SChand CLASS 9 Chapter 3 Expansions Exercise 3(A)

 Exercise 3(A)

Question 1

Write down the products for each of the following:

(i) (x + 4) (x + 2)

(ii) (4a – 5) (5a + 6)

(iii) (xy + 6) (xy – 5)

(iv) (7x² – 5y) (x² – 3y)

Sol :

(i) (x+4)(x+2)

$(x+a)(x+b)=x^{2}+x(a+b)+a \times b$

$=x^{2}+x(4+2)+4 \times 2$

$=x^{2}+x(6)+8$

$=x^{2}+6 x+8$

(ii) (4a-5)(5a+6)

=4a(5a+6)-5(5a+6)

$=20 a^{2}+24 a-25 a-30$

$=20 a^{2}-a-30$

(iii) (xy+6)(xy-5)

$=(x y)^{2}+(6-5) x y+6(-5)$

$=x^{2} y^{2}+x y-30$

(iv) $\left(7 x^{2}-5 y\right)\left(x^{2}-3 y\right)$

$=7 x^{2} \times x^{2}-7 x^{2} \times 3 y-5 y \times x^{2}+5 y \times 3 y$

$=7 x^{4}-21 x^{2} y-5 x^{2} y+15 y^{2}$

$=7 x^{4}-26 x^{2} y+15 y^{2}$

Question 2

Write down the squares of the following expressions

(i) 3x + 5y

(ii) 5y – 2z

(iii) 5p – $\frac{1}{4}$q

(iv) (5x + 3y + z)²

(v) (- 3m – 5n + 2p)²

(v) (2x – $\frac{1}{3}$p + 3q)²

Sol :

(i) $(3 x+5 y)^{2}$

We know that $(a+b)^{2}=\left(a^{2}+(b)^{2}+2 a b\right.$

$=(3 x)^{2}+(5 y)^{2}+2 \times 3 x \times 5 y$

$=9 x^{2}+25 y^{2}+30 x y$

(ii) $(5 y-2 z)^{2}$

We know that $(a-b)^{2}=\left(a^{2}+(b)^{2}+2 a b\right.$

$=(5 y)^{2}+(2 z)^{2}-2 \times 5 y \times 2 z$

$=25 y^{2}+4 z^{2}-20 y z$

Question 3

Simplify:

(2x – p + c)² – (2x + p – c)²

Sol :

$(2 x-p+c)^{2}-(2 x+p-c)^{2}$

$=\left(4 x^{2}+p^{2}+c^{2}-4 x p-2 p c+4 c x\right)$

$=\left(4 x^{2}+p^{2}+c^{2}+4 p x-2 p c-4cx)\right.$

$=4 x^{2}+p^{2}+c^{2}-4 x p-2 p c+4 c x-4 x^{2}-p^{2}-c^{2}-4 p x+2pc+4cx$

=-8xp+8cx

Question 4

Write down the following products :

(i) (3b + 7) (3b – 7)

(ii) $(\frac{1}{3}−5x)(\frac{1}{3}+5x)$

(iii) (x³ – 3) (x³ + 3)

(iv) $(a^4−\frac{1}{5y})(a^4+\frac{1}{5y})$

Sol :

(i) (3b+7)(3b-7)

we know that $(a+b)(a-b)=(a)^{2}-(b)^{2}$

$=(3 b)^{2}-(7)^{2}$

$=9 b^{2}-49$

(ii) $\left(\frac{1}{3}-5 x\right)\left(\frac{1}{3}+5 x\right)$

we know that $(a-b)(a+b)=(a)^{2}-(b)^{2}$

$=\left(\frac{1}{3}\right)^{2}-(5 x)^{2}$

$=\frac{1}{9}-25 x^{2}$

(iii) $\left(x^{3}-3\right)\left(x^{3}+3\right)$

We know that $(a-b)(a+b)=(a)^{2}-(b)^{2}$

$=\left(x^{3}\right)^{2}-(3)^{2}$

$=x^{6}-9$

(iv) $\left(a^{4}-\frac{1}{5 y}\right)\left(a^{4}+\frac{1}{5 y}\right)$

We know that $(a-b)(a+b)=(a)^{2}-(b)^{2}$

$=\left(a^{4}\right)^{2}-\left(\frac{1}{5 y}\right)^{2}$

$=a^{8}-\frac{1}{25 y^{2}}$

Question 5

Find the products :

(i) (x + y) (x – y) (x² + y²)

(ii) (a² + b²) (a4 + b4) (a + b) (a – b)

Sol :

(i) $(x+y)(x-y)\left(x^{2}+y^{2}\right)$

$=\left[(x)^{2}-(y)^{2}\right]\left(x^{2}+y^{2}\right)$

$=\left(x^{2}-y^{2}\right)\left(x^{2}+y^{2}\right)$

$=\left(x^{2}\right)^{2}-\left(y^{2}\right)^{2}$

$=x^{4}-y^{4}$

(ii) $\left(a^{2}+b^{2}\right)\left(a^{4}+b^{4}\right)(a+b)(a-b)$

$=\left(a^{2}+b^{2}\right)\left(a^{4}+b^{4}\right)\left[(a)^{2}-(b)^{2}\right]$

$=\left(a^{2}+b^{2}\right)\left(a^{4}+b^{4}\right)\left(a^{2}+b^{2}\right)$

$=\left(a^{2}+b^{2}\right)\left(a^{2}-b^{2}\right)\left(a^{4}+b^{4}\right)$

$=\left[\left(a^{2}\right)^{2}+\left(b^{2}\right)^{2}\right]\left(a^{4}+b^{4}\right)$

$=\left(a^{4}-b^{4}\right)\left(a^{4}+b^{4}\right)$

$=\left(a^{4}\right)^{2}-\left(b^{4}\right)^{2}$

$=a^{8}-b^{8}$

Question 6

State which of the following expressions is a perfect square :

(i) x² + 8x + 16

(ii) y² + 3y + 9

(iii) 4m² + 4m + 1

(iv) 4x² – 2 + $\frac{1}{4x²}$

(v) m² – 6m + 4

Sol :

(i) $x^{2}+8 x+16$

$=(x)^{2}+8 \times x \times 16+(16)^{2}$

$=(x+16)^{2}$ It is a perfect square of $(x+16)$

(ii) $y^{2}+3 y+9$

$=(y)^{2}+3 y+(3)^{2}$

It is not a perfect square because the second term of 3y is not twice the product of y and 3

(iii) $4 m^{2}+4 m+7$

$=(2 m)^2+2\times 2 m+1+(1)^{2}$

$=(2 m+1)^{2}$

Hence , it is a perfect square of (2m+1)

(iv) $4 x^{2}-2+\frac{1}{4 x^{2}}$

$=\left[2 x-\frac{1}{2 x}\right]^{2}$

Hence , it is a perfect square of $\left(2 x-\frac{1}{2 x}\right)$

(v) $m^{2}-6 m+4$

$=(m)^{2}-6 m+(2)^{2}$

It is not a perfect square because the second term 6m is not twice the product of m and 2

Question 7

If 4x² – 12x + k is a perfect square, find the numerical value of k.

Sol :

$=4 x^{2}-12 x+k$

$=(2 x)^{2}-2 \times 2 x \times 3+\left(30)^{2}\right.$

By comparing we get K=9

Question 8

What term should be added to each of the following expression to make it a perfect square?

(i) 4a² + 28a

(ii) 36a² + 49b²

(iii) 4a² + 81

(iv) 9a² + 2ab + b²

(v) $49a^4+ 50a^2b^2 + 16b^4$

Sol :

(i) $4 a^{2}+28 a$

$=(2 a)^{2}+2 \times 2 a \times 7+(7)^{2}$

To complete it in perfect square

We have to add $(7)^{2}=49$

By adding 49, we get $(2 a+7)^{2}$

(ii) $\left(36 a^{2}+49 b^{2}\right)$

$=(6 a)^{2}+(7 b)^{2}+2 \times 69 \times 7$

To complete it in a perfect square 

We have to add 2×6a×7b=84

On adding 84ab, we get $(6a-7b)^2$

(iii) $4 a^{2}+81$

Sol :

$=(2 a)^{2}+(9)^{2}+2 \times 2 a \times 9$

To complete it in a perfect square 

We have to add 2×2a×9=36a

On adding 36a, we get $(2a+9)^2$

(iv) $9 a^{2}+2 a b+b^{2}$

$=(3 a)^{2}+(b)^{2}+2 \times 3 a \times b$

$=(3 a)^{2}+(b)^{2}+6 a b$

To complete it in a perfect square 

We have to add 6ab-2ab=4ab

By adding 4ab

We get $(3a+b)^2$

(v) $49 a^{4}+50 a^{2} b^{2}+16 b^{4}$

$=(7 a)^{2}+2 \times 7 a^{2} \times 4 b^{2}+\left(4 \mathrm{~b}^{2}\right)^{2}$

$=\left(7 a^{2}\right)^{2}+56 a^{2} b^{2}+\left(4 b^{2}\right)^{2}$

To complete it in a perfect square

We have to add $56 a^{2} b^{2}-50 a^{2} b^{2}=6 a^{2} b^{2}$

By adding $6 a^{2} b^{2}$

We get $\left(7 a^{2}+4 b^{2}\right)^{2}$

Question 9

Write down the expansion of the following

(i) (a + 1)³

(ii) (3x – 2y)³

(iii) (x² + y)³

(iv) $\left(2x – \frac{1}{3x}\right)^3$

(v) $\left(\frac{a}{5}+\frac{b}{2}\right)^3$

Sol :

(i) $(a+1)^{3}$

$\Rightarrow(a+1)^{2} \Rightarrow(9)^{3}+3 a^{2} \times 1+3 a \times\left(1)^{2}+(1)^{3}\right.$

$\Rightarrow a^{3}+3 a^{2}+3 a+1$

(ii) $(3 x-2 y)^{3}$

$=(3 x)^{3}-3(3 x)^{2}(2 y)+3(3 x)(2 y)^{2}-(2 y)^{3}$

$= 27x^{3}- 3 \times 9 x^{2} \times 2 y+9 x \times 4 y^{2}-8 y^{3}$

$=27 x^{3}-54 x^{2} y+36 x y^{2}-8 y^{3}$

(iii) $\left(x^{2}+y\right)^{3}$

$=\left(x^{2}\right)^{3}+3\left(x^{2}\right)^{2}(y)+3\left(x^{2}\right)(y)^{2}+(y)^{3}$

$= x^{6}+3 \times x^{4} \times y+3 x^{2} y^{2}+y^{3}$

$= x^{6}+3 x^{4} y+3 x^{2} y^{2}+y^{3}$

(iv) $\left(2 x-\frac{1}{3 x}\right)^{3}$

$=(2 x)^{3}-3 \times (2 x)^{2} \times \left(\frac{1}{3 x}\right)+3(2 x)\left(\frac{1}{3 x}\right)^{2}-\left(\frac{1}{3 x}\right)^{3}$

$=8x^{3}-3 \times 4 x^{2} \times \frac{1}{3x}+3 \times 2 x \times \frac{1}{9x^2}-\frac{1}{27 x^{3}}$

$=8 x^{3}-4 x+\frac{2}{3 x}-\frac{1}{27 x^{3}}$

(v) $\left(\frac{a}{5}+\frac{b}{2}\right)^{3}$

$=\left(\frac{a}{5}\right)^{3}+3\left(\frac{a}{5}\right)^{2}\left(\frac{b}{2}\right)+3\left(\frac{a}{5}\right)\left(\frac{b}{2}\right)^{2}+\left(\frac{b}{2}\right)^{3}$

$=\frac{a^{3}}{125}+3 \times \frac{a^{2}}{25} \times \frac{b}{2}+3 \times \frac{9}{5} \times \frac{b^{2}}{4}+\frac{b^{3}}{8}$

$=\frac{a^{3}}{125}+\frac{3 a^{2} b}{50}+\frac{3 a b^{2}}{20}+\frac{b^{3}}{8}$