Expansions — Exercise 3(A)

Using standard formulae, expand each of the following:

(i) (4a + 9)²

(ii) (3x + 10y)²

(iii) $(\sqrt{2}m + \sqrt{3}n)^2$

Answer

We know that,

⇒ (a + b)² = a² + b² + 2ab.

(i) Given,

⇒ (4a + 9)²

⇒ (4a)² + (9)² + 2 × 4a × 9

⇒ 16a² + 81 + 72a.

Hence, (4a + 9)² = 16a² + 81 + 72a.

(ii) Given,

⇒ (3x + 10y)²

⇒ (3x)² + (10y)² + 2 × 3x × 10y

⇒ 9x² + 100y² + 60xy.

Hence, (3x + 10y)² = 9x² + 100y² + 60xy.

(iii) Given,

$\Rightarrow (\sqrt{2}m + \sqrt{3}n)^2 \\[1em] \Rightarrow (\sqrt{2}m)^2 + (\sqrt{3}n)^2 + 2 \times \sqrt{2}m \times \sqrt{3}n \\[1em] \Rightarrow 2m^2 + 3n^2 + 2\sqrt{6}mn.$

Hence, $(\sqrt{2}m + \sqrt{3}n)^2 = 2m^2 + 3n^2 + 2\sqrt{6}mn$.

Using standard formulae, expand each of the following:

(i) (2a² + 3b)²

(ii) (3x²y + z)²

(iii) $\Big(2x + \dfrac{1}{3x}\Big)^2$

Answer

We know that,

⇒ (a + b)² = a² + b² + 2ab.

(i) Given,

⇒ (2a² + 3b)²

⇒ (2a²)² + (3b)² + 2 × 2a² × 3b

⇒ 4a⁴ + 9b² + 12a²b

Hence, (2a² + 3b)² = 4a⁴ + 9b² + 12a²b .

(ii) Given,

⇒ (3x²y + z)²

⇒ (3x²y)² + (z)² + 2 × 3x²y × z

⇒ 9x⁴y² + z² + 6x²yz

Hence, (3x²y + z)² = 9x⁴y² + z² + 6x²yz .

(iii) Given,

$\Rightarrow \Big(2x + \dfrac{1}{3x}\Big)^2 \\[1em] \Rightarrow (2x)^2 + \Big(\dfrac{1}{3x}\Big)^2 + 2 \times 2x \times \dfrac{1}{3x} \\[1em] \Rightarrow 4x^2 + \dfrac{1}{9x^2} + \dfrac{4}{3}$

Hence, $\Big(2x + \dfrac{1}{3x}\Big)^2 = 4x^2 + \dfrac{1}{9x^2} + \dfrac{4}{3}$.

Using standard formulae, expand each of the following:

(i) $\Big(\dfrac{2}{5}x + \dfrac{5}{6}y\Big)^2$

(ii) $\Big(\dfrac{x}{3} + \dfrac{6}{x}\Big)^2$

(iii) $\Big(6 + \dfrac{5}{x}\Big)^2$

Answer

We know that,

⇒ (a + b)² = a² + b² + 2ab.

(i) Given,

$\Rightarrow \Big(\dfrac{2}{5}x + \dfrac{5}{6}y\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{2}{5}x\Big)^2 + \Big(\dfrac{5}{6}y\Big)^2 + 2 \times \dfrac{2}{5}x \times \dfrac{5}{6}y \\[1em] \Rightarrow \dfrac{4}{25}x^2 + \dfrac{25}{36}y^2 + \dfrac{4}{6}xy \\[1em] \Rightarrow \dfrac{4}{25}x^2 + \dfrac{25}{36}y^2 + \dfrac{2}{3}xy$

Hence, $\Big(\dfrac{2}{5}x + \dfrac{5}{6}y\Big)^2 = \dfrac{4}{25}x^2 + \dfrac{25}{36}y^2 + \dfrac{2}{3}xy$.

(ii) Given,

$\Rightarrow \Big(\dfrac{x}{3} + \dfrac{6}{x}\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{x}{3}\Big)^2 + \Big(\dfrac{6}{x}\Big)^2 + 2 \times \dfrac{x}{3} \times \dfrac{6}{x} \\[1em] \Rightarrow \dfrac{x^2}{9} + \dfrac{36}{x^2} + 4$

Hence, $\Big(\dfrac{x}{3} + \dfrac{6}{x}\Big)^2 = \dfrac{x^2}{9} + \dfrac{36}{x^2} + 4$.

(iii) Given,

$\Rightarrow \Big(6 + \dfrac{5}{x}\Big)^2 \\[1em] \Rightarrow (6)^2 + \Big(\dfrac{5}{x}\Big)^2 + 2 \times 6 \times \dfrac{5}{x} \\[1em] \Rightarrow 36 + \dfrac{25}{x^2} + \dfrac{60}{x}$

Hence, $\Big(6 + \dfrac{5}{x}\Big)^2 = 36 + \dfrac{25}{x^2} + \dfrac{60}{x}$.

Using standard formulae, expand each of the following:

(i) (5x - 3y)²

(ii) (3a - 7b)²

(iii) $\Big(\dfrac{1}{2}x - \dfrac{3}{2}y\Big)^2$

Answer

We know that,

⇒ (a - b)² = a² + b² - 2ab.

(i) Given,

⇒ (5x - 3y)²

⇒ (5x)² + (3y)² - 2 × 5x × 3y

⇒ 25x² + 9y² - 30xy

Hence, (5x - 3y)² = 25x² + 9y² - 30xy .

(ii) Given,

⇒ (3a - 7b)²

⇒ (3a)² + (7b)² - 2 × 3a × 7b

⇒ 9a² + 49b² - 42ab

Hence, (3a - 7b)² = 9a² + 49b² - 42ab.

(iii) Given,

$\Rightarrow \Big(\dfrac{1}{2}x\ -\ \dfrac{3}{2}y\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{1}{2}x\Big)^2 + \Big(\dfrac{3}{2}y\Big)^2 - 2 \times \dfrac{1}{2}x \times \dfrac{3}{2}y \\[1em] \Rightarrow \dfrac{x^2}{4} + \dfrac{9}{4}y^2 - \dfrac{3}{2}xy$

Hence, $\Big(\dfrac{1}{2}x - \dfrac{3}{2}y\Big)^2 = \dfrac{x^2}{4} + \dfrac{9}{4}y^2 - \dfrac{3}{2}xy$.

Using standard formulae, expand each of the following:

(i) $\Big(a^2 - \dfrac{b}{2}\Big)^2$

(ii) $\Big(\dfrac{3a}{2b} - \dfrac{2b}{3a}\Big)^2$

(iii) $\Big(5x - \dfrac{2}{3x}\Big)^2$

Answer

We know that,

⇒ (a - b)² = a² + b² - 2ab.

(i) Given,

$\Rightarrow \Big(a^2 - \dfrac{b}{2}\Big)^2 \\[1em] \Rightarrow (a^2)^2 + \Big(\dfrac{b}{2}\Big)^2 - 2 \times a^2 \times \dfrac{b}{2} \\[1em] \Rightarrow a^4 + \dfrac{b^2}{4} - a^2b$

Hence, $\Big(a^2 - \dfrac{b}{2}\Big)^2 = a^4 + \dfrac{b^2}{4} - a^2b$.

(ii) Given,

$\Rightarrow \Big(\dfrac{3a}{2b} - \dfrac{2b}{3a}\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{3a}{2b}\Big)^2 + \Big(\dfrac{2b}{3a}\Big)^2 - 2 \times \dfrac{3a}{2b} \times \dfrac{2b}{3a} \\[1em] \Rightarrow \dfrac{9a^2}{4b^2} + \dfrac{4b^2}{9a^2} - 2$

Hence, $\Big(\dfrac{3a}{2b} - \dfrac{2b}{3a}\Big)^2 = \dfrac{9a^2}{4b^2} + \dfrac{4b^2}{9a^2} - 2$.

(iii) Given,

$\Rightarrow \Big(5x - \dfrac{2}{3x}\Big)^2 \\[1em] \Rightarrow (5x)^2 + \Big(\dfrac{2}{3x}\Big)^2 - 2 \times 5x \times \dfrac{2}{3x} \\[1em] \Rightarrow 25x^2 + \dfrac{4}{9x^2} - \dfrac{20}{3}$

Hence, $\Big(5x - \dfrac{2}{3x}\Big)^2 = 25x^2 + \dfrac{4}{9x^2} - \dfrac{20}{3}$.

Using standard formulae, expand each of the following:

(i) (a + 2b + 3c)²

(ii) (3x + 5y - 2z)²

(iii) (2x - 3y + 7z)²

Answer

We know that,

⇒ (a + b + c)² = a² + b² + c² + 2(ab + bc + ca).

(i) Given,

⇒ (a + 2b + 3c)²

⇒ (a)² + (2b)² + (3c)² + 2 × (a × 2b + 2b × 3c + 3c × a)

⇒ a² + 4b² + 9c² + 2 × (2ab + 6bc + 3ca)

⇒ a² + 4b² + 9c² + 4ab + 12bc + 6ac

Hence, (a + 2b + 3c)² = a² + 4b² + 9c² + 4ab + 12bc + 6ac.

(ii) Given,

⇒ (3x + 5y - 2z)²

⇒ [3x + 5y + (-2z)]²

⇒ (3x)² + (5y)² + (-2z)² + 2 × [3x × 5y + 5y × (-2z) + (-2z) × 3x]

⇒ 9x² + 25y² + 4z² + 2 × (15xy - 10yz - 6xz)

⇒ 9x² + 25y² + 4z² + 30xy - 20yz - 12xz

Hence, (3x + 5y - 2z) = 9x² + 25y² + 4z² + 30xy - 20yz - 12xz.

(iii) Given,

⇒ (2x - 3y + 7z)²

⇒ [2x + (-3y) + 7z]²

⇒ (2x)² + (-3y)² + (7z)² + 2 × [2x × (-3y) + (-3y) × (7z) + 7z × 2x]

⇒ 4x² + 9y² + 49z² + 2 × [-6xy - 21yz + 14xz]

⇒ 4x² + 9y² + 49z² - 12xy - 42yz + 28xz.

Hence, (2x - 3y + 7z) = 4x² + 9y² + 49z² - 12xy - 42yz + 28xz.

Using standard formulae, expand each of the following:

(i) (6 - 2y + 4z)²

(ii) (4x - 3y + z)²

(iii) (7 - 2x - 3y)²

Answer

We know that,

⇒ (a + b + c)² = a² + b² + c² + 2(ab + bc + ca).

(i) Given,

⇒ (6 - 2y + 4z)²

⇒ [6 + (-2y) + 4z]²

⇒ (6)² + (-2y)² + (4z)² + 2 × [6 × (-2y) + (-2y) × (4z) + 4z × 6]

⇒ 36 + 4y² + 16z² + 2 × [-12y - 8yz + 24z]

⇒ 36 + 4y² + 16z² - 24y - 16yz + 48z

Hence, (6 - 2y + 4z)² = 36 + 4y² + 16z² - 24y - 16yz + 48z.

(ii) Given,

⇒ (4x - 3y + z)²

⇒ [4x + (-3y) + z]²

⇒ (4x)² + (-3y)² + (z)² + 2 × [4x × (-3y) + (-3y) × (z) + z × 4x]

⇒ (16x)² + 9y² + z² + 2 × [-12xy - 3yz + 4xz]

⇒ 16x² + 9y² + z² - 24xy - 6yz + 8xz

Hence, (4x - 3y + z)² = 16x² + 9y² + z² - 24xy - 6yz + 8xz.

(iii) Given,

⇒ (7 - 2x - 3y)²

⇒ [7 + (-2x) + (-3y)]²

⇒ (7)² + (-2x)² + (-3y)² + 2 × [7 × (-2x) + (-2x) × (-3y) + (-3y) × 7]

⇒ 49 + 4x² + 9y² + 2 × (-14x + 6xy - 21y)

⇒ 49 + 4x² + 9y² - 28x + 12xy - 42y.

Hence, (7 - 2x - 3y)² = 49 + 4x² + 9y² - 28x + 12xy - 42y.

Using standard formulae, expand each of the following:

(i) $\Big(\dfrac{a}{2} + \dfrac{b}{3} + \dfrac{c}{4}\Big)^2$

(ii) $\Big(\dfrac{2x}{3} + \dfrac{3}{2y} - 2\Big)^2$

(iii) $\Big(2x + \dfrac{3}{x} - 1 \Big)^2$

Answer

We know that,

⇒ (a + b + c)² = a² + b² + c² + 2(ab + bc + ca).

(i) Given,

$\Rightarrow \Big(\dfrac{a}{2} + \dfrac{b}{3} + \dfrac{c}{4}\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{a}{2}\Big)^2 +\Big(\dfrac{b}{3}\Big)^2 + \Big(\dfrac{c}{4}\Big)^2 + 2 × \Big[\big(\dfrac{a}{2}\big) × \big(\dfrac{b}{3}\big) + \big(\dfrac{b}{3}\big) × \big(\dfrac{c}{4}\big) + \big(\dfrac{c}{4}\big) × \big(\dfrac{a}{2}\big)\Big] \\[1em] \Rightarrow \dfrac{a^2}{4} + \dfrac{b^2}{9} + \dfrac{c^2}{16} + 2 × \Big[\big(\dfrac{ab}{6}\big) + \big(\dfrac{bc}{12}\big) + \big(\dfrac{ca}{8}\big)\Big] \\[1em] \Rightarrow \dfrac{a^2}{4} + \dfrac{b^2}{9} + \dfrac{c^2}{16} + \dfrac{ab}{3} + \dfrac{bc}{6} + \dfrac{ca}{4} \\[1em]$

Hence, $\Big(\dfrac{a}{2} + \dfrac{b}{3} + \dfrac{c}{4}\Big)^2 = \dfrac{a^2}{4} + \dfrac{b^2}{9} + \dfrac{c^2}{16} + \dfrac{ab}{3} + \dfrac{bc}{6} + \dfrac{ca}{4}$.

(ii) Given,

$\Rightarrow \Big(\dfrac{2x}{3} + \dfrac{3}{2y} - 2\Big)^2 \\[1em] \Rightarrow \Big[\dfrac{2x}{3} + \dfrac{3}{2y} + (-2)\Big]^2 \\[1em] \Rightarrow \Big(\dfrac{2x}{3}\Big)^2 +\Big(\dfrac{3}{2y}\Big)^2 + (-2)^2 + 2 × \Big[\big(\dfrac{2x}{3}\big) × \big(\dfrac{3}{2y}\big) + \big(\dfrac{3}{2y}\big) × (-2) + (-2) × \big(\dfrac{2x}{3}\big)\Big] \\[1em] \Rightarrow \dfrac{4x^2}{9} + \dfrac{9}{4y^2} + 4 + 2 × \Big[\dfrac{6x}{6y} - \dfrac{6}{2y} - \dfrac{4x}{3}\Big] \\[1em] \Rightarrow \dfrac{4x^2}{9} + \dfrac{9}{4y^2} + 4 + 2 × \Big[\dfrac{x}{y} - \dfrac{3}{y} - \dfrac{4x}{3}\Big]\\[1em] \Rightarrow \dfrac{4x^2}{9} + \dfrac{9}{4y^2} + 4 + \dfrac{2x}{y} - \dfrac{6}{y} - \dfrac{8x}{3} \\[1em]$

Hence, $\Big(\dfrac{2x}{3} + \dfrac{3}{2y} - 2\Big)^2 = \dfrac{4x^2}{9} + \dfrac{9}{4y^2} + 4 + \dfrac{2x}{y} - \dfrac{6}{y} - \dfrac{8x}{3}$.

(iii) Given,

$\Rightarrow \Big(2x + \dfrac{3}{x} - 1 \Big)^2 \\[1em] \Rightarrow \Big[2x + \dfrac{3}{x} + (-1) \Big]^2\\[1em] \Rightarrow (2x)^2 +\Big(\dfrac{3}{x}\Big)^2 + (-1)^2 + 2 × \Big[2x × \Big(\dfrac{3}{x}\Big) + \Big(\dfrac{3}{x}\Big) × (-1) + (-1) × (2x)\Big] \\[1em] \Rightarrow 4x^2 + \dfrac{9}{x^2} + 1 + 2 × \Big[6 - \dfrac{3}{x} - 2x\Big] \\[1em] \Rightarrow 4x^2 + \dfrac{9}{x^2} + 1 + 12 - \dfrac{6}{x} - 4x \\[1em] \Rightarrow 4x^2 + \dfrac{9}{x^2} + 13 - \dfrac{6}{x} - 4x \\[1em]$

Hence, $\Big(2x + \dfrac{3}{x} - 1 \Big)^2 = 4x^2 + \dfrac{9}{x^2} + 13 - \dfrac{6}{x} - 4x$.

Using standard formulae, expand each of the following:

(i) (x + 7)(x + 4)

(ii) (a + 13)(a - 8)

(iii) (y - 6)(y - 4)

Answer

(i) Given,

⇒ (x + 7)(x + 4)

⇒ x² + 4x + 7x + 28

⇒ x² + 11x + 28.

Hence, (x + 7)(x + 4) = x² + 11x + 28.

(ii) Given,

⇒ (a + 13)(a - 8)

⇒ a² - 8a + 13a - 104

⇒ a² - 5a - 104.

Hence, (a + 13)(a - 8) = a² + 5a - 104.

(iii) Given,

⇒ (y - 6)(y - 4)

⇒ y² - 4y - 6y + 24

⇒ y² - 10y + 24.

Hence, (y - 6)(y - 4) = y² - 10y + 24.

Using standard formulae, expand each of the following:

(i) (9 + 2x)(9 - 3x)

(ii) (5x - 4y)(5x + 3y)

(iii) (3 - 7a)(3 + 4a)

Answer

(i) Given,

⇒ (9 + 2x)(9 - 3x)

⇒ 81 - 27x + 18x - 6x²

⇒ 81 - 9x - 6x².

Hence, (9 + 2x)(9 - 3x) = 81 - 9x - 6x².

(ii) Given,

⇒ (5x - 4y)(5x + 3y)

⇒ 25x² + 15xy - 20xy - 12y²

⇒ 25x² - 5xy - 12y².

Hence, (5x - 4y)(5x + 3y) = 25x² - 5xy - 12y².

(iii) Given,

⇒ (3 - 7a)(3 + 4a)

⇒ 9 + 12a - 21a - 28a²

⇒ 9 - 9a - 28a².

Hence, (3 - 7a)(3 + 4a) = 9 - 9a - 28a².

Using standard formulae, expand each of the following:

(i) (3a + 2b)(3a - 2b)

(ii) $\Big(5x + \dfrac{1}{5x}\Big)\Big(5x - \dfrac{1}{5x}\Big)$

(iii) $\Big(2x^2 + \dfrac{3}{x^2}\Big)\Big(2x^2 - \dfrac{3}{x^2}\Big)$

Answer

We know that,

(a + b)(a - b) = a² - b²

(i) Given,

⇒ (3a + 2b)(3a - 2b)

⇒ (3a)² - (2b)²

⇒ 9a² - 4b².

Hence, (3a + 2b)(3a - 2b) = 9a² - 4b².

(ii) Given,

$\Rightarrow \Big(5x + \dfrac{1}{5x}\Big)\Big(5x - \dfrac{1}{5x}\Big)\\[1em] \Rightarrow (5x)^2 - \Big(\dfrac{1}{5x}\Big)^2 \\[1em] \Rightarrow 25x^2 - \dfrac{1}{25x^2}.$

Hence, $\Big(5x + \dfrac{1}{5x}\Big)\Big(5x - \dfrac{1}{5x}\Big) = 25x^2 - \dfrac{1}{25x^2}$.

(iii) Given,

$\Rightarrow \Big(2x^2 + \dfrac{3}{x^2}\Big)\Big(2x^2 - \dfrac{3}{x^2}\Big) \\[1em] \Rightarrow (2x^2)^2 - \Big(\dfrac{3}{x^2}\Big)^2 \\[1em] \Rightarrow 4x^4 - \dfrac{9}{x^4} \\[1em]$

Hence, $\Big(2x^2 + \dfrac{3}{x^2}\Big)\Big(2x^2 - \dfrac{3}{x^2}\Big) = 4x^4 - \dfrac{9}{x^4}$.

Using standard formulae, expand each of the following:

(i) (2 - x)(2 + x)(4 + x²)

(ii) (x + y)(x - y)(x² + y²)

Answer

We know that,

(a + b)(a - b) = a² - b²

(i) Given,

⇒ (2 - x)(2 + x)(4 + x²)

⇒ [(2)² - (x)²](4 + x²)

⇒ (4 - x²)(4 + x²)

⇒ (4)² - (x²)²

⇒ 16 - x⁴.

Hence, (2 - x)(2 + x)(4 + x²) = 16 - x⁴.

(ii) Given,

⇒ (x + y)(x - y)(x² + y²)

⇒ [(x)² - (y)²](x² + y²)

⇒ (x² - y²)(x² + y²)

⇒ (x²)² - (y²)²

⇒ (x⁴ - y⁴).

Hence, (x + y)(x - y)(x² + y²) = (x⁴ - y⁴).

Using standard formulae, expand each of the following:

(i) (x - 2)(x - 3)(x + 4)

(ii) (x - 5)(2x - 1)(2x + 3)

Answer

(i) Given,

⇒ (x - 2)(x - 3)(x + 4)

⇒ (x² - 3x - 2x + 6)(x + 4)

⇒ (x² - 5x + 6)(x + 4)

⇒ x²(x + 4) - 5x(x + 4) + 6(x + 4)

⇒ (x³ + 4x² - 5x² - 20x + 6x + 24)

⇒ (x³ - x² - 14x + 24)

Hence, (x - 2)(x - 3)(x + 4) = x³ - x² - 14x + 24.

(ii) Given,

⇒ (x - 5)(2x - 1)(2x + 3)

⇒ (2x² - x - 10x + 5)(2x + 3)

⇒ (2x² - 11x + 5)(2x + 3)

⇒ 2x²(2x + 3) - 11x(2x + 3) + 5(2x + 3)

⇒ (4x³ + 6x² - 22x² - 33x + 10x + 15)

⇒ (4x³ - 16x² - 23x + 15)

Hence, (x - 5)(2x - 1)(2x + 3) = 4x³ - 16x² - 23x + 15.

Simplify:

(i) (a + b)² + (a - b)²

(ii) (a + b)² - (a - b)²

(iii) $\Big(x + \dfrac{1}{x}\Big)^2 + \Big(x - \dfrac{1}{x}\Big)^2$

(iv) $\Big(x + \dfrac{1}{x}\Big)^2 - \Big(x - \dfrac{1}{x}\Big)^2$

(v) $\Big(\dfrac{a}{2b} + \dfrac{2b}{a}\Big)^2 - \Big(\dfrac{2b}{a} - \dfrac{a}{2b}\Big)^2$

(vi) $\Big(3x - \dfrac{1}{3x}\Big)^2 - \Big(3x + \dfrac{1}{3x}\Big)\Big(3x - \dfrac{1}{3x}\Big)$

(vii) (5a + 3b)² - (5a - 3b)² - 60ab

(viii) (3x + 1)² - (3x + 2)(3x - 1)

Answer

(i) Given,

⇒ (a + b)² + (a - b)²

⇒ a² + b² + 2ab + a² + b² - 2ab

⇒ 2a² + 2b²

⇒ 2(a² + b²)

Hence, (a + b)² + (a - b)² = 2(a² + b²).

(ii) Given,

⇒ (a + b)² - (a - b)²

⇒ (a² + b² + 2ab) - (a² + b² - 2ab)

⇒ a² + b² + 2ab - a² - b² + 2ab

⇒ 4ab

Hence, (a + b)² - (a - b)² = 4ab.

(iii) Given,

$\Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 + \Big(x - \dfrac{1}{x}\Big)^2 \\[1em] \Rightarrow \Big[x^2 + \Big(\dfrac{1}{x^2}\Big) + 2 \times x \times \Big(\dfrac{1}{x}\Big) + x^2 + \Big(\dfrac{1}{x^2}\Big) - 2 \times x \times \Big(\dfrac{1}{x}\Big)\Big] \\[1em] \Rightarrow \Big(x^2 + \dfrac{1}{x^2} + 2\Big) + \Big(x^2 + \dfrac{1}{x^2} - 2\Big) \\[1em] \Rightarrow 2x^2 + \dfrac{2}{x^2} \\[1em] \Rightarrow 2\Big(x^2 + \dfrac{1}{x^2}\Big) \\[1em]$

Hence, $\Big(x + \dfrac{1}{x}\Big)^2 + \Big(x - \dfrac{1}{x}\Big)^2 = 2\Big(x^2 + \dfrac{1}{x^2}\Big)$.

(iv) Given,

$\Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 - \Big(x - \dfrac{1}{x}\Big)^2 \\[1em] \Rightarrow \Big[x^2 + \Big(\dfrac{1}{x^2}\Big) + 2 \times x \times \Big(\dfrac{1}{x}\Big)\Big] - \Big[x^2 + \Big(\dfrac{1}{x^2}\Big) - 2 \times x \times \Big(\dfrac{1}{x}\Big)\Big] \\[1em] \Rightarrow \Big(x^2 + \dfrac{1}{x^2} + 2\Big) - \Big(x^2 + \dfrac{1}{x^2} - 2\Big) \\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} + 2 - x^2 - \dfrac{1}{x^2} + 2 \\[1em] \Rightarrow 4.$

Hence, $\Big(x + \dfrac{1}{x}\Big)^2 - \Big(x - \dfrac{1}{x}\Big)^2 = 4$.

(v) Given,

$\Rightarrow \Big(\dfrac{a}{2b} + \dfrac{2b}{a}\Big)^2 - \Big(\dfrac{2b}{a} - \dfrac{a}{2b}\Big)^2 \\[1em] \Rightarrow \Big[\Big(\dfrac{a}{2b}\Big)^2 + \Big(\dfrac{2b}{a}\Big)^2 + 2 \times \dfrac{a}{2b} \times \dfrac{2b}{a}\Big] - \Big[\Big(\dfrac{2b}{a}\Big)^2 + \Big(\dfrac{a}{2b}\Big)^2 - 2 \times \dfrac{a}{2b} \times \dfrac{2b}{a}\Big] \\[1em] \Rightarrow \Big(\dfrac{a^2}{4b^2} + \dfrac{4b^2}{a^2} + 2\Big) - \Big(\dfrac{4b^2}{a^2} + \dfrac{a^2}{4b^2} - 2\Big) \\[1em] \Rightarrow \dfrac{a^2}{4b^2} + \dfrac{4b^2}{a^2} + 2 - \dfrac{a^2}{4b^2} - \dfrac{4b^2}{a} + 2 \\[1em] \Rightarrow 4.$

Hence, $\Big(\dfrac{a}{2b} + \dfrac{2b}{a}\Big)^2 - \Big(\dfrac{2b}{a} - \dfrac{a}{2b}\Big)^2 = 4$.

(vi) Given,

$\Rightarrow \Big(3x - \dfrac{1}{3x}\Big)^2 - \Big(3x + \dfrac{1}{3x}\Big)\Big(3x - \dfrac{1}{3x}\Big) \\[1em] \Rightarrow \Big[(3x)^2 + \Big(\dfrac{1}{3x}\Big)^2 - 2 \times 3x \times \dfrac{1}{3x}\Big] - \Big[(3x)^2 - \Big(\dfrac{1}{3x}\Big)^2\Big] \\[1em] \Rightarrow \Big(9x^2 + \dfrac{1}{9x^2} - 2 \Big) - \Big(9x^2 - \dfrac{1}{9x^2}\Big) \\[1em] \Rightarrow 9x^2 + \dfrac{1}{9x^2} - 2 - 9x^2 + \dfrac{1}{9x^2} \\[1em] \Rightarrow \dfrac{2}{9x^2} - 2 \\[1em] \Rightarrow 2\Big(\dfrac{1}{9x^2} - 1\Big)$

Hence, $\Big(3x - \dfrac{1}{3x}\Big)^2 - \Big(3x + \dfrac{1}{3x})\Big(3x - \dfrac{1}{3x}\Big) = 2\Big(\dfrac{1}{9x^2} - 1\Big)$.

(vii) Given,

⇒ (5a + 3b)² - (5a - 3b)² - 60ab

⇒ [(5a)² + (3b)² + 2 × 5a × 3b] - [(5a)² + (3b)² - 2 × 5a × 3b] - 60ab

⇒ [25a² + 9b² + 2 × 5a × 3b] - [25a² + 9b² - 2 × 5a × 3b] - 60ab

⇒ 25a² + 9b² + 30ab - 25a² - 9b² + 30ab - 60ab

⇒ 60ab - 60ab

⇒ 0

Hence, (5a + 3b)² - (5a - 3b)² - 60ab = 0.

(viii) Given,

⇒ (3x + 1)² - [(3x + 2)(3x - 1)]

⇒ (3x)² + (1)² + 2 × 3x × 1 - (9x² - 3x + 6x - 2)

⇒ 9x² + 1 + 6x - (9x² + 3x - 2)

⇒ 9x² + 1 + 6x - 9x² - 3x + 2

⇒ 1 + 3x + 2

⇒ 3x + 3

⇒ 3(x + 1).

Hence, (3x + 1)² - (3x + 2)(3x - 1) = 3(x + 1).

(i) If (a + b) = 7 and ab = 10, find the value of (a - b).

(ii) If (x - y) = 5 and xy = 24, find the value of (x + y).

Answer

(i) Given,

(a + b) = 7 and ab = 10

Using identity,

⇒ (a + b)² - (a - b)² = 4ab

Substituting values we get :

⇒ (7)² - (a - b)² = 4 × 10

⇒ 49 - (a - b)² = 40

⇒ (a - b)² = 49 - 40

⇒ (a - b)² = 9

⇒ (a - b)² = $\sqrt{9}$

⇒ (a - b) = $\pm 3$

Hence, (a - b) = $\pm 3$.

(ii) Given,

(x - y) = 5 and xy = 24

Using identity,

⇒ (x + y)² - (x - y)² = 4xy

⇒ (x + y)² = 4xy + (x - y)²

⇒ (x + y)² = 4 × 24 + (5)²

⇒ (x + y)² = 96 + 25

⇒ (x + y) = $\sqrt{121}$

⇒ (x + y) = $\pm 11$

Hence, (x + y) = $\pm 11$.

If (3a + 4b) = 16 and ab = 4, find the value of (9a² + 16b²).

Answer

⇒ (3a + 4b)² = (3a)² + (4b)² + 2 × 3a × 4b

⇒ (3a + 4b)² = 9a² + 16b² + 24ab

⇒ 9a² + 16b² = (3a + 4b)² - 24ab

Given,

(3a + 4b) = 16 and ab = 4

Substituting values we get :

⇒ 9a² + 16b² = (16)² - 24 × 4

⇒ 9a² + 16b² = 256 - 96

⇒ 9a² + 16b² = 160.

Hence, 9a² + 16b² = 160.

If (a + b) = 2 and (a - b) = 10, find the values of :

(i) (a² + b²)

(ii) ab

Answer

(i) Given,

(a + b) = 2 and (a - b) = 10

Using identity,

⇒ (a + b)² + (a - b)² = 2(a² + b²)

⇒ (2)² + (10)² = 2(a² + b²)

⇒ 2(a² + b²) = 4 + 100

⇒ 2(a² + b²) = 104

⇒ a² + b² = 52.

Hence, a² + b² = 52.

(ii) Given,

(a + b) = 2 and (a - b) = 10

Using identity,

⇒ (a + b)² - (a - b)² = 4ab

⇒ (2)² - (10)² = 4ab

⇒ 4ab = 4 - 100

⇒ 4ab = -96

⇒ ab = -24

Hence, ab = -24.

If (a - b) = 0.9 and ab = 0.36, find the values of :

(i) (a + b).

(ii) (a² - b²).

Answer

(i) Given,

(a - b) = 0.9 and ab = 0.36

Using identity,

⇒ (a + b)² - (a - b)² = 4ab

⇒ (a + b)² = 4ab + (a - b)²

⇒ (a + b)² = 4 × 0.36 + (0.9)²

⇒ (a + b)² = 1.44 + 0.81

⇒ (a + b)² = 2.25

⇒ (a + b) = $\sqrt{2.25}$

⇒ (a + b) = $\pm 1.5$

Hence, (a + b) = $\pm 1.5$

(ii) Using identity,

⇒ a² − b² = (a − b)(a + b)

⇒ a² − b² = 0.9 × $\pm 1.5$

⇒ a² − b² = $\pm 1.35$

Hence, a² − b² = $\pm 1.35$

If $\Big(x + \dfrac{1}{x}\Big) = 5$, find the values of :

(i) $\Big(x^2 + \dfrac{1}{x^2}\Big)$

(ii) $\Big(x^4 + \dfrac{1}{x^4}\Big)$

Answer

(i) Given,

$\Big(x + \dfrac{1}{x}\Big) = 5$

$\Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 = x^2 + \Big(\dfrac{1}{x}\Big)^2 + 2 \times x \times \dfrac{1}{x} \\[1em] \Rightarrow (5)^2 = x^2 + \dfrac{1}{x^2} + 2 \times x \times \dfrac{1}{x} \\[1em] \Rightarrow 25 = x^2 + \dfrac{1}{x^2} + 2 \\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} = 25 - 2 \\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} = 23.$

Hence, $x^2 + \dfrac{1}{x^2} = 23$.

(ii) Given,

$\Big(x + \dfrac{1}{x}\Big) = 5$

From part (i),

$x^2 + \dfrac{1}{x^2} = 23$

Using identity,

$\Rightarrow \Big(x^2 + \dfrac{1}{x^2}\Big)^2 = (x^2)^2 + \Big(\dfrac{1}{x^2}\Big)^2 + 2 \times x^2 \times \dfrac{1}{x^2} \\[1em] \Rightarrow (23)^2 = (x^2)^2 + \Big(\dfrac{1}{x^2}\Big)^2 + 2 \times x^2 \times \dfrac{1}{x^2} \\[1em] \Rightarrow 529 = x^4 + \dfrac{1}{x^4} + 2 \\[1em] \Rightarrow x^4 + \dfrac{1}{x^4} = 529 - 2 \\[1em] \Rightarrow x^4 + \dfrac{1}{x^4} = 527 \\[1em]$

Hence, $x^4 + \dfrac{1}{x^4} = 527$.

If $\Big(x - \dfrac{1}{x}\Big) = 4$, find the values of :

(i) $\Big(x^2 + \dfrac{1}{x^2}\Big)$

(ii) $\Big(x^4 + \dfrac{1}{x^4}\Big)$.

Answer

(i) Given,

$\Big(x - \dfrac{1}{x}\Big) = 4$

$\Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = x^2 + \Big(\dfrac{1}{x}\Big)^2 - 2 \times x \times \dfrac{1}{x} \\[1em] \Rightarrow (4)^2 = x^2 + \Big(\dfrac{1}{x}\Big)^2 - 2 \times x \times \dfrac{1}{x} \\[1em] \Rightarrow 16 = x^2 + \dfrac{1}{x^2} - 2 \\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} = 16 + 2 \\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} = 18$

Hence, $x^2 + \dfrac{1}{x^2} = 18$.

(ii) Given,

$\Big(x - \dfrac{1}{x}\Big) = 4$

From part (i),

$x^2 + \dfrac{1}{x^2} = 18$

$\Rightarrow \Big(x^2 + \dfrac{1}{x^2}\Big)^2 = (x^2)^2 + \Big(\dfrac{1}{x^2}\Big)^2 + 2 \times x^2 \times \dfrac{1}{x^2} \\[1em] \Rightarrow (18)^2 = (x^2)^2 + \Big(\dfrac{1}{x^2}\Big)^2 + 2 \times x^2 \times \dfrac{1}{x^2} \\[1em] \Rightarrow 324 = x^4 + \dfrac{1}{x^4} + 2 \\[1em] \Rightarrow x^4 + \dfrac{1}{x^4} = 324 - 2 \\[1em] \Rightarrow x^4 + \dfrac{1}{x^4} = 322.$

Hence, $x^4 + \dfrac{1}{x^4} = 322$.

If $x - 2 = \dfrac{1}{3x}$, find the values of :

(i) $\Big(x^2 + \dfrac{1}{9x^2}\Big)$

(ii) $\Big(x^4 + \dfrac{1}{81x^4}\Big)$.

Answer

(i) Given,

$\Rightarrow x - 2 = \dfrac{1}{3x} \\[1em] \Rightarrow x - \dfrac{1}{3x} = 2$

We know that,

$\Rightarrow \Big(x - \dfrac{1}{3x}\Big)^2 = x^2 + \Big(\dfrac{1}{3x}\Big)^2 - 2 \times x \times \dfrac{1}{3x} \\[1em] \Rightarrow (2)^2 = x^2 + \Big(\dfrac{1}{3x}\Big)^2 - 2 \times x \times \dfrac{1}{3x} \\[1em] \Rightarrow 4 = x^2 + \dfrac{1}{9x^2} - \dfrac{2}{3} \\[1em] \Rightarrow x^2 + \dfrac{1}{9x^2} = 4 + \dfrac{2}{3} \\[1em] \Rightarrow x^2 + \dfrac{1}{9x^2} = \dfrac{12 + 2}{3} \\[1em] \Rightarrow x^2 + \dfrac{1}{9x^2} = \dfrac{14}{3}$

Hence, $x^2 + \dfrac{1}{9x^2} = \dfrac{14}{3}$.

(ii) From part (i),

$x^2 + \dfrac{1}{9x^2} = \dfrac{14}{3}$

Using identity,

$\Rightarrow \Big(x^2 + \dfrac{1}{9x^2}\Big)^2 = (x^2)^2 + \Big(\dfrac{1}{9x^2}\Big)^2 + 2 \times x^2 \times \dfrac{1}{9x^2} \\[1em] \Rightarrow \Big(\dfrac{14}{3}\Big)^2 = (x^2)^2 + \Big(\dfrac{1}{9x^2}\Big)^2 + 2 \times x^2 \times \dfrac{1}{9x^2} \\[1em] \Rightarrow \dfrac{196}{9} = x^4 + \dfrac{1}{81x^4} + \dfrac{2}{9} \\[1em] \Rightarrow x^4 + \dfrac{1}{81x^4} = \dfrac{196}{9} - \dfrac{2}{9} \\[1em] \Rightarrow x^4 + \dfrac{1}{81x^4} = \dfrac{196 - 2}{9} \\[1em] \Rightarrow x^4 + \dfrac{1}{81x^4} = \dfrac{194}{9}$

Hence, $x^4 + \dfrac{1}{81x^4} = \dfrac{194}{9}$.

If $\Big(x + \dfrac{1}{x}\Big) = 6$, find the values of :

(i) $\Big(x - \dfrac{1}{x}\Big)$.

(ii) $\Big(x^2 - \dfrac{1}{x^2}\Big)$

Answer

(i) Given,

$\Big(x + \dfrac{1}{x}\Big) = 6$

We know that,

$\Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 - \Big(x - \dfrac{1}{x}\Big)^2 = 4 \\[1em] \Rightarrow (6)^2 - \Big(x - \dfrac{1}{x}\Big)^2 = 4 \\[1em] \Rightarrow 36 - 4 = \Big(x - \dfrac{1}{x}\Big)^2 \\[1em] \Rightarrow 32 = \Big(x - \dfrac{1}{x}\Big)^2 \\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big) = \sqrt{32} \\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big) = \pm 4\sqrt{2}.$

Hence, $\Big(x - \dfrac{1}{x}\Big) = \pm 4\sqrt{2}$.

(ii) Given,

$\Big(x + \dfrac{1}{x}\Big) = 6$

From part (i),

$\Rightarrow \Big(x - \dfrac{1}{x}\Big) = ±4\sqrt{2}$

We know that,

$\Rightarrow \Big(x^2 - \dfrac{1}{x^2}\Big) = \Big(x + \dfrac{1}{x}\Big)\Big(x - \dfrac{1}{x}\Big) \\[1em] \Rightarrow \Big(x^2 - \dfrac{1}{x^2}\Big) = 6 \times \pm 4\sqrt{2} \\[1em] \Rightarrow \Big(x^2 - \dfrac{1}{x^2}\Big) = \pm 24\sqrt{2}$

Hence, $x^2 - \dfrac{1}{x^2} = \pm 24\sqrt{2}$.

If $\Big(x - \dfrac{1}{x}\Big) = 8$, find the values of :

(i) $\Big(x + \dfrac{1}{x}\Big)$

(ii) $\Big(x^2 - \dfrac{1}{x^2}\Big)$

Answer

(i) Given,

$\Big(x - \dfrac{1}{x}\Big) = 8$

We know that,

$\Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 - \Big(x - \dfrac{1}{x}\Big)^2 = 4 \\[1em] \Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 - (8)^2 = 4 \\[1em] \Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 - 64 = 4 \\[1em] \Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 = 64 + 4 \\[1em] \Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 = 68 \\[1em] \Rightarrow \Big(x + \dfrac{1}{x}\Big) = \sqrt{68} \\[1em] \Rightarrow \Big(x + \dfrac{1}{x}\Big) = \pm 2\sqrt{17} \\[1em]$

Hence, $\Big(x + \dfrac{1}{x}\Big) = \pm 2\sqrt{17}$.

(ii) Given,

$\Big(x - \dfrac{1}{x}\Big) = 8$

From (i),

$\Big(x + \dfrac{1}{x}\Big) = \pm 2\sqrt{17}$

Using identity,

$\Rightarrow \Big(x^2 - \dfrac{1}{x^2}\Big) = \Big(x + \dfrac{1}{x}\Big)\Big(x - \dfrac{1}{x}\Big) \\[1em] \Rightarrow \Big(x^2 - \dfrac{1}{x^2}\Big) = (\pm 2\sqrt{17})\times 8 \\[1em] \Rightarrow \Big(x^2 - \dfrac{1}{x^2}\Big) = \pm 16\sqrt{17}$

Hence, $x^2 - \dfrac{1}{x^2} = \pm 16\sqrt{17}$.

If $\Big(x^2 + \dfrac{1}{x^2}\Big) = 7$, find the values of :

(i) $\Big(x + \dfrac{1}{x}\Big)$

(ii) $\Big(x - \dfrac{1}{x}\Big)$

(iii) $\Big(2x^2 - \dfrac{2}{x^2}\Big)$.

Answer

(i) Given,

$\Big(x^2 + \dfrac{1}{x^2}\Big) = 7$

Using identity,

$\Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 = x^2 + \dfrac{1}{x^2} + 2 \\[1em] \Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 = 7 + 2 \\[1em] \Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 = 9 \\[1em] \Rightarrow \Big(x + \dfrac{1}{x}\Big) = \pm \sqrt{9} \\[1em] \Rightarrow \Big(x + \dfrac{1}{x}\Big) = \pm 3.$

Hence, $\Big(x + \dfrac{1}{x}\Big) = \pm 3$.

(ii) Given,

$\Big(x^2 + \dfrac{1}{x^2}\Big) = 7$

Using identity,

$\Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = x^2 + \dfrac{1}{x^2} - 2 \\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = 7 - 2 \\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = 5 \\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big) = \pm \sqrt{5} \\[1em]$

Hence, $\Big(x - \dfrac{1}{x}\Big) = \pm \sqrt{5}$.

(iii) Given,

$\Big(x^2 + \dfrac{1}{x^2}\Big) = 7$

From part (i) and (ii),

$\Big(x + \dfrac{1}{x}\Big) = \pm 3 \text{ and } \Big(x - \dfrac{1}{x}\Big) = \pm \sqrt{5}$

Using identity,

$\Rightarrow \Big(x^2 - \dfrac{1}{x^2}\Big) = \Big(x + \dfrac{1}{x}\Big)\Big(x - \dfrac{1}{x}\Big) \\[1em] \Rightarrow \Big(x^2 - \dfrac{1}{x^2}\Big) = (\pm 3) \times (\pm \sqrt{5}) \\[1em] \Rightarrow \Big(x^2 - \dfrac{1}{x^2}\Big) = \pm 3\sqrt{5} \\[1em] \Rightarrow 2\Big(x^2 - \dfrac{1}{x^2}\Big) = 2 \times \pm 3\sqrt{5} \\[1em] \Rightarrow \Big(2x^2 - \dfrac{2}{x^2}\Big) = 2 \times (\pm 3\sqrt{5}) \\[1em] \Rightarrow \Big(2x^2 - \dfrac{2}{x^2}\Big) = \pm 6\sqrt{5}$

Hence, $\Big(2x^2 - \dfrac{2}{x^2}\Big) = \pm 6\sqrt{5}$.

If $\Big(x^2 + \dfrac{1}{25x^2}\Big) = 9\dfrac{2}{5}$, find the value of $\Big(x - \dfrac{1}{5x}\Big)$.

Answer

$\Rightarrow \Big(x - \dfrac{1}{5x}\Big)^2 = \Big[x^2 + \Big(\dfrac{1}{5x}\Big)^2 - 2 \times x \times \dfrac{1}{5x}\Big] \\[1em] \Rightarrow \Big(x - \dfrac{1}{5x}\Big)^2 = \Big[x^2 + \dfrac{1}{25x^2} - \dfrac{2}{5}\Big] \\[1em] \Rightarrow \Big(x - \dfrac{1}{5x}\Big)^2 = 9\dfrac{2}{5} - \dfrac{2}{5} \\[1em] \Rightarrow \Big(x - \dfrac{1}{5x}\Big)^2 = \dfrac{47}{5} - \dfrac{2}{5} \\[1em] \Rightarrow \Big(x - \dfrac{1}{5x}\Big)^2 = \dfrac{47 - 2}{5} \\[1em] \Rightarrow \Big(x - \dfrac{1}{5x}\Big)^2 = \dfrac{45}{5} \\[1em] \Rightarrow \Big(x - \dfrac{1}{5x}\Big)^2 = 9 \\[1em] \Rightarrow (x - \dfrac{1}{5x}\Big) = \sqrt{9} \\[1em] \Rightarrow \Big(x - \dfrac{1}{5x}\Big) = \pm 3 \\[1em]$

Hence, $\Big(x - \dfrac{1}{5x}\Big) = \pm 3$.

If $a^2 - 4a - 1 = 0$ and $a \neq 0$, find the values of:

(i) $\Big(a - \dfrac{1}{a}\Big)$

(ii) $\Big(a + \dfrac{1}{a}\Big)$

(iii) $\Big(a^2 - \dfrac{1}{a^2}\Big)$

(iv) $\Big(a^2 + \dfrac{1}{a^2}\Big)$

Answer

(i) Solving,

$\Rightarrow a^2 - 4a - 1 = 0 \\[1em] \Rightarrow a^2 - 4a = 1 \\[1em] \Rightarrow a(a - 4) = 1 \\[1em] \Rightarrow a - 4 = \dfrac{1}{a} \\[1em] \Rightarrow a - \dfrac{1}{a} = 4 \\[1em]$

Hence, $\Big(a - \dfrac{1}{a}\Big) = 4$

(ii) Using identity,

$\Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 - \Big(a - \dfrac{1}{a}\Big)^2 = 4 \\[1em] \Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 - (4)^2 = 4 \\[1em] \Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 - 16 = 4 \\[1em] \Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 = 4 + 16\\[1em] \Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 = 20 \\[1em] \Rightarrow a + \dfrac{1}{a} = \sqrt{20} \\[1em] \Rightarrow a + \dfrac{1}{a} = \pm 2\sqrt{5}.$

Hence, $a + \dfrac{1}{a} = \pm 2\sqrt{5}$

(iii) From part (i) and (ii),

$\Rightarrow a - \dfrac{1}{a} = 4 \\[1em] \Rightarrow a + \dfrac{1}{a} = \pm 2\sqrt{5}$

Case 1:

$\Rightarrow a + \dfrac{1}{a} = 2\sqrt{5}$

Using identity,

$\Rightarrow \Big(a + \dfrac{1}{a}\Big)\Big(a - \dfrac{1}{a}\Big) = \Big( a^2 - \dfrac{1}{a^2}\Big) \\[1em] \Rightarrow (2\sqrt{5}) \times (4) = \Big( a^2 - \dfrac{1}{a^2}\Big) \\[1em] \Rightarrow \Big( a^2 - \dfrac{1}{a^2}\Big) = 8\sqrt{5} \\[1em]$

Case 2:

$\Rightarrow a + \dfrac{1}{a} = -2\sqrt{5}$

Using identity,

$\Rightarrow \Big(a + \dfrac{1}{a}\Big)\Big(a - \dfrac{1}{a}\Big) = \Big( a^2 - \dfrac{1}{a^2}\Big) \\[1em] \Rightarrow (-2\sqrt{5}) \times (4) = \Big( a^2 - \dfrac{1}{a^2}\Big) \\[1em] \Rightarrow \Big( a^2 - \dfrac{1}{a^2}\Big) = -8\sqrt{5} \\[1em]$

Hence, $\Big( a^2 - \dfrac{1}{a^2}\Big) = \pm 8\sqrt{5}$.

(iv) From part (i) and (ii),

$\Rightarrow a - \dfrac{1}{a} = 4 \\[1em] \Rightarrow a + \dfrac{1}{a} = \pm 2\sqrt{5}$

Using identity,

$\Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 + \Big(a - \dfrac{1}{a}\Big)^2 = 2\Big( a^2 + \dfrac{1}{a^2}\Big) \\[1em] \Rightarrow (\pm 2\sqrt{5})^2 + (4)^2 = 2\Big( a^2 + \dfrac{1}{a^2}\Big) \\[1em] \Rightarrow 2\Big( a^2 + \dfrac{1}{a^2}\Big) = 20 + 16 \\[1em] \Rightarrow 2\Big( a^2 + \dfrac{1}{a^2}\Big) = 36 \\[1em] \Rightarrow \Big( a^2 + \dfrac{1}{a^2}\Big) = \dfrac{36}{2} \\[1em] \Rightarrow \Big( a^2 + \dfrac{1}{a^2}\Big) = 18 \\[1em]$