Expansions — Exercise 4(D)

If $a + \dfrac{1}{a} = 2.5 \text{ and } a - \dfrac{1}{a} = 1.5$, the value of $\Big(a + \dfrac{1}{a}\Big)^2 - \Big(a - \dfrac{1}{a}\Big)^2$ is :

1. 4

2. 2

3. 1

4. 8.5

Answer

Solving,

$\Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 - \Big(a - \dfrac{1}{a}\Big)^2 \\[1em] \Rightarrow (2.5)^2 - (1.5)^2 \\[1em] \Rightarrow 6.25 - 2.25 \\[1em] \Rightarrow 4.$

Hence, Option 1 is the correct option.

In (3x + 2)(x - 4), the coefficient of x is :

1. -14

2. 8

3. -10

4. 3

Answer

Given,

⇒ (3x + 2)(x - 4)

⇒ 3x² - 12x + 2x - 8

⇒ 3x² - 10x - 8.

∴ Coefficient of x is -10.

Hence, Option 3 is the correct option.

(x + y - z)(x - y + z) is equal to :

1. x² - y² - z² - 2yz

2. x² - y² - z² + 2yz

3. x² - y² + z² - 2yz

4. x² - y² + z² + 2yz

Answer

Given,

⇒ (x + y - z)(x - y + z)

Expanding,

⇒ x² - xy + xz + xy - y² + yz - zx + yz - z²

⇒ x² - y² - z² + 2yz.

Hence, Option 2 is the correct option.

If (3x - 4y)² = 9x² + axy + 16y²; the value of a is :

1. -24

2. 24

3. -12

4. 12

Answer

Solving,

⇒ (3x - 4y)² = 9x² + axy + 16y²

⇒ 9x² + 16y² - 24xy = 9x² + axy + 16y²

From above equation,

axy = -24xy

a = $-\dfrac{24xy}{xy}$ = -24.

Hence, Option 1 is the correct option.

If x + 2y + 3z = 0 and x³ + 4y³ + 9z³ = 18xyz; evaluate :

$\dfrac{(x + 2y)^2}{xy} + \dfrac{(2y + 3z)^2}{yz} + \dfrac{(3z + x)^2}{zx}$

Answer

Given,

x + 2y + 3z = 0

∴ x + 2y = -3z, 2y + 3z = -x and 3z + x = -2y

Solving,

$\Rightarrow \dfrac{(x + 2y)^2}{xy} + \dfrac{(2y + 3z)^2}{yz} + \dfrac{(3z + x)^2}{zx} \\[1em] \Rightarrow \dfrac{(-3z)^2}{xy} + \dfrac{(-x)^2}{yz} + \dfrac{(-2y)^2}{zx}\\[1em] \Rightarrow \dfrac{9z^2}{xy} + \dfrac{x^2}{yz} + \dfrac{4y^2}{zx} \\[1em] \Rightarrow \dfrac{9z^3 + x^3 + 4y^3}{xyz} \\[1em] \Rightarrow \dfrac{x^3 + 4y^3 + 9z^3}{xyz} \\[1em] \Rightarrow \dfrac{18xyz}{xyz} \\[1em] \Rightarrow 18.$

Hence, $\dfrac{(x + 2y)^2}{xy} + \dfrac{(2y + 3z)^2}{yz} + \dfrac{(3z + x)^2}{zx}$ = 18.

If $a + \dfrac{1}{a} = m$ and a ≠ 0; find in terms of 'm'; the value of :

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

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

Answer

(i) By formula,

$\Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 - \Big(a - \dfrac{1}{a}\Big)^2 = 4$

Substituting values we get :

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

Hence, $a - \dfrac{1}{a} = \pm \sqrt{m^2 - 4}.$

(ii) By formula,

$\Rightarrow a^2 - \dfrac{1}{a^2} = \Big(a - \dfrac{1}{a}\Big)\Big(a + \dfrac{1}{a}\Big)$

Substituting values we get :

$\Rightarrow \Rightarrow a^2 - \dfrac{1}{a^2} = \pm \sqrt{m^2 - 4} \times m \\[1em] = \pm m\sqrt{m^2 - 4}.$

Hence, $a^2 - \dfrac{1}{a^2} = \pm m\sqrt{m^2 - 4}.$

In the expansion of (2x² - 8)(x - 4)²; find the value of :

(i) coefficient of x³

(ii) coefficient of x²

(iii) constant term.

Answer

Given,

⇒ (2x² - 8)(x - 4)²

Expanding,

⇒ (2x² - 8)(x² + 16 - 8x)

⇒ 2x⁴ + 32x² - 16x³ - 8x² - 128 + 64x

⇒ 2x⁴ - 16x³ + 24x² + 64x - 128.

(i) Hence, coefficient of x³ is -16.

(ii) Hence, coefficient of x² is 24.

(iii) Hence, constant term = -128.

If x > 0 and $x^2 + \dfrac{1}{9x^2} = \dfrac{25}{36}$, find : $x^3 + \dfrac{1}{27x^3}$.

Answer

Given,

$\Rightarrow x^2 + \dfrac{1}{9x^2} = \dfrac{25}{36} \\[1em] \Rightarrow \Big(x + \dfrac{1}{3x}\Big)^2 - \dfrac{2}{3} = \dfrac{25}{36} \\[1em] \Rightarrow \Big(x + \dfrac{1}{3x}\Big)^2 = \dfrac{25}{36} + \dfrac{2}{3} \\[1em] \Rightarrow \Big(x + \dfrac{1}{3x}\Big)^2 = \dfrac{25 + 24}{36} \\[1em] \Rightarrow \Big(x + \dfrac{1}{3x}\Big)^2 = \dfrac{49}{36}\\[1em] \Rightarrow x + \dfrac{1}{3x} = \sqrt{\dfrac{49}{36}} \\[1em] \Rightarrow x + \dfrac{1}{3x} = \pm \dfrac{7}{6}.$

Since, x is > 0,

∴ $x + \dfrac{1}{3x} = \dfrac{7}{6}$

By formula,

$\Rightarrow \Big(x^3 + \dfrac{1}{27x^3}\Big) = \Big(x + \dfrac{1}{3x}\Big)^3 - \Big(x + \dfrac{1}{3x}\Big)$ .......(1)

Substituting $x + \dfrac{1}{3x} = \dfrac{7}{6}$ in equation (1), we get :

$\Rightarrow \Big(x^3 + \dfrac{1}{27x^3}\Big) = \Big(\dfrac{7}{6}\Big)^3 - \Big(\dfrac{7}{6}\Big) \\[1em] = \dfrac{343}{216} - \dfrac{7}{6} \\[1em] = \dfrac{343 - 252}{216} \\[1em] = \dfrac{91}{216}.$

Hence, $x^3 + \dfrac{1}{27x^3} = \dfrac{91}{216}$.

If 2(x² + 1) = 5x, find :

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

(ii) $x^3 - \dfrac{1}{x^3}$

Answer

(i) Given,

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

By formula,

$\Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 - \Big(x - \dfrac{1}{x}\Big)^2 = 4 \\[1em] \Rightarrow \Big(\dfrac{5}{2}\Big)^2 - \Big(x - \dfrac{1}{x}\Big)^2 = 4 \\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = \dfrac{25}{4} - 4\\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = \dfrac{25 - 16}{4} \\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = \dfrac{9}{4} \\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big) = \sqrt{\dfrac{9}{4}} \\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big) = \pm \dfrac{3}{2}.$

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

(ii) By formula,

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

Substituting $\Big(x - \dfrac{1}{x}\Big) = \dfrac{3}{2}$

$\Rightarrow \Big(x^3 - \dfrac{1}{x^3}\Big) = \Big(\dfrac{3}{2}\Big)^3 + 3 \times \dfrac{3}{2} \\[1em] = \dfrac{27}{8} + \dfrac{9}{2} \\[1em] = \dfrac{27 + 36}{8} \\[1em] = \dfrac{63}{8}.$

Substituting $\Big(x - \dfrac{1}{x}\Big) = -\dfrac{3}{2}$

$\Rightarrow \Big(x^3 - \dfrac{1}{x^3}\Big) = \Big(-\dfrac{3}{2}\Big)^3 + 3 \times -\dfrac{3}{2} \\[1em] = -\dfrac{27}{8} - \dfrac{9}{2} \\[1em] = \dfrac{-27 - 36}{8} \\[1em] = \dfrac{-63}{8}.$

Hence, $\Big(x^3 - \dfrac{1}{x^3}\Big) = \pm \dfrac{63}{8}$.

If a² + b² = 34 and ab = 12; find :

(i) 3(a + b)² + 5(a - b)²

(ii) 7(a - b)² - 2(a + b)²

Answer

(i) Expanding,

⇒ 3(a + b)² + 5(a - b)²

⇒ 3(a² + b² + 2ab) + 5(a² + b² - 2ab)

⇒ 3(34 + 2 × 12) + 5(34 - 2 × 12)

⇒ 3(34 + 24) + 5(34 - 24)

⇒ 3 × 58 + 5 × 10

⇒ 174 + 50

⇒ 224.

Hence, 3(a + b)² + 5(a - b)² = 224.

(ii) Expanding,

⇒ 7(a - b)² - 2(a + b)²

⇒ 7(a² + b² - 2ab) - 2(a² + b² + 2ab)

⇒ 7(34 - 2 × 12) - 2(34 + 2 × 12)

⇒ 7(34 - 24) - 2(34 + 24)

⇒ 7 × 10 - 2 × 58

⇒ 70 - 116

⇒ -46.

Hence, 7(a - b)² - 2(a + b)² = -46.

If 3x - $\dfrac{4}{x}$ = 4 and x ≠ 0; find : $27x^3 - \dfrac{64}{x^3}$.

Answer

Given,

$\Rightarrow 3x - \dfrac{4}{x} = 4$

Cubing both sides we get :

$\Rightarrow \Big(3x - \dfrac{4}{x}\Big)^3 = 4^3 \\[1em] \Rightarrow (3x)^3 - \Big(\dfrac{4}{x}\Big)^3 - 3 \times 3x \times \dfrac{4}{x} \times \Big(3x - \dfrac{4}{x}\Big) = 64 \\[1em] \Rightarrow (3x)^3 - \Big(\dfrac{4}{x}\Big)^3 - 36 \times 4 = 64 \\[1em] \Rightarrow 27x^3 - \dfrac{64}{x^3} - 144 = 64 \\[1em] \Rightarrow 27x^3 - \dfrac{64}{x^3} = 64 + 144 \\[1em] \Rightarrow 27x^3 - \dfrac{64}{x^3} = 208.$

Hence, $27x^3 - \dfrac{64}{x^3} = 208.$

If $x^2 + \dfrac{1}{x^2}$ = 7 and x≠ 0; find the value of :

7x³ + 8x - $\dfrac{7}{x^3} - \dfrac{8}{x}$.

Answer

By formula,

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

By formula,

$\Rightarrow \Big(x - \dfrac{1}{x}\Big)^3 = x^3 - \dfrac{1}{x^3} - 3\Big(x - \dfrac{1}{x}\Big)$

Substituting $x - \dfrac{1}{x} = \sqrt{5}$ in above equation, we get :

$\Rightarrow (\sqrt{5})^3 = x^3 - \dfrac{1}{x^3} - 3 \times \sqrt{5} \\[1em] \Rightarrow 5\sqrt{5} = x^3 - \dfrac{1}{x^3} - 3\sqrt{5} \\[1em] \Rightarrow x^3 - \dfrac{1}{x^3} = 8\sqrt{5}.$

Substituting $x - \dfrac{1}{x} = -\sqrt{5}$ in above equation, we get :

$\Rightarrow (-\sqrt{5})^3 = x^3 - \dfrac{1}{x^3} - 3 \times -\sqrt{5} \\[1em] \Rightarrow -5\sqrt{5} = x^3 - \dfrac{1}{x^3} + 3\sqrt{5} \\[1em] \Rightarrow x^3 - \dfrac{1}{x^3} = -8\sqrt{5}.$

Solving given equation,

$\Rightarrow 7x^3 + 8x - \dfrac{7}{x^3} - \dfrac{8}{x} \\[1em] \Rightarrow 7x^3 - \dfrac{7}{x^3} + 8x - \dfrac{8}{x} \\[1em] \Rightarrow 7\Big(x^3 - \dfrac{1}{x^3}\Big) + 8\Big(x - \dfrac{1}{x}\Big)$

Substituting $x^3 - \dfrac{1}{x^3} = 8\sqrt{5}\text{ and } x - \dfrac{1}{x} = \sqrt{5}$, we get :

$\Rightarrow 7 \times 8\sqrt{5} + 8 \times \sqrt{5} \\[1em] \Rightarrow 56\sqrt{5} + 8\sqrt{5} \\[1em] \Rightarrow 64\sqrt{5}.$

Substituting $x^3 - \dfrac{1}{x^3} = -8\sqrt{5}\text{ and } x - \dfrac{1}{x} = -\sqrt{5}$, we get :

$\Rightarrow 7 \times -8\sqrt{5} + 8 \times -\sqrt{5} \\[1em] \Rightarrow -56\sqrt{5} - 8\sqrt{5} \\[1em] \Rightarrow -64\sqrt{5}.$

Hence, 7x³ + 8x - $\dfrac{7}{x^3} - \dfrac{8}{x} = \pm 64\sqrt{5}$.

If $x = \dfrac{1}{x - 5}$ and x≠ 5, find : $x^2 - \dfrac{1}{x^2}$.

Answer

Given,

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

By formula,

$\Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 - \Big(x - \dfrac{1}{x}\Big)^2 = 4 \\[1em] \Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 - 5^2 = 4 \\[1em] \Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 - 25 = 4 \\[1em] \Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 = 25 + 4 \\[1em] \Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 = 29 \\[1em] \Rightarrow x + \dfrac{1}{x} = \pm \sqrt{29}.$

By formula,

$\Rightarrow x^2 - \dfrac{1}{x^2} = \Big(x - \dfrac{1}{x}\Big)\Big(x + \dfrac{1}{x}\Big)\\[1em] = 5 \times \pm \sqrt{29} \\[1em] = \pm 5\sqrt{29}.$

Hence, $x^2 - \dfrac{1}{x^2} = \pm 5\sqrt{29}$.

If x = $\dfrac{1}{5 - x}$ and x ≠ 5, find : $x^3 + \dfrac{1}{x^3}$.

Answer

Given,

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

By formula,

$\Rightarrow \Big(x + \dfrac{1}{x}\Big)^3 = x^3 + \dfrac{1}{x^3} + 3\Big(x + \dfrac{1}{x}\Big) \\[1em] \Rightarrow 5^3 = x^3 + \dfrac{1}{x^3} + 3 \times 5 \\[1em] \Rightarrow 125 = x^3 + \dfrac{1}{x^3} + 15 \\[1em] \Rightarrow x^3 + \dfrac{1}{x^3} = 125 - 15 = 110.$

Hence, $x^3 + \dfrac{1}{x^3} = 110.$

If 3a + 5b + 4c = 0, show that :

27a³ + 125b³ + 64c³ = 180abc.

Answer

By property,

If x + y + z = 0, then :

x³ + y³ + z³ = 3xyz ........(1)

Comparing,

3a + 5b + 4c = 0 with x + y + z = 0, we get :

x = 3a, y = 5b and z = 4c.

Substituting values in equation (1), we get :

⇒ (3a)³ + (5b)³ + (4c)³ = 3 × 3a × 5b × 4c

⇒ 27a³ + 125b³ + 64c³ = 180abc.

Hence, proved that 27a³ + 125b³ + 64c³ = 180abc.

The sum of two Whole numbers is 7 and the sum of their cubes is 133, find the sum of their squares.

Answer

Let two numbers be x and y.

Given,

The sum of two numbers is 7 and the sum of their cubes is 133.

x + y = 7 and x³ + y³ = 133

By formula,

⇒ x³ + y³ = (x + y)³ - 3xy(x + y)

⇒ 133 = 7³ - 3xy × 7

⇒ 133 = 343 - 21xy

⇒ 21xy = 343 - 133

⇒ 21xy = 210

⇒ xy = 10.

By formula,

⇒ (x + y)² = x² + y² + 2xy

⇒ 7² = x² + y² + 2 × 10

⇒ 49 = x² + y² + 20

⇒ x² + y² = 49 - 20

⇒ x² + y² = 29.

Hence, sum of squares of numbers = 29.

In each of the following, find the value of 'a' :

(i) 4x² + ax + 9 = (2x + 3)²

(ii) 4x² + ax + 9 = (2x - 3)²

(iii) 9x² + (7a - 5)x + 25 = (3x + 5)²

Answer

(i) Given,

⇒ 4x² + ax + 9 = (2x + 3)²

⇒ 4x² + ax + 9 = (2x)² + 3² + 2 × 2x × 3

⇒ 4x² + ax + 9 = 4x² + 9 + 12x

⇒ ax = 12x

⇒ a = 12.

Hence, a = 12.

(ii) Given,

⇒ 4x² + ax + 9 = (2x - 3)²

⇒ 4x² + ax + 9 = (2x)² + 3² - 2 × 2x × 3

⇒ 4x² + ax + 9 = 4x² + 9 - 12x

⇒ ax = -12x

⇒ a = -12.

Hence, a = -12.

(iii) Given,

⇒ 9x² + (7a - 5)x + 25 = (3x + 5)²

⇒ 9x² + (7a - 5)x + 25 = (3x)² + 5² + 2 × 3x × 5

⇒ 9x² + (7a - 5)x + 25 = 9x² + 30x + 25

⇒ 7a - 5 = 30

⇒ 7a = 35

⇒ a = $\dfrac{35}{7}$ = 5.

Hence, a = 5.

If $\dfrac{x^2 + 1}{x} = 3\dfrac{1}{3}$ and x > 1; find :

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

(ii) $x^3 - \dfrac{1}{x^3}$

Answer

(i) Given,

$\Rightarrow \dfrac{x^2 + 1}{x} = 3\dfrac{1}{3} \\[1em] \Rightarrow x + \dfrac{1}{x} = \dfrac{10}{3}.$

By formula,

$\Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 - \Big(x - \dfrac{1}{x}\Big)^2 = 4 \\[1em] \Rightarrow \Big(\dfrac{10}{3}\Big)^2 - \Big(x - \dfrac{1}{x}\Big)^2 = 4 \\[1em] \Rightarrow \dfrac{100}{9} - \Big(x - \dfrac{1}{x}\Big)^2 = 4 \\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = \dfrac{100}{9} - 4 \\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = \dfrac{100 - 36}{9} \\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = \dfrac{64}{9} \\[1em] \Rightarrow x - \dfrac{1}{x} = \sqrt{\dfrac{64}{9}} \\[1em] \Rightarrow x - \dfrac{1}{x} = \dfrac{8}{3} = 2\dfrac{2}{3}.$

Hence, $x - \dfrac{1}{x} = 2\dfrac{2}{3}$.

(ii) By formula,

$\Rightarrow x^3 - \dfrac{1}{x^3} = \Big(x - \dfrac{1}{x}\Big)^3 + 3\Big(x - \dfrac{1}{x}\Big) \\[1em] = \Big( \dfrac{8}{3}\Big)^3 + 3 \times \dfrac{8}{3} \\[1em] = \dfrac{512}{27} + \dfrac{24}{3} \\[1em] = \dfrac{512 + 216}{27} \\[1em] = \dfrac{728}{27} \\[1em] = 26\dfrac{26}{27}.$

Hence, $x^3 - \dfrac{1}{x^3} = 26\dfrac{26}{27}$.

The difference between two positive numbers is 4 and the difference between their cubes is 316. Find :

(i) their product

(ii) the sum of their squares.

Answer

Let two positive numbers be x and y with x > y.

Given,

The difference between two positive numbers is 4 and the difference between their cubes is 316.

x - y = 4 and x³ - y³ = 316.

(i) Given,

⇒ x - y = 4

Cubing both sides we get :

⇒ (x - y)³ = 4³

⇒ x³ - y³ - 3xy(x - y) = 64

⇒ 316 - 3xy × 4 = 64

⇒ 316 - 12xy = 64

⇒ 12xy = 316 - 64

⇒ 12xy = 252

⇒ xy = $\dfrac{252}{12}$ = 21.

Hence, product of numbers = 21.

(ii) Given,

x - y = 4

Squaring both sides we get :

⇒ (x - y)² = 4²

⇒ x² + y² - 2xy = 16

⇒ x² + y² - 2 × 21 = 16

⇒ x² + y² - 42 = 16

⇒ x² + y² = 16 + 42

⇒ x² + y² = 58.

Hence, sum of squares = 58.