Chapter 4 End-of-Chapter Solutions
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Overview
This page provides comprehensive step-by-step solutions for Ch 4: Algebraic Identities – End-of-Chapter Exercises. Master algebraic factorization, cubic expansions, rational simplifications, factor theorems, and real-world playground and swimming pool dimensions word problems.
Q1: Expand Using Suitable Identities
Use suitable identities to find the following products:
(i) $(-3x + 4)^2$ (ii) $(2s + 7)(2s - 7)$ (iii) $\left(p^2 + \dfrac{1}{2}\right)\left(p^2 - \dfrac{1}{2}\right)$
(iv) $(2n + 7)(2n - 7)$ (v) $(s - 2t)(s^2 + 2st + 4t^2)$ (vi) $\left(\dfrac{r}{2} - 4\right)\left(\dfrac{r^2}{4} + 2r + 16\right)$
(vii) $(-3m + 4k - l)^2$ (viii) $\left(x - \dfrac{1}{3}y\right)^3$ (ix) $\left(\dfrac{7}{2}k^2 - \dfrac{2}{3}m^3\right)^3$
(i) $(-3x + 4)^2$ (ii) $(2s + 7)(2s - 7)$ (iii) $\left(p^2 + \dfrac{1}{2}\right)\left(p^2 - \dfrac{1}{2}\right)$
(iv) $(2n + 7)(2n - 7)$ (v) $(s - 2t)(s^2 + 2st + 4t^2)$ (vi) $\left(\dfrac{r}{2} - 4\right)\left(\dfrac{r^2}{4} + 2r + 16\right)$
(vii) $(-3m + 4k - l)^2$ (viii) $\left(x - \dfrac{1}{3}y\right)^3$ (ix) $\left(\dfrac{7}{2}k^2 - \dfrac{2}{3}m^3\right)^3$
(i) $(-3x + 4)^2$:
$$= (-3x)^2 + 2(-3x)(4) + 4^2 = \mathbf{9x^2 - 24x + 16}$$
$$= (-3x)^2 + 2(-3x)(4) + 4^2 = \mathbf{9x^2 - 24x + 16}$$
(ii) $(2s + 7)(2s - 7)$:
Using $(a+b)(a-b) = a^2 - b^2$:
$$= (2s)^2 - 7^2 = \mathbf{4s^2 - 49}$$
Using $(a+b)(a-b) = a^2 - b^2$:
$$= (2s)^2 - 7^2 = \mathbf{4s^2 - 49}$$
(iii) $\left(p^2 + \dfrac{1}{2}\right)\left(p^2 - \dfrac{1}{2}\right)$:
$$= (p^2)^2 - \left(\frac{1}{2}\right)^2 = \mathbf{p^4 - \frac{1}{4}}$$
$$= (p^2)^2 - \left(\frac{1}{2}\right)^2 = \mathbf{p^4 - \frac{1}{4}}$$
(iv) $(2n + 7)(2n - 7)$:
$$= (2n)^2 - 7^2 = \mathbf{4n^2 - 49}$$
$$= (2n)^2 - 7^2 = \mathbf{4n^2 - 49}$$
(v) $(s - 2t)(s^2 + 2st + 4t^2)$:
Using difference of cubes $(a - b)(a^2 + ab + b^2) = a^3 - b^3$:
$$= s^3 - (2t)^3 = \mathbf{s^3 - 8t^3}$$
Using difference of cubes $(a - b)(a^2 + ab + b^2) = a^3 - b^3$:
$$= s^3 - (2t)^3 = \mathbf{s^3 - 8t^3}$$
(vi) $\left(\dfrac{r}{2} - 4\right)\left(\dfrac{r^2}{4} + 2r + 16\right)$:
Here $a = \frac{r}{2}$, $b = 4$.
$$= \left(\frac{r}{2}\right)^3 - 4^3 = \mathbf{\frac{r^3}{8} - 64}$$
Here $a = \frac{r}{2}$, $b = 4$.
$$= \left(\frac{r}{2}\right)^3 - 4^3 = \mathbf{\frac{r^3}{8} - 64}$$
(vii) $(-3m + 4k - l)^2$:
$$= (-3m)^2 + (4k)^2 + (-l)^2 + 2(-3m)(4k) + 2(4k)(-l) + 2(-l)(-3m)$$
$$= \mathbf{9m^2 + 16k^2 + l^2 - 24mk - 8kl + 6lm}$$
$$= (-3m)^2 + (4k)^2 + (-l)^2 + 2(-3m)(4k) + 2(4k)(-l) + 2(-l)(-3m)$$
$$= \mathbf{9m^2 + 16k^2 + l^2 - 24mk - 8kl + 6lm}$$
(viii) $\left(x - \dfrac{1}{3}y\right)^3$:
Using $(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$:
$$= x^3 - 3x^2\left(\frac{1}{3}y\right) + 3x\left(\frac{1}{9}y^2\right) - \frac{1}{27}y^3 = \mathbf{x^3 - x^2y + \frac{1}{3}xy^2 - \frac{1}{27}y^3}$$
Using $(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$:
$$= x^3 - 3x^2\left(\frac{1}{3}y\right) + 3x\left(\frac{1}{9}y^2\right) - \frac{1}{27}y^3 = \mathbf{x^3 - x^2y + \frac{1}{3}xy^2 - \frac{1}{27}y^3}$$
(ix) $\left(\dfrac{7}{2}k^2 - \dfrac{2}{3}m^3\right)^3$:
$$= \left(\frac{7}{2}k^2\right)^3 - 3\left(\frac{7}{2}k^2\right)^2\left(\frac{2}{3}m^3\right) + 3\left(\frac{7}{2}k^2\right)\left(\frac{2}{3}m^3\right)^2 - \left(\frac{2}{3}m^3\right)^3$$
$$= \mathbf{\frac{343}{8}k^6 - \frac{49}{2}k^4m^3 + \frac{14}{3}k^2m^6 - \frac{8}{27}m^9}$$
$$= \left(\frac{7}{2}k^2\right)^3 - 3\left(\frac{7}{2}k^2\right)^2\left(\frac{2}{3}m^3\right) + 3\left(\frac{7}{2}k^2\right)\left(\frac{2}{3}m^3\right)^2 - \left(\frac{2}{3}m^3\right)^3$$
$$= \mathbf{\frac{343}{8}k^6 - \frac{49}{2}k^4m^3 + \frac{14}{3}k^2m^6 - \frac{8}{27}m^9}$$
Q2: Evaluate Numerical Expressions
Find the values using suitable identities:
(i) $17 \times 21$ (ii) $104 \times 96$ (iii) $24 \times 16$ (iv) $147^3$
(v) $199^3$ (vi) $127^3$ (vii) $(-107)^3$ (viii) $(-299)^3$
(i) $17 \times 21$ (ii) $104 \times 96$ (iii) $24 \times 16$ (iv) $147^3$
(v) $199^3$ (vi) $127^3$ (vii) $(-107)^3$ (viii) $(-299)^3$
(i) $17 \times 21$:
$$= (19 - 2)(19 + 2) = 19^2 - 2^2 = 361 - 4 = \mathbf{357}$$
$$= (19 - 2)(19 + 2) = 19^2 - 2^2 = 361 - 4 = \mathbf{357}$$
(ii) $104 \times 96$:
$$= (100 + 4)(100 - 4) = 100^2 - 4^2 = 10000 - 16 = \mathbf{9984}$$
$$= (100 + 4)(100 - 4) = 100^2 - 4^2 = 10000 - 16 = \mathbf{9984}$$
(iii) $24 \times 16$:
$$= (20 + 4)(20 - 4) = 20^2 - 4^2 = 400 - 16 = \mathbf{384}$$
$$= (20 + 4)(20 - 4) = 20^2 - 4^2 = 400 - 16 = \mathbf{384}$$
(iv) $147^3$:
$$= (150 - 3)^3 = 150^3 - 3(150)^2(3) + 3(150)(3)^2 - 3^3$$
$$= 3375000 - 202500 + 4050 - 27 = \mathbf{3176523}$$
$$= (150 - 3)^3 = 150^3 - 3(150)^2(3) + 3(150)(3)^2 - 3^3$$
$$= 3375000 - 202500 + 4050 - 27 = \mathbf{3176523}$$
(v) $199^3$:
$$= (200 - 1)^3 = 200^3 - 3(200)^2(1) + 3(200)(1)^2 - 1^3$$
$$= 8000000 - 120000 + 600 - 1 = \mathbf{7880599}$$
$$= (200 - 1)^3 = 200^3 - 3(200)^2(1) + 3(200)(1)^2 - 1^3$$
$$= 8000000 - 120000 + 600 - 1 = \mathbf{7880599}$$
(vi) $127^3$:
$$= (130 - 3)^3 = 130^3 - 3(130)^2(3) + 3(130)(3)^2 - 3^3$$
$$= 2197000 - 152100 + 3510 - 27 = \mathbf{2048383}$$
$$= (130 - 3)^3 = 130^3 - 3(130)^2(3) + 3(130)(3)^2 - 3^3$$
$$= 2197000 - 152100 + 3510 - 27 = \mathbf{2048383}$$
(vii) $(-107)^3$:
$$= -(100 + 7)^3 = -[100^3 + 3(100)^2(7) + 3(100)(7)^2 + 7^3]$$
$$= -[1000000 + 210000 + 14700 + 343] = \mathbf{-1225043}$$
$$= -(100 + 7)^3 = -[100^3 + 3(100)^2(7) + 3(100)(7)^2 + 7^3]$$
$$= -[1000000 + 210000 + 14700 + 343] = \mathbf{-1225043}$$
(viii) $(-299)^3$:
$$= -(300 - 1)^3 = -[300^3 - 3(300)^2(1) + 3(300)(1)^2 - 1^3]$$
$$= -[27000000 - 270000 + 900 - 1] = \mathbf{-26730899}$$
$$= -(300 - 1)^3 = -[300^3 - 3(300)^2(1) + 3(300)(1)^2 - 1^3]$$
$$= -[27000000 - 270000 + 900 - 1] = \mathbf{-26730899}$$
Q3: Factorise Completely
Factor the following algebraic expressions:
(i) $4y^2 + 1 + \dfrac{1}{16y^2}$ (ii) $\dfrac{9}{25}m^2 - \dfrac{1}{9}n^2$ (iii) $27b^3 - \dfrac{1}{64b^3}$
(iv) $x^2 + \dfrac{5}{6}x + \dfrac{1}{6}$ (v) $27u^3 - \dfrac{27}{5}u^2 + \dfrac{9}{25}u - \dfrac{1}{125}$ (vi) $64y^3 + \dfrac{1}{125}z^3$
(vii) $p^3 + q^3 + 27r^3 - 9pqr$ (viii) $9 - 12m + 4m^2$
(ix) $x^3 - 8y^3 + z^3 + 6xyz$ (x) $9x^2 + 36y^2 + 4z^2 + 36xy - 24yz - 12zx$
(xi) $27u^3 - \dfrac{9}{2}u^2 + \dfrac{1}{4}u - \dfrac{1}{216}$
(i) $4y^2 + 1 + \dfrac{1}{16y^2}$ (ii) $\dfrac{9}{25}m^2 - \dfrac{1}{9}n^2$ (iii) $27b^3 - \dfrac{1}{64b^3}$
(iv) $x^2 + \dfrac{5}{6}x + \dfrac{1}{6}$ (v) $27u^3 - \dfrac{27}{5}u^2 + \dfrac{9}{25}u - \dfrac{1}{125}$ (vi) $64y^3 + \dfrac{1}{125}z^3$
(vii) $p^3 + q^3 + 27r^3 - 9pqr$ (viii) $9 - 12m + 4m^2$
(ix) $x^3 - 8y^3 + z^3 + 6xyz$ (x) $9x^2 + 36y^2 + 4z^2 + 36xy - 24yz - 12zx$
(xi) $27u^3 - \dfrac{9}{2}u^2 + \dfrac{1}{4}u - \dfrac{1}{216}$
(i) $4y^2 + 1 + \dfrac{1}{16y^2}$:
$$= (2y)^2 + 2(2y)\left(\frac{1}{4y}\right) + \left(\frac{1}{4y}\right)^2 = \mathbf{\left(2y + \frac{1}{4y}\right)^2}$$
$$= (2y)^2 + 2(2y)\left(\frac{1}{4y}\right) + \left(\frac{1}{4y}\right)^2 = \mathbf{\left(2y + \frac{1}{4y}\right)^2}$$
(ii) $\dfrac{9}{25}m^2 - \dfrac{1}{9}n^2$:
$$= \left(\frac{3}{5}m\right)^2 - \left(\frac{1}{3}n\right)^2 = \mathbf{\left(\frac{3}{5}m - \frac{1}{3}n\right)\left(\frac{3}{5}m + \frac{1}{3}n\right)}$$
$$= \left(\frac{3}{5}m\right)^2 - \left(\frac{1}{3}n\right)^2 = \mathbf{\left(\frac{3}{5}m - \frac{1}{3}n\right)\left(\frac{3}{5}m + \frac{1}{3}n\right)}$$
(iii) $27b^3 - \dfrac{1}{64b^3}$:
$$= (3b)^3 - \left(\frac{1}{4b}\right)^3 = \mathbf{\left(3b - \frac{1}{4b}\right)\left(9b^2 + \frac{3}{4} + \frac{1}{16b^2}\right)}$$
$$= (3b)^3 - \left(\frac{1}{4b}\right)^3 = \mathbf{\left(3b - \frac{1}{4b}\right)\left(9b^2 + \frac{3}{4} + \frac{1}{16b^2}\right)}$$
(iv) $x^2 + \dfrac{5}{6}x + \dfrac{1}{6}$:
Find two numbers adding to $\frac{5}{6}$ and multiplying to $\frac{1}{6}$: $\frac{1}{2}$ and $\frac{1}{3}$.
$$= x^2 + \frac{1}{2}x + \frac{1}{3}x + \frac{1}{6} = x\left(x + \frac{1}{2}\right) + \frac{1}{3}\left(x + \frac{1}{2}\right) = \mathbf{\left(x + \frac{1}{2}\right)\left(x + \frac{1}{3}\right)}$$
Find two numbers adding to $\frac{5}{6}$ and multiplying to $\frac{1}{6}$: $\frac{1}{2}$ and $\frac{1}{3}$.
$$= x^2 + \frac{1}{2}x + \frac{1}{3}x + \frac{1}{6} = x\left(x + \frac{1}{2}\right) + \frac{1}{3}\left(x + \frac{1}{2}\right) = \mathbf{\left(x + \frac{1}{2}\right)\left(x + \frac{1}{3}\right)}$$
(v) $27u^3 - \dfrac{27}{5}u^2 + \dfrac{9}{25}u - \dfrac{1}{125}$:
This matches the expansion of $(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$ where $a = 3u$ and $b = \frac{1}{5}$.
$$= \mathbf{\left(3u - \frac{1}{5}\right)^3}$$
This matches the expansion of $(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$ where $a = 3u$ and $b = \frac{1}{5}$.
$$= \mathbf{\left(3u - \frac{1}{5}\right)^3}$$
(vi) $64y^3 + \dfrac{1}{125}z^3$:
$$= (4y)^3 + \left(\frac{1}{5}z\right)^3 = \mathbf{\left(4y + \frac{1}{5}z\right)\left(16y^2 - \frac{4}{5}yz + \frac{1}{25}z^2\right)}$$
$$= (4y)^3 + \left(\frac{1}{5}z\right)^3 = \mathbf{\left(4y + \frac{1}{5}z\right)\left(16y^2 - \frac{4}{5}yz + \frac{1}{25}z^2\right)}$$
(vii) $p^3 + q^3 + 27r^3 - 9pqr$:
$$= p^3 + q^3 + (3r)^3 - 3(p)(q)(3r)$$
$$= \mathbf{(p + q + 3r)(p^2 + q^2 + 9r^2 - pq - 3qr - 3rp)}$$
$$= p^3 + q^3 + (3r)^3 - 3(p)(q)(3r)$$
$$= \mathbf{(p + q + 3r)(p^2 + q^2 + 9r^2 - pq - 3qr - 3rp)}$$
(viii) $9 - 12m + 4m^2$:
$$= 3^2 - 2(3)(2m) + (2m)^2 = \mathbf{(3 - 2m)^2}$$
$$= 3^2 - 2(3)(2m) + (2m)^2 = \mathbf{(3 - 2m)^2}$$
(ix) $x^3 - 8y^3 + z^3 + 6xyz$:
$$= x^3 + (-2y)^3 + z^3 - 3(x)(-2y)(z)$$
$$= \mathbf{(x - 2y + z)(x^2 + 4y^2 + z^2 + 2xy + 2yz - zx)}$$
$$= x^3 + (-2y)^3 + z^3 - 3(x)(-2y)(z)$$
$$= \mathbf{(x - 2y + z)(x^2 + 4y^2 + z^2 + 2xy + 2yz - zx)}$$
(x) $9x^2 + 36y^2 + 4z^2 + 36xy - 24yz - 12zx$:
$$= (3x)^2 + (6y)^2 + (-2z)^2 + 2(3x)(6y) + 2(6y)(-2z) + 2(-2z)(3x)$$
$$= \mathbf{(3x + 6y - 2z)^2}$$
$$= (3x)^2 + (6y)^2 + (-2z)^2 + 2(3x)(6y) + 2(6y)(-2z) + 2(-2z)(3x)$$
$$= \mathbf{(3x + 6y - 2z)^2}$$
(xi) $27u^3 - \dfrac{9}{2}u^2 + \dfrac{1}{4}u - \dfrac{1}{216}$:
This is $(a - b)^3$ where $a = 3u$ and $b = \frac{1}{6}$.
$$= \mathbf{\left(3u - \frac{1}{6}\right)^3}$$
This is $(a - b)^3$ where $a = 3u$ and $b = \frac{1}{6}$.
$$= \mathbf{\left(3u - \frac{1}{6}\right)^3}$$
Q4: Simplify Expressions
Simplify the following (assume denominators are not equal to 0):
(i) $\dfrac{4x^2 + 4x + 1}{4x^2 - 1}$ (ii) $\dfrac{27a^3 - 8b^3}{9a^2 - 4b^2}$ (iii) $\dfrac{s^3 + t^3 + 125 - 15st}{s^2 + t^2 + 25 - st - 5s - 5t}$
(i) $\dfrac{4x^2 + 4x + 1}{4x^2 - 1}$ (ii) $\dfrac{27a^3 - 8b^3}{9a^2 - 4b^2}$ (iii) $\dfrac{s^3 + t^3 + 125 - 15st}{s^2 + t^2 + 25 - st - 5s - 5t}$
(i) $\dfrac{4x^2 + 4x + 1}{4x^2 - 1}$:
Numerator $= (2x + 1)^2$.
Denominator $= (2x - 1)(2x + 1)$.
$$= \frac{(2x + 1)^2}{(2x - 1)(2x + 1)} = \mathbf{\frac{2x + 1}{2x - 1}}$$
Numerator $= (2x + 1)^2$.
Denominator $= (2x - 1)(2x + 1)$.
$$= \frac{(2x + 1)^2}{(2x - 1)(2x + 1)} = \mathbf{\frac{2x + 1}{2x - 1}}$$
(ii) $\dfrac{27a^3 - 8b^3}{9a^2 - 4b^2}$:
Numerator $= (3a - 2b)(9a^2 + 6ab + 4b^2)$.
Denominator $= (3a - 2b)(3a + 2b)$.
$$= \frac{(3a - 2b)(9a^2 + 6ab + 4b^2)}{(3a - 2b)(3a + 2b)} = \mathbf{\frac{9a^2 + 6ab + 4b^2}{3a + 2b}}$$
Numerator $= (3a - 2b)(9a^2 + 6ab + 4b^2)$.
Denominator $= (3a - 2b)(3a + 2b)$.
$$= \frac{(3a - 2b)(9a^2 + 6ab + 4b^2)}{(3a - 2b)(3a + 2b)} = \mathbf{\frac{9a^2 + 6ab + 4b^2}{3a + 2b}}$$
(iii) $\dfrac{s^3 + t^3 + 125 - 15st}{s^2 + t^2 + 25 - st - 5s - 5t}$:
Numerator $= (s + t + 5)(s^2 + t^2 + 25 - st - 5s - 5t)$ by the master identity.
$$= \frac{(s + t + 5)(s^2 + t^2 + 25 - st - 5s - 5t)}{s^2 + t^2 + 25 - st - 5s - 5t} = \mathbf{s + t + 5}$$
Numerator $= (s + t + 5)(s^2 + t^2 + 25 - st - 5s - 5t)$ by the master identity.
$$= \frac{(s + t + 5)(s^2 + t^2 + 25 - st - 5s - 5t)}{s^2 + t^2 + 25 - st - 5s - 5t} = \mathbf{s + t + 5}$$
Q5: Rectangle Dimensions
Find possible expressions for the length and breadth of each of the following rectangles whose areas are given by:
(i) $25a^2 - 30ab + 9b^2$ (ii) $36s^2 - 49t^2$
(i) $25a^2 - 30ab + 9b^2$ (ii) $36s^2 - 49t^2$
(i) $25a^2 - 30ab + 9b^2$:
This is a perfect square trinomial:
$$25a^2 - 30ab + 9b^2 = (5a - 3b)^2 = (5a - 3b)(5a - 3b)$$
Possible expressions are:
$$\mathbf{\text{Length} = 5a - 3b, \quad \text{Breadth} = 5a - 3b}$$
This is a perfect square trinomial:
$$25a^2 - 30ab + 9b^2 = (5a - 3b)^2 = (5a - 3b)(5a - 3b)$$
Possible expressions are:
$$\mathbf{\text{Length} = 5a - 3b, \quad \text{Breadth} = 5a - 3b}$$
(ii) $36s^2 - 49t^2$:
Using difference of squares:
$$36s^2 - 49t^2 = (6s - 7t)(6s + 7t)$$
Possible expressions are:
$$\mathbf{\text{Length} = 6s + 7t, \quad \text{Breadth} = 6s - 7t} \quad (\text{assuming } 6s > 7t)$$
Using difference of squares:
$$36s^2 - 49t^2 = (6s - 7t)(6s + 7t)$$
Possible expressions are:
$$\mathbf{\text{Length} = 6s + 7t, \quad \text{Breadth} = 6s - 7t} \quad (\text{assuming } 6s > 7t)$$
Q6: Cuboid Dimensions
Find possible expressions for the length, breadth, and height of each of the following cuboids whose volumes are given by:
(i) $6a^2 - 24b^2$ (ii) $3ps^2 - 15ps + 12p$
(i) $6a^2 - 24b^2$ (ii) $3ps^2 - 15ps + 12p$
(i) $6a^2 - 24b^2$:
Factorise the expression:
$$6a^2 - 24b^2 = 6(a^2 - 4b^2) = 6(a - 2b)(a + 2b)$$
Possible dimensions are:
$$\mathbf{\text{Length} = 6, \quad \text{Breadth} = a - 2b, \quad \text{Height} = a + 2b}$$
Factorise the expression:
$$6a^2 - 24b^2 = 6(a^2 - 4b^2) = 6(a - 2b)(a + 2b)$$
Possible dimensions are:
$$\mathbf{\text{Length} = 6, \quad \text{Breadth} = a - 2b, \quad \text{Height} = a + 2b}$$
(ii) $3ps^2 - 15ps + 12p$:
Factorise:
$$3p(s^2 - 5s + 4) = 3p(s - 1)(s - 4)$$
Possible dimensions are:
$$\mathbf{\text{Length} = 3p, \quad \text{Breadth} = s - 1, \quad \text{Height} = s - 4}$$
Factorise:
$$3p(s^2 - 5s + 4) = 3p(s - 1)(s - 4)$$
Possible dimensions are:
$$\mathbf{\text{Length} = 3p, \quad \text{Breadth} = s - 1, \quad \text{Height} = s - 4}$$
Q7: Playground Path Area
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 $s$.
Case 1: The path is created outside the playground.
• Inner playground side $= 40\text{ m}$.
• Outer total square side including path of width $s$ on both sides $= 40 + 2s\text{ m}$.
• Area of path $= \text{Outer Area} - \text{Inner Area}$:
$$\text{Area} = (40 + 2s)^2 - 40^2 = (1600 + 160s + 4s^2) - 1600 = \mathbf{4s^2 + 160s \text{ m}^2}$$
• Inner playground side $= 40\text{ m}$.
• Outer total square side including path of width $s$ on both sides $= 40 + 2s\text{ m}$.
• Area of path $= \text{Outer Area} - \text{Inner Area}$:
$$\text{Area} = (40 + 2s)^2 - 40^2 = (1600 + 160s + 4s^2) - 1600 = \mathbf{4s^2 + 160s \text{ m}^2}$$
Case 2: The path is created inside the playground boundaries.
• Outer playground side $= 40\text{ m}$.
• Inner play area side $= 40 - 2s\text{ m}$.
• Area of path $= \text{Outer Area} - \text{Inner Area}$:
$$\text{Area} = 40^2 - (40 - 2s)^2 = 1600 - (1600 - 160s + 4s^2) = \mathbf{160s - 4s^2 \text{ m}^2}$$
• Outer playground side $= 40\text{ m}$.
• Inner play area side $= 40 - 2s\text{ m}$.
• Area of path $= \text{Outer Area} - \text{Inner Area}$:
$$\text{Area} = 40^2 - (40 - 2s)^2 = 1600 - (1600 - 160s + 4s^2) = \mathbf{160s - 4s^2 \text{ m}^2}$$
Q8: Number Reciprocal Problem
If a number plus its reciprocal equals $\dfrac{10}{3}$, find the number.
Let the number be $x$.
$$x + \frac{1}{x} = \frac{10}{3}$$
Multiply by $3x$:
$$3x^2 + 3 = 10x \implies 3x^2 - 10x + 3 = 0$$
$$x + \frac{1}{x} = \frac{10}{3}$$
Multiply by $3x$:
$$3x^2 + 3 = 10x \implies 3x^2 - 10x + 3 = 0$$
Split the middle term:
$$3x^2 - 9x - x + 3 = 0 \implies 3x(x - 3) - 1(x - 3) = 0$$
$$(3x - 1)(x - 3) = 0$$
Therefore, $x = 3$ or $x = \dfrac{1}{3}$.
$$3x^2 - 9x - x + 3 = 0 \implies 3x(x - 3) - 1(x - 3) = 0$$
$$(3x - 1)(x - 3) = 0$$
Therefore, $x = 3$ or $x = \dfrac{1}{3}$.
The number is either 3 or 1/3.
Q9: Rectangular Pool Dimensions
A rectangular pool has area $2x^2 + 7x + 3$ square hastas. If its width is $2x + 1$ hastas, find its length.
We know: $\text{Area} = \text{Length} \times \text{Width}$.
$$\text{Length} = \frac{\text{Area}}{\text{Width}} = \frac{2x^2 + 7x + 3}{2x + 1}$$
$$\text{Length} = \frac{\text{Area}}{\text{Width}} = \frac{2x^2 + 7x + 3}{2x + 1}$$
Factorise the numerator:
$$2x^2 + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3)$$
Substitute back:
$$\text{Length} = \frac{(2x + 1)(x + 3)}{2x + 1} = \mathbf{x + 3}$$
$$2x^2 + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3)$$
Substitute back:
$$\text{Length} = \frac{(2x + 1)(x + 3)}{2x + 1} = \mathbf{x + 3}$$
Length is x + 3 hastas.
Q10: Proof: Equal Coefficients
If both $x - 2$ and $x - \dfrac{1}{2}$ are factors of $px^2 + 5x + r$, show that $p = r$.
Let $f(x) = px^2 + 5x + r$.
Since $x - 2$ is a factor, by the Factor Theorem: $f(2) = 0$.
$$p(2)^2 + 5(2) + r = 0 \implies 4p + r = -10 \quad \text{--- (Equation 1)}$$
Since $x - 2$ is a factor, by the Factor Theorem: $f(2) = 0$.
$$p(2)^2 + 5(2) + r = 0 \implies 4p + r = -10 \quad \text{--- (Equation 1)}$$
Since $x - \dfrac{1}{2}$ is a factor, $f\left(\frac{1}{2}\right) = 0$.
$$p\left(\frac{1}{2}\right)^2 + 5\left(\frac{1}{2}\right) + r = 0 \implies \frac{p}{4} + \frac{5}{2} + r = 0 \implies p + 4r = -10 \quad \text{--- (Equation 2)}$$
$$p\left(\frac{1}{2}\right)^2 + 5\left(\frac{1}{2}\right) + r = 0 \implies \frac{p}{4} + \frac{5}{2} + r = 0 \implies p + 4r = -10 \quad \text{--- (Equation 2)}$$
Equate Equation 1 and Equation 2:
$$4p + r = p + 4r \implies 3p = 3r \implies \mathbf{p = r}$$
Hence Proved.
$$4p + r = p + 4r \implies 3p = 3r \implies \mathbf{p = r}$$
Hence Proved.
p = r is proved.
Q11: Proof: Master Identity Evaluation
If $a + b + c = 5$ and $ab + bc + ca = 10$, then prove that $a^3 + b^3 + c^3 - 3abc = -25$.
We know:
$$(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$$
$$5^2 = a^2 + b^2 + c^2 + 2(10)$$
$$25 = a^2 + b^2 + c^2 + 20 \implies a^2 + b^2 + c^2 = 5$$
$$(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$$
$$5^2 = a^2 + b^2 + c^2 + 2(10)$$
$$25 = a^2 + b^2 + c^2 + 20 \implies a^2 + b^2 + c^2 = 5$$
Apply the Master Cubic Identity:
$$a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$$
$$= 5 \times (5 - 10)$$
$$= 5 \times (-5) = \mathbf{-25}$$
Hence Proved.
$$a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$$
$$= 5 \times (5 - 10)$$
$$= 5 \times (-5) = \mathbf{-25}$$
Hence Proved.
-25 is proved.
Q12: Divisibility of n³ − n by 6
By factoring the expression, check that $n^3 - n$ is always divisible by 6 for all natural numbers $n$. Give reasons.
Factorise $n^3 - n$:
$$n^3 - n = n(n^2 - 1) = (n - 1)n(n + 1)$$
This is the product of three consecutive integers.
$$n^3 - n = n(n^2 - 1) = (n - 1)n(n + 1)$$
This is the product of three consecutive integers.
Reasons for divisibility by 6:
1. **Divisibility by 2:** Out of any two consecutive numbers, one must be even. Thus, the product is always divisible by 2.
2. **Divisibility by 3:** Out of three consecutive numbers, exactly one must be a multiple of 3.
3. Since 2 and 3 are coprime, a number divisible by both 2 and 3 is always divisible by their product: $2 \times 3 = 6$.
1. **Divisibility by 2:** Out of any two consecutive numbers, one must be even. Thus, the product is always divisible by 2.
2. **Divisibility by 3:** Out of three consecutive numbers, exactly one must be a multiple of 3.
3. Since 2 and 3 are coprime, a number divisible by both 2 and 3 is always divisible by their product: $2 \times 3 = 6$.
n³ − n is always divisible by 6.
Q13: Evaluate Conditional Cubic Expressions
Find the value of:
(i) $x^3 + y^3 - 12xy + 64$, when $x + y = -4$
(ii) $x^3 - 8y^3 - 36xy - 216$, when $x = 2y + 6$
(i) $x^3 + y^3 - 12xy + 64$, when $x + y = -4$
(ii) $x^3 - 8y^3 - 36xy - 216$, when $x = 2y + 6$
(i) When $x + y = -4 \implies x + y + 4 = 0$:
Using the identity if $a + b + c = 0$, then $a^3 + b^3 + c^3 = 3abc$.
Let $a = x$, $b = y$, $c = 4$.
$$x^3 + y^3 + 4^3 = 3(x)(y)(4) \implies x^3 + y^3 + 64 = 12xy$$
$$x^3 + y^3 - 12xy + 64 = \mathbf{0}$$
Using the identity if $a + b + c = 0$, then $a^3 + b^3 + c^3 = 3abc$.
Let $a = x$, $b = y$, $c = 4$.
$$x^3 + y^3 + 4^3 = 3(x)(y)(4) \implies x^3 + y^3 + 64 = 12xy$$
$$x^3 + y^3 - 12xy + 64 = \mathbf{0}$$
(ii) When $x = 2y + 6 \implies x - 2y - 6 = 0$:
Let $a = x$, $b = -2y$, $c = -6$.
Since $a + b + c = 0$:
$$x^3 + (-2y)^3 + (-6)^3 = 3(x)(-2y)(-6)$$
$$x^3 - 8y^3 - 216 = 36xy$$
$$x^3 - 8y^3 - 36xy - 216 = \mathbf{0}$$
Let $a = x$, $b = -2y$, $c = -6$.
Since $a + b + c = 0$:
$$x^3 + (-2y)^3 + (-6)^3 = 3(x)(-2y)(-6)$$
$$x^3 - 8y^3 - 216 = 36xy$$
$$x^3 - 8y^3 - 36xy - 216 = \mathbf{0}$$
(i) 0 (ii) 0