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Exercise Set 4.1, Q1: Using the identity $(a+b)^2 = a^2 + 2ab + b^2$, expand the following:
(i) $(7x+4y)^2$
(ii) $(\frac{7}{5}x + \frac{3}{2}y)^2$
(iii) $(2.5p + 1.5q)^2$
(iv) $(\frac{3}{4}s + 8t)^2$
(v) $(x + \frac{1}{2y})^2$
(vi) $(\frac{1}{x} + \frac{1}{y})^2$
Answer:
(i) $(7x)^2 + 2(7x)(4y) + (4y)^2 = \mathbf{49x^2 + 56xy + 16y^2}$
(ii) $(\frac{7}{5}x)^2 + 2(\frac{7}{5}x)(\frac{3}{2}y) + (\frac{3}{2}y)^2 = \mathbf{\frac{49}{25}x^2 + \frac{21}{5}xy + \frac{9}{4}y^2}$
(iii) $(2.5p)^2 + 2(2.5p)(1.5q) + (1.5q)^2 = \mathbf{6.25p^2 + 7.5pq + 2.25q^2}$
(iv) $(\frac{3}{4}s)^2 + 2(\frac{3}{4}s)(8t) + (8t)^2 = \mathbf{\frac{9}{16}s^2 + 12st + 64t^2}$
(v) $x^2 + 2(x)(\frac{1}{2y}) + (\frac{1}{2y})^2 = \mathbf{x^2 + \frac{x}{y} + \frac{1}{4y^2}}$
(vi) $(\frac{1}{x})^2 + 2(\frac{1}{x})(\frac{1}{y}) + (\frac{1}{y})^2 = \mathbf{\frac{1}{x^2} + \frac{2}{xy} + \frac{1}{y^2}}$
Exercise Set 4.1, Q2: Using the same identity, find the values of the following:
(i) $(64)^2$
(ii) $(105)^2$
(iii) $(205)^2$
Answer:
(i) $(60 + 4)^2 = 60^2 + 2(60)(4) + 4^2 = 3600 + 480 + 16 = \mathbf{4096}$
(ii) $(100 + 5)^2 = 100^2 + 2(100)(5) + 5^2 = 10000 + 1000 + 25 = \mathbf{11025}$
(iii) $(200 + 5)^2 = 200^2 + 2(200)(5) + 5^2 = 40000 + 2000 + 25 = \mathbf{42025}$
Exercise Set 4.2, Q1: Factor completely:
(i) $9x^2 + 24xy + 16y^2$
(ii) $4s^2 + 20st + 25t^2$
(iii) $\frac{m^2}{9} + \frac{mk}{3} + \frac{k^2}{4} + 3nk + 2mn + 9n^2$
(iv) $\frac{p^2}{16} - 2 + \frac{16}{p^2}$
(v) $9a^2 + 4b^2 + c^2 - 12ab + 6ac - 4bc$
Answer:
(i) $(3x)^2 + 2(3x)(4y) + (4y)^2 = \mathbf{(3x + 4y)^2}$
(ii) $(2s)^2 + 2(2s)(5t) + (5t)^2 = \mathbf{(2s + 5t)^2}$
(iii) This is of the form $(a+b+c)^2$.
$\frac{m^2}{9} = (\frac{m}{3})^2$, $\frac{k^2}{4} = (\frac{k}{2})^2$, $9n^2 = (3n)^2$.
$2(\frac{m}{3})(\frac{k}{2}) = \frac{mk}{3}$, $2(\frac{m}{3})(3n) = 2mn$, $2(\frac{k}{2})(3n) = 3nk$.
Therefore, it factors to $\mathbf{(\frac{m}{3} + \frac{k}{2} + 3n)^2}$
(iv) $(\frac{p}{4})^2 - 2(\frac{p}{4})(\frac{4}{p}) + (\frac{4}{p})^2 = \mathbf{(\frac{p}{4} - \frac{4}{p})^2}$
(v) This is $(a+b+c)^2$. Since $-12ab$ and $-4bc$ are negative, but $6ac$ is positive, the term with $b$ must be negative. $(3a)^2 + (-2b)^2 + (c)^2 + 2(3a)(-2b) + 2(3a)(c) + 2(-2b)(c)$.
Factors to $\mathbf{(3a - 2b + c)^2}$
Exercise Set 4.2, Q2: Find the values of the following using the identity $(a-b)^2 = a^2 - 2ab + b^2$.
(i) $(79)^2$
(ii) $(193)^2$
Answer:
(i) $(80 - 1)^2 = 80^2 - 2(80)(1) + 1^2 = 6400 - 160 + 1 = \mathbf{6241}$
(ii) $(200 - 7)^2 = 200^2 - 2(200)(7) + 7^2 = 40000 - 2800 + 49 = \mathbf{37249}$
Exercise Set 4.3, Q1: Find the following squares using one of the above identities. Determine which of these identities will make these calculations easier.
(i) $117^2$
(ii) $78^2$
(iii) $198^2$
(iv) $214^2$
(v) $1104^2$
(vi) $1120^2$
Answer:
(i) $117^2 = (120 - 3)^2 = 14400 - 720 + 9 = \mathbf{13689}$ (Using $(a-b)^2$)
(ii) $78^2 = (80 - 2)^2 = 6400 - 320 + 4 = \mathbf{6084}$ (Using $(a-b)^2$)
(iii) $198^2 = (200 - 2)^2 = 40000 - 800 + 4 = \mathbf{39204}$ (Using $(a-b)^2$)
(iv) $214^2 = (200 + 14)^2 = 40000 + 5600 + 196 = \mathbf{45796}$ (Using $(a+b)^2$)
(v) $1104^2 = (1000 + 100 + 4)^2 = 1000^2 + 100^2 + 4^2 + 2(1000)(100) + 2(1000)(4) + 2(100)(4) = 1000000 + 10000 + 16 + 200000 + 8000 + 800 = \mathbf{1218816}$ (Using $(a+b+c)^2$)
(vi) $1120^2 = (1100 + 20)^2 = 1210000 + 44000 + 400 = \mathbf{1254400}$ (Using $(a+b)^2$)
Exercise Set 4.4, Q1: Fill in the blanks to complete the following identities:
(i) $s^2 - 11s + 24 = (\_\_)(\_\_)$
(ii) $(\_\_)(x + 1) = (3x^2 - 4x - 7)$
(iii) $10x^2 - 11x - 6 = (2x - \_\_)(\_\_ + 2)$
(iv) $6x^2 + 7x + 2 = (\_\_)(\_\_)$
Answer:
(i) Find factors of 24 that sum to -11 (-8 and -3). $\mathbf{(s - 8)(s - 3)}$
(ii) Divide $(3x^2 - 4x - 7)$ by $(x + 1)$. The missing factor is $\mathbf{(3x - 7)}$.
(iii) $10x^2 - 11x - 6 = (2x - 3)(5x + 2)$. So the blanks are $\mathbf{3}$ and $\mathbf{5x}$.
(iv) Splitting the middle term: $6x^2 + 4x + 3x + 2 = 2x(3x + 2) + 1(3x + 2)$. So, $\mathbf{(2x + 1)(3x + 2)}$.
Exercise Set 4.4, Q2: Select and use the identity that will help you to find the following products without multiplying directly:
(i) $(41)^2$
(ii) $(27)^2$
(iii) $(23 \times 17)$
(iv) $(135)^2$
(v) $(97)^2$
(vi) $(18 \times 29)$
(vii) $(34 \times 43)$
(viii) $(205)^2$
Answer:
(i) $(40 + 1)^2 = 1600 + 80 + 1 = \mathbf{1681}$
(ii) $(30 - 3)^2 = 900 - 180 + 9 = \mathbf{729}$
(iii) $(20 + 3)(20 - 3) = 20^2 - 3^2 = 400 - 9 = \mathbf{391}$
(iv) $(130 + 5)^2 = 16900 + 1300 + 25 = \mathbf{18225}$
(v) $(100 - 3)^2 = 10000 - 600 + 9 = \mathbf{9409}$
(vi) $(20 - 2)(30 - 1) = 600 - 20 - 60 + 2 = \mathbf{522}$ (Using $(x+a)(x+b)$ or distributive)
(vii) $(40 - 6)(40 + 3) = 1600 + 120 - 240 - 18 = \mathbf{1462}$
(viii) $(200 + 5)^2 = 40000 + 2000 + 25 = \mathbf{42025}$
Exercise Set 4.5, Q1: Simplify the following rational expressions assuming that the expressions in the denominators are not equal to zero:
(i) $\frac{3p^2 - 3pq - 18q^2}{p^2 + 3pq - 10q^2}$
(ii) $\frac{n^3 - 3n^2m + 3nm^2 - m^3}{5m^2 - 10mn + 5n^2}$
(iv) $\frac{4y^2 - 20yz + 25z^2}{25z^2 - 4y^2}$
(v) $\frac{(x^2 + x - 6)(x^2 - 7x + 12)}{(x^2 - 6x + 8)(x^2 - 9)}$
(vi) $\frac{p^4 - 16}{p^2 - 4p + 4}$
Answer:
(i) $\frac{3(p^2 - pq - 6q^2)}{(p+5q)(p-2q)} = \frac{3(p-3q)(p+2q)}{(p+5q)(p-2q)}$
(ii) $\frac{(n-m)^3}{5(m^2 - 2mn + n^2)} = \frac{(n-m)^3}{5(m-n)^2} = \frac{-(m-n)^3}{5(m-n)^2} = \mathbf{\frac{n-m}{5}}$
(iv) $\frac{(2y - 5z)^2}{(5z - 2y)(5z + 2y)} = \frac{(5z - 2y)^2}{(5z - 2y)(5z + 2y)} = \mathbf{\frac{5z - 2y}{5z + 2y}}$
(v) $\frac{(x+3)(x-2)(x-3)(x-4)}{(x-2)(x-4)(x-3)(x+3)} = \mathbf{1}$
(vi) $\frac{(p^2-4)(p^2+4)}{(p-2)^2} = \frac{(p-2)(p+2)(p^2+4)}{(p-2)^2} = \mathbf{\frac{(p+2)(p^2+4)}{p-2}}$
End-of-Chapter Exercises, Q1: Use suitable identities to find the following products:
(i) $(-3x + 4)^2$
(ii) $(2s + 7)(2s - 7)$
(iii) $(p^2 + \frac{1}{2})(p^2 - \frac{1}{2})$
(v) $(s - 2t)(s^2 + 2st + 4t^2)$
(vi) $(\frac{1}{2r} - 4r)^2$
(viii) $(x - \frac{1}{3}y)^3$
Answer:
(i) $(-3x)^2 + 2(-3x)(4) + 4^2 = \mathbf{9x^2 - 24x + 16}$
(ii) $(2s)^2 - 7^2 = \mathbf{4s^2 - 49}$
(iii) $(p^2)^2 - (\frac{1}{2})^2 = \mathbf{p^4 - \frac{1}{4}}$
(v) This is $(a-b)(a^2+ab+b^2) = a^3 - b^3$. So, $s^3 - (2t)^3 = \mathbf{s^3 - 8t^3}$
(vi) $(\frac{1}{2r})^2 - 2(\frac{1}{2r})(4r) + (4r)^2 = \mathbf{\frac{1}{4r^2} - 4 + 16r^2}$
(viii) $x^3 - 3(x^2)(\frac{1}{3}y) + 3(x)(\frac{1}{9}y^2) - (\frac{1}{3}y)^3 = \mathbf{x^3 - x^2y + \frac{1}{3}xy^2 - \frac{1}{27}y^3}$
End-of-Chapter Exercises, Q4: Simplify the following:
(i) $\frac{4x^2 + 4x + 1}{4x^2 - 1}$
(ii) $\frac{9(3a^3 - 24b^3)}{9a^2 - 36b^2}$
(iii) $\frac{s^3 + 125t^3}{s^2 - 2st - 35t^2}$
Answer:
(i) $\frac{(2x+1)^2}{(2x-1)(2x+1)} = \mathbf{\frac{2x+1}{2x-1}}$
(ii) $\frac{27(a^3 - 8b^3)}{9(a^2 - 4b^2)} = \frac{3(a-2b)(a^2+2ab+4b^2)}{(a-2b)(a+2b)} = \mathbf{\frac{3(a^2+2ab+4b^2)}{a+2b}}$
(iii) $\frac{(s+5t)(s^2-5st+25t^2)}{(s-7t)(s+5t)} = \mathbf{\frac{s^2-5st+25t^2}{s-7t}}$
End-of-Chapter Exercises, Q5: Find possible expressions for the length and breadth of each of the following rectangles whose areas are given by the following expressions in square units.
(i) $25a^2 - 30ab + 9b^2$
(ii) $36s^2 - 49t^2$
Answer:
(i) This factors to $(5a - 3b)^2$. Length and breadth can both be $\mathbf{5a - 3b}$ (making it a square).
(ii) This is a difference of squares: $(6s)^2 - (7t)^2$. Length and breadth are $\mathbf{(6s - 7t)}$ and $\mathbf{(6s + 7t)}$.
End-of-Chapter Exercises, Q6: Find possible expressions for the length, breadth, and heights of each of the following cuboids whose volumes are given by the following expressions in cubic units.
(i) $6a^2 - 24b^2$
(ii) $3ps^2 - 15ps + 12p$
Answer:
(i) $6(a^2 - 4b^2) = 6(a - 2b)(a + 2b)$. Dimensions: $\mathbf{6}$, $\mathbf{(a - 2b)}$, and $\mathbf{(a + 2b)}$.
(ii) $3p(s^2 - 5s + 4) = 3p(s - 1)(s - 4)$. Dimensions: $\mathbf{3p}$, $\mathbf{(s - 1)}$, and $\mathbf{(s - 4)}$.
End-of-Chapter Exercises, Q7: The village playground is shaped as a square of side 40 metres. A path of width $s$ metres is created around the playground for people to walk. Find an expression for the area of the path in terms of $s$.
Answer: The outer dimensions of the path + playground will be $(40 + 2s)$.
The area of the path is the outer square area minus the playground area.
Area = $(40 + 2s)^2 - 40^2 = (40 + 2s - 40)(40 + 2s + 40) = 2s(80 + 2s) = \mathbf{160s + 4s^2}$ sq. meters.
End-of-Chapter Exercises, Q11: If $a + b + c = 5$ and $ab + bc + ca = 10$, then prove that $a^3 + b^3 + c^3 - 3abc = -25$.
Answer: We know the identity: $a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2+b^2+c^2 - (ab+bc+ca))$.
First find $a^2+b^2+c^2$.
$(a+b+c)^2 = a^2+b^2+c^2 + 2(ab+bc+ca)$.
$5^2 = a^2+b^2+c^2 + 2(10) \implies 25 = a^2+b^2+c^2 + 20 \implies a^2+b^2+c^2 = 5$.
Substitute back into the main identity:
$a^3 + b^3 + c^3 - 3abc = (5)(5 - 10) = 5(-5) = \mathbf{-25}$.
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