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 :
We know that $(a+b)^{2}=\left(a^{2}+(b)^{2}+2 a b\right.$
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 :
$=\left(4 x^{2}+p^{2}+c^{2}-4 x p-2 p c+4 c x\right)$
$=\left(4 x^{2}+p^{2}+c^{2}+4 p x-2 p c-4cx)\right.$
$=4 x^{2}+p^{2}+c^{2}-4 x p-2 p c+4 c x-4 x^{2}-p^{2}-c^{2}-4 p x+2pc+4cx$
=-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 :
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 :
$=\left[(x)^{2}-(y)^{2}\right]\left(x^{2}+y^{2}\right)$
$=\left(x^{2}-y^{2}\right)\left(x^{2}+y^{2}\right)$
$=\left(x^{2}\right)^{2}-\left(y^{2}\right)^{2}$
$=x^{4}-y^{4}$
(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 :
$=(x)^{2}+8 \times x \times 16+(16)^{2}$
$=(x+16)^{2}$ 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 :
$=(2 x)^{2}-2 \times 2 x \times 3+\left(30)^{2}\right.$
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 :
$=(2 a)^{2}+2 \times 2 a \times 7+(7)^{2}$
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 :
$\Rightarrow(a+1)^{2} \Rightarrow(9)^{3}+3 a^{2} \times 1+3 a \times\left(1)^{2}+(1)^{3}\right.$
$\Rightarrow a^{3}+3 a^{2}+3 a+1$
(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) $\left(\frac{a}{5}+\frac{b}{2}\right)^{3}$
$=\left(\frac{a}{5}\right)^{3}+3\left(\frac{a}{5}\right)^{2}\left(\frac{b}{2}\right)+3\left(\frac{a}{5}\right)\left(\frac{b}{2}\right)^{2}+\left(\frac{b}{2}\right)^{3}$
$=\frac{a^{3}}{125}+3 \times \frac{a^{2}}{25} \times \frac{b}{2}+3 \times \frac{9}{5} \times \frac{b^{2}}{4}+\frac{b^{3}}{8}$
$=\frac{a^{3}}{125}+\frac{3 a^{2} b}{50}+\frac{3 a b^{2}}{20}+\frac{b^{3}}{8}$
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