Mathematics Solutions Solutions for Class 7 Math Chapter 15 - Algebraic Formulae Expansion Of Squares

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Page No 93:

Question 1:

Expand.

(i) ( 5a + 6b)² (ii) ${(\frac{a}{2} + \frac{b}{3})}^{2}$ (iii) ${(2 p - 3 q)}^{2}$ (iv) ${(x - \frac{2}{x})}^{2}$

(v) ${(a x + b y)}^{2}$ (vi) ${(7 m - 4)}^{2}$ (vii) ${(x + \frac{1}{2})}^{2}$ (viii) ${(a - \frac{1}{a})}^{2}$

Answer:

It is known that, (a + b)² = a² + 2ab + b² and (a − b)² = a² − 2ab + b².
$(i) {(5 a + 6 b)}^{2} = {(5 a)}^{2} + 2 \times (5 a) (6 b) + {(6 b)}^{2} = 25 a^{2} + 60 a b + 36 b^{2}$
$(ii) {(\frac{a}{2} + \frac{b}{3})}^{2} = {(\frac{a}{2})}^{2} + 2 \times (\frac{a}{2}) \times (\frac{b}{3}) + {(\frac{b}{3})}^{2} = \frac{a^{2}}{4} + \frac{a b}{3} + \frac{b^{2}}{9}$
$(iii) {(2 p - 3 q)}^{2} = {(2 p)}^{2} - 2 \times (2 p) \times (3 q) + {(3 q)}^{2} = 4 p^{2} - 12 p q + 9 q^{2}$
$(iv) {(x - \frac{2}{x})}^{2} = {(x)}^{2} - 2 \times (x) \times (\frac{2}{x}) + {(\frac{2}{x})}^{2} = x^{2} - 4 + \frac{4}{x^{2}}$
$(v) {(a x + b y)}^{2} = {(a x)}^{2} + 2 \times (a x) \times (b y) + {(b y)}^{2} = a^{2} x^{2} + 2 a x b y + b^{2} y^{2}$
$(vi) {(7 m - 4)}^{2} = {(7 m)}^{2} - 2 \times (7 m) \times (4) + {(4)}^{2} = 49 m^{2} - 56 m + 16$
$(vii) {(x + \frac{1}{2})}^{2} = {(x)}^{2} + 2 \times (x) \times (\frac{1}{2}) + {(\frac{1}{2})}^{2} = x^{2} + x + \frac{1}{4}$
$(viii) {(a - \frac{1}{a})}^{2} = {(a)}^{2} - 2 \times (a) \times (\frac{1}{a}) + {(\frac{1}{a})}^{2} = a^{2} - 2 + \frac{1}{a^{2}}$

Page No 93:

Question 2:

Which of the options given below is the square of the binomial $(8 - \frac{1}{x})$ ?

(i) $64 - \frac{1}{x^{2}}$ (ii) $64 + \frac{1}{x^{2}}$ (iii) $64 - \frac{16}{x} + \frac{1}{x^{2}}$ (iv) $64 + \frac{16}{x} + \frac{1}{x^{2}}$

Answer:

The given binomial is $(8 - \frac{1}{x})$ .
${(8 - \frac{1}{x})}^{2} = {(8)}^{2} - 2 \times (8) \times (\frac{1}{x}) + {(\frac{1}{x})}^{2} [∵ {(a - b)}^{2} = a^{2} - 2 a b + b^{2}] = 64 - \frac{16}{x} + \frac{1}{x^{2}}$
Hence, the correct answer is option (iii).

Page No 93:

Question 3:

Of which of the binomials given below is m²n²+ 14mnpq + 49p²q²the expansion?

(i) (m + n) (p + q) (ii) (mn- pq) (iii) (7mn + pq) (iv) (mn + 7pq)

Answer:

Let us check each of the given options.
(i) (m + n)(p + q)
= m(p + q) + n(p + q)
= mp + mq + np + nq
So, it is not a correct option.

(ii) (mn − pq)²
= (mn)² − 2 × (mn) × (pq) + (pq)² [âˆµ (a − b)² = a² − 2ab + b²]
= m²n² − 2mnpq + p²q²
So, it is not a correct option.

(iii) (7mn + pq)²
= (7mn)² + 2 × (7mn) × (pq) + (pq)² [âˆµ (a + b)² = a² + 2ab + b²]
= 49m²n² + 14mnpq + p²q²
So, it is not a correct option.

(iv) (mn + 7pq)²
= (mn)² + 2 × (mn) × (7pq) + (7pq)² [âˆµ (a + b)² = a² + 2ab + b²]
= m²n² + 14mnpq + 49p²q²
So, it is a correct option.

Hence, the correct answer is option (iv).

Page No 93:

Question 4:

Use an expansion formula to find the values.

(i) (997)² (ii) (102)² (iii) (97)²(iv) (1005)²

Answer:

It is known that, (a + b)² = a² + 2ab + b² and (a − b)² = a² − 2ab + b²

(i) (997)²
= (1000 − 3)²
= (1000)² − 2 × 1000 × 3 + (3)²
= 1000000 − 6000 + 9
= 994009

(ii) (102)²
= (100 + 2)²
= (100)² + 2 × 100 × 2 + (2)²
= 10000 + 400 + 4
= 10404

(iii) (97)²
= (100 − 3)²
= (100)² − 2 × 100 × 3 + (3)²
= 10000 − 600 + 9
= 9409

(iv) (1005)²
= (1000 + 5)²
= (1000)² + 2 × 1000 × 5 + (5)²
= 1000000 + 10000 + 25
= 1010025

Page No 93:

Question 1:

Use the formula to multiply the following.

(i) (x + y) (x

-

y) (ii) (3x

-

5) (3x + 5) (iii) (a + 6) (a

-

6) (iv)

(\frac{x}{5} + 6) (\frac{x}{5} - 6)

Answer:

It is known that, (a + b) (a − b) = a² − b².
$(i) (x + y) (x - y) = {(x)}^{2} - {(y)}^{2} = x^{2} - y^{2}$
$(ii) (3 x - 5) (3 x + 5) = {(3 x)}^{2} - {(5)}^{2} = 9 x^{2} - 25$
$(iii) (a + 6) (a - 6) = {(a)}^{2} - {(6)}^{2} = a^{2} - 36$
$(iv) (\frac{x}{5} + 6) (\frac{x}{5} - 6) = {(\frac{x}{5})}^{2} - {(6)}^{2} = \frac{x^{2}}{25} - 36$

Page No 93:

Question 2:

Use the formula to find the values.

(i) 502 × 498 (ii) 97 × 103 (iii) 54 × 46 (iv) 98 × 102

Answer:

It is known that, (a + b) (a − b) = a² − b².

(i) 502 × 498
= (500 + 2) × (500 − 2)
= (500)² − (2)²
= 250000 − 4
= 249996

(ii) 97 × 103
= (100 − 3) × (100 + 3)
= (100)² − (3)²
= 10000 − 9
= 9991

(iii) 54 × 46
= (50 + 4) × (50 − 4)
= (50)² − (4)²
= 2500 − 16
= 2484

(iv) 98 × 102
= (100 − 2) × (100 + 2)
= (100)² − (2)²
= 10000 − 4
= 9996

Page No 94:

Question 1:

Factorise the following expressions and write them in the product form.

(i) 201 a³b², (ii) 91 xyt² , (iii) 24 a²b² , (iv) tr² s³

Answer:

(i) 201a³b²
= 3 × 67 × a × a × a × b × b

(ii) 91xyt²
= 7 × 13 × x × y × t × t

(iii) 24a²b²
= 2 × 2 × 2 × 3 × a × a × b × b

(iv) tr²s³
= t × r × r × s × s × s

Page No 94:

Question 1:

Factorise the following expressions.

(i) p ²

-

q ² (ii) 4x²

-

25y²(iii) y²

-

4 (iv) p²

-

\frac{1}{25}

(v) 9x²

- \frac{1}{16} y^{2}

(vi) x²

-

\frac{1}{x^{2}}

(vii) a²b

-

ab (viii) 4 x²y

-

6 x²

(ix)

\frac{1}{2} y^{2}

- 8 z^{2}

(x)

2 x^{2} - 8 y^{2}

Answer:

$(i) p^{2} - q^{2} = {(p)}^{2} - {(q)}^{2} = (p + q) (p - q) [∵ (a + b) (a - b) = a^{2} - b^{2}]$
$(ii) 4 x^{2} - 25 y^{2} = {(2 x)}^{2} - {(5 y)}^{2} = (2 x + 5 y) (2 x - 5 y) [∵ a^{2} - b^{2} = (a + b) (a - b)]$
$(iii) y^{2} - 4 = {(y)}^{2} - {(2)}^{2} = (y + 2) (y - 2) [∵ a^{2} - b^{2} = (a + b) (a - b)]$
$(iv) p^{2} - \frac{1}{25} = {(p)}^{2} - {(\frac{1}{5})}^{2} = (p + \frac{1}{5}) (p - \frac{1}{5}) [∵ a^{2} - b^{2} = (a + b) (a - b)]$
$(v) 9 x^{2} - \frac{1}{16} y^{2} = {(3 x)}^{2} - {(\frac{1}{4} y)}^{2} = (3 x + \frac{1}{4} y) (3 x - \frac{1}{4} y) [∵ a^{2} - b^{2} = (a + b) (a - b)]$
$(vi) x^{2} - \frac{1}{x^{2}} = {(x)}^{2} - {(\frac{1}{x})}^{2} = (x + \frac{1}{x}) (x - \frac{1}{x}) [∵ a^{2} - b^{2} = (a + b) (a - b)]$
$(vii) a^{2} b - a b = a b (a - 1)$
$(viii) 4 x^{2} y - 6 x^{2} = 2 x^{2} (2 y - 3)$
$(ix) \frac{1}{2} y^{2} - 8 z^{2} = \frac{1}{2} (y^{2} - 16 z^{2}) = \frac{1}{2} [{(y)}^{2} - {(4 z)}^{2}] = \frac{1}{2} (y + 4 z) (y - 4 z) [∵ a^{2} - b^{2} = (a + b) (a - b)]$
$(x) 2 x^{2} - 8 y^{2} = 2 (x^{2} - 4 y^{2}) = 2 [{(x)}^{2} - {(2 y)}^{2}] = 2 (x + 2 y) (x - 2 y)$

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