \frac {x}{5}\) – 6\()\) = \(\large \frac {x²}{25}\) – 36
Ans: \((\large \frac {x}{5}\) + 6\()\) \((\large \frac {x}{5}\) – 6\()\) = \(\large \frac {x²}{25}\) – 36
2. Use the formula to find the values.
(i) 502 × 498
Solution:
We can write 502 as (500 + 2) and 498 as (500 – 2)
∴ 502 × 498 = (500 + 2) (500 – 2)
Here a = 500 and b = 2
(a + b)(a – b) = a² – b²
∴ (500 + 2) (500 – 2) = (500)² – (2)²
∴ (500 + 2) (500 – 2) = 250000 – 4
∴ (500 + 2) (500 – 2) = 249996
Ans: 502 × 498 = 249996
(ii) 97 × 103
Solution:
We can write 97 as (100 – 3) and 103 as (100 + 3)
∴ 97 × 103 = (100 – 3) (100 + 3)
Here a = 100 and b = 3
(a + b)(a – b) = a² – b²
∴ (100 – 3) (100 + 3) = (100)² – (3)²
∴ (100 – 3) (100 + 3) = 10000 – 9
∴ (100 – 3) (100 + 3) = 9991
Ans: 97 × 103 = 9991
(iii) 54 × 46
Solution:
We can write 54 as (50 + 4) and 46 as (50 – 4)
∴ 54 × 46 = (50 + 4) (50 – 4)
Here a = 50 and b = 4
(a + b)(a – b) = a² – b²
∴ (50 + 4) (50 – 4) = (50)² – (4)²
∴ (50 + 4) (50 – 4) = 2500 – 16
∴ (50 + 4) (50 – 4) = 2484
Ans: 54 × 46 = 2484
(iv) 98 × 102
Solution:
We can write 98 as (100 – 2) and 102 as (100 + 2)
∴ 98 × 102 = (100 – 2) (100 + 2)
Here a = 100 and b = 2
(a + b)(a – b) = a² – b²
∴ (100 – 2) (100 + 2) = (100)² – (2)²
∴ (100 – 2) (100 + 2) = 10000 – 4
∴ (100 – 2) (100 + 2) = 9996
Ans: 98 × 102 = 9996
Practice set 52
Factorise the following expressions and write them in the product form.
(i) 201 a³b²
Solution:
201 a³b² = 201 × a³ × b²
∴ 201 a³b² = 3 × 67 × a × a × a × b × b
Ans: 201 a³b² = 3 × 67 × a × a × a × b × b
(ii) 91 xyt²
Solution:
91 xyt² = 91 × x × y × t²
∴ 91 xyt² = 7 × 13 × x × y × t × t
Ans: 91 xyt² = 7 × 13 × x × y × t × t
(iii) 24 a²b²
Solution:
24 a²b² = 24 × a² × b²
∴ 24 a²b² = 2 × 2 × 2 × 3 × a × a × b × b
Ans: 24 a²b² = 2 × 2 × 2 × 3 × a × a × b × b
(iv) tr²s³
Solution:
tr²s³ = t × r² × s³
∴ tr²s³ = t × r × r × s × s × s
Ans: tr²s³ = t × r × r × s × s × s
Practice set 53
Factorise the following expressions.
(i) p² – q²
Solution:
In p² – q²
a = p and b = q
We know that,
(a² – b²) = (a + b)(a – b)
∴ p² – q² = (p + q)(p – q)
Ans: p² – q² = (p + q)(p – q)
(ii) 4x² – 25y²
Solution:
4x² – 25y² = (2x)² – (5y)²
In (2x)² – (5y)²
a = 2x and b = 5y
We know that,
(a² – b²) = (a + b)(a – b)
∴ (2x)² – (5y)² = (2x + 5y)(2x – 5y)
Ans: 4x² – 25y² = (2x + 5y)(2x – 5y)
(iii) y² – 4
Solution:
y² – 4 = y² – 2²
In y² – 2²
a = y and b = 2
We know that,
(a² – b²) = (a + b)(a – b)
∴ y² – 2² = (y + 2)(y – 2)
Ans: y² – 2² = (y + 2)(y – 2)
(iv) p² – \(\large \frac {1}{25}\)
Solution:
p² – \(\large \frac {1}{25}\) = p² – (\(\large \frac {1}{5}\))²
In p² – (\(\large \frac {1}{5}\))²
a = p and b = \(\large \frac {1}{5}\)
We know that,
(a² – b²) = (a + b)(a – b)
∴ p² – (\(\large \frac {1}{5}\))² = (p + \(\large \frac {1}{5}\))(p – \(\large \frac {1}{5}\))
Ans: p² – \(\large \frac {1}{25}\) = (p + \(\large \frac {1}{5}\))(p – \(\large \frac {1}{5}\))
(v) 9x² – \(\large \frac {1}{16}\) y²
Solution:
9x² – \(\large \frac {1}{16}\) y² = (3x)² – (\(\large \frac {1}{4}\) y)²
In (3x)² – (\(\large \frac {1}{4}\) y)²,
a = 3x and b = \(\large \frac {1}{4}\) y
We know that,
(a² – b²) = (a + b)(a – b)
∴ (3x)² – (\(\large \frac {1}{4}\) y)² = (3x + \(\large \frac {1}{4}\) y)(3x – \(\large \frac {1}{4}\) y)
Ans: 9x² – \(\large \frac {1}{16}\) y² = (3x + \(\large \frac {1}{4}\) y)(3x – \(\large \frac {1}{4}\) y)
(vi) x² – \(\large \frac {1}{x²}\)
Solution:
x² – \(\large \frac {1}{x²}\) = x² – \((\large \frac {1}{x})\)²
In x² – \((\large \frac {1}{x})\)²,
a = x and b = \(\large \frac {1}{x}\)
We know that,
(a² – b²) = (a + b)(a – b)
∴ x² – \((\large \frac {1}{x})\)² = (x + \(\large \frac {1}{x})\)(x – \(\large \frac {1}{x})\)
Ans: x² – \(\large \frac {1}{x²}\) = (x + \(\large \frac {1}{x})\)(x – \(\large \frac {1}{x})\)
(vii) a²b – ab
Solution:
a²b – ab = a (ab – b)
∴ a²b – ab = ab (a – 1)
Ans: a²b – ab = ab (a – 1)
(viii) 4x²y – 6x²
Solution:
4x²y – 6x² = 2 (2x²y – 3x²)
∴ 4x²y – 6x² = 2x² (2y – 3)
Ans: 4x²y – 6x² = 2x² (2y – 3)
(ix) \(\large \frac {1}{2}\)y² – 8z²
Solution:
\(\large \frac {1}{2}\)y² – 8z² = \(\large \frac {1}{2}\)y² – \(\large \frac {1}{2}\) × 16z²
∴ \(\large \frac {1}{2}\)y² – 8z² = \(\large \frac {1}{2}\) [y² – 16z²]
∴ \(\large \frac {1}{2}\)y² – 8z² = \(\large \frac {1}{2}\) [y² – (4z)²]
We know that,
(a² – b²) = (a + b)(a – b)
∴ \(\large \frac {1}{2}\)y² – 8z² = \(\large \frac {1}{2}\) [(y + 4z)(y – 4z)]
Ans: \(\large \frac {1}{2}\)y² – 8z² = \(\large \frac {1}{2}\) [(y + 4z)(y – 4z)]
(x) 2x² – 8y²
Solution:
2x² – 8y² = 2 (x² – 4y²)
∴ 2x² – 8y² = 2 [x² – (2y)²]
We know that,
(a² – b²) = (a + b)(a – b)
∴ 2x² – 8y² = 2 [(x + 2y)(x – 2y)]
Ans: 2x² – 8y² = 2 [(x + 2y)(x – 2y)]