Expansions — Exercise 4(A)

For x = 9 and y = 4, the value of x² + 2xy + y² - 3 is :

1. 172

2. 100

3. 166

4. 169

Answer

Given,

⇒ x² + 2xy + y² - 3

⇒ (x + y)² - 3

Substituting values we get :

⇒ (9 + 4)² - 3

⇒ 13² - 3

⇒ 169 - 3

⇒ 166.

Hence, Option 3 is the correct option.

For x = 5 and y = 3, the value of x² + y² - 2xy + 7 is :

1. 11

2. 169

3. 71

4. 1

Answer

Given,

⇒ x² + y² - 2xy + 7

⇒ (x - y)² + 7

Substituting values we get :

⇒ (5 - 3)² + 7

⇒ 2² + 7

⇒ 4 + 7

⇒ 11.

Hence, Option 1 is the correct option.

If $x + \dfrac{1}{x} = 2$, the value of $x^2 + \dfrac{1}{x^2} + 5$ is :

1. -1

2. 2

3. 9

4. 7

Answer

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

Given,

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

$\therefore x^2 + \dfrac{1}{x^2} = 2^2 - 2 \\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} = 4 - 2 \\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} = 2$

Adding 5 on both sides,

$x^2 + \dfrac{1}{x^2} + 5 = 2 + 5 \\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} + 5 = 7$

Hence, Option 4 is the correct option.

If $x - \dfrac{1}{x} = 8$, the value of $x^2 + \dfrac{1}{x^2} - 8$ is :

1. 56

2. 58

3. 70

4. -4

Answer

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

Subtracting 8 on both sides,

$\Rightarrow x^2 + \dfrac{1}{x^2} - 8 = \Big(x - \dfrac{1}{x}\Big)^2 + 2 - 8 \\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} - 8 = 8^2 - 6 \\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} - 8 = 64 - 6 \\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} - 8 = 58.$

Hence, Option 2 is the correct option.

If $x^2 + \dfrac{1}{x^2} = 9$, the value of $x^4 + \dfrac{1}{x^4} + 5$ is :

1. 78

2. 86

3. 84

4. 81

Answer

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

Adding 5 on both sides,

$x^4 + \dfrac{1}{x^4} + 5 = \Big(x^2 + \dfrac{1}{x^2}\Big)^2 - 2 + 5\\[1em] \Rightarrow x^4 + \dfrac{1}{x^4} + 5 = 9^2 + 3 \\[1em] \Rightarrow x^4 + \dfrac{1}{x^4} + 5 = 84.$

Hence, Option 3 is the correct option.

If x² - 3x + 1 = 0, the value of

$x^2 + \dfrac{1}{x^2} + 1$ is :

1. 8

2. 10

3. 5

4. $\dfrac{10}{9}$

Answer

Given,

⇒ x² - 3x + 1 = 0

⇒ x² + 1 = 3x

Dividing above equation by x, we get :

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

Squaring both sides we get :

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

Hence, Option 1 is the correct option.

Evaluate $\Big(\dfrac{7}{8}x + \dfrac{4}{5}y\Big)^2$

Answer

Solving,

$\Rightarrow \Big(\dfrac{7}{8}x + \dfrac{4}{5}y\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{7}{8}x\Big)^2 + \Big(\dfrac{4}{5}y\Big)^2 + 2 \times \dfrac{7}{8}x \times \dfrac{4}{5}y \\[1em] \Rightarrow \dfrac{49}{64}x^2 + \dfrac{16}{25}y^2 + \dfrac{7}{5}xy.$

Hence, $\Big(\dfrac{7}{8}x + \dfrac{4}{5}y\Big)^2 = \dfrac{49}{64}x^2 + \dfrac{16}{25}y^2 + \dfrac{7}{5}xy.$

Evaluate $\Big(\dfrac{2x}{7} - \dfrac{7y}{4}\Big)^2$

Answer

Solving,

$\Rightarrow \Big(\dfrac{2x}{7} - \dfrac{7y}{4}\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{2x}{7}\Big)^2 + \Big(\dfrac{7y}{4}\Big)^2 - 2 \times \dfrac{2x}{7} \times \dfrac{7y}{4} \\[1em] \Rightarrow \dfrac{4x^2}{49} + \dfrac{49y^2}{16} - xy.$

Hence, $\Big(\dfrac{2x}{7} - \dfrac{7y}{4}\Big)^2 = \dfrac{4x^2}{49} + \dfrac{49y^2}{16} - xy.$

Evaluate $\Big(\dfrac{a}{2b} + \dfrac{2b}{a}\Big)^2 - \Big(\dfrac{a}{2b} - \dfrac{2b}{a}\Big)^2 - 4$

Answer

Solving,

$\Rightarrow \Big(\dfrac{a}{2b} + \dfrac{2b}{a}\Big)^2 - \Big(\dfrac{a}{2b} - \dfrac{2b}{a}\Big)^2 - 4 \\[1em] \Rightarrow \Big(\dfrac{a}{2b}\Big)^2 + \Big(\dfrac{2b}{a}\Big)^2 + 2 \times \dfrac{a}{2b} \times \dfrac{2b}{a} - \Big[\Big(\dfrac{a}{2b}\Big)^2 + \Big(\dfrac{2b}{a}\Big)^2 - 2 \times \dfrac{a}{2b} \times \dfrac{2b}{a}\Big] - 4 \\[1em] \Rightarrow \dfrac{a^2}{4b^2} + \dfrac{4b^2}{a^2} + 2 - \Big[\dfrac{a^2}{4b^2} + \dfrac{4b^2}{a^2} - 2\Big] - 4 \\[1em] \Rightarrow \dfrac{a^2}{4b^2} + \dfrac{4b^2}{a^2} + 2 - \dfrac{a^2}{4b^2} - \dfrac{4b^2}{a^2} + 2 - 4 \\[1em] \Rightarrow 4 - 4 \\[1em] \Rightarrow 0.$

Hence, $\Big(\dfrac{a}{2b} + \dfrac{2b}{a}\Big)^2 - \Big(\dfrac{a}{2b} - \dfrac{2b}{a}\Big)^2 - 4$ = 0.

If x + y = $\dfrac{7}{2}$ and xy = $\dfrac{5}{2}$; find:

(i) x - y

(ii) x² - y²

Answer

(i) By formula,

(x - y)² = (x + y)² - 4xy

$\Rightarrow (x - y)^2 = \Big(\dfrac{7}{2}\Big)^2 - 4 \times \dfrac{5}{2} \\[1em] \Rightarrow (x - y)^2 = \dfrac{49}{4} - 10 \\[1em] \Rightarrow (x - y)^2 = \dfrac{49 - 40}{4} \\[1em] \Rightarrow (x - y)^2 = \dfrac{9}{4} \\[1em] \Rightarrow (x - y) = \sqrt{\dfrac{9}{4}} \\[1em] \Rightarrow x - y = \pm\dfrac{3}{2}.$

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

(ii) Solving,

$\Rightarrow x^2 - y^2 \\[1em] \Rightarrow (x - y)(x + y) \\[1em] \Rightarrow \pm \dfrac{3}{2} \times \dfrac{7}{2} \\[1em] \Rightarrow \pm \dfrac{21}{4}.$

Hence, x² - y² = $\pm \dfrac{21}{4}$.

If a - b = 0.9 and ab = 0.36; find :

(i) a + b

(ii) a² - b²

Answer

(i) By formula,

⇒ (a + b)² = (a - b)² + 4ab

⇒ (a + b)² = (0.9)² + 4 × 0.36

⇒ (a + b)² = 0.81 + 1.44

⇒ (a + b)² = 2.25

⇒ a + b = $\sqrt{2.25} = \pm1.5$

Hence, a + b = $\pm 1.5$

(ii) Solving,

⇒ a² - b²

⇒ (a - b)(a + b)

⇒ 0.9 × $\pm 1.5$

⇒ $\pm 1.35$

Hence, a² - b² = $\pm 1.35$

If a - b = 4 and a + b = 6; find :

(i) a² + b²

(ii) ab

Answer

(i) We know that,

(a - b)² = a² + b² - 2ab .............(1)

(a + b)² = a² + b² + 2ab ..............(2)

Adding equation (1) and (2), we get :

(a - b)² + (a + b)² = 2(a² + b²)

(a² + b²) = $\dfrac{(a - b)^2 + (a + b)^2}{2}$

Substituting values we get :

$\Rightarrow a^2 + b^2 = \dfrac{4^2 + 6^2}{2} \\[1em] \Rightarrow a^2 + b^2 = \dfrac{16 + 36}{2} \\[1em] \Rightarrow a^2 + b^2 = \dfrac{52}{2} \\[1em] \Rightarrow a^2 + b^2 = 26.$

Hence, a² + b² = 26.

(ii) We know that,

(a - b)² = a² + b² - 2ab .............(1)

(a + b)² = a² + b² + 2ab ..............(2)

Subtracting equation (1) from (2), we get :

⇒ (a + b)² - (a - b)² = a² + b² + 2ab - (a² + b² - 2ab)

⇒ (a + b)² - (a - b)² = 4ab

⇒ ab = $\dfrac{(a + b)^2 - (a - b)^2}{4}$

Substituting values we get :

$\Rightarrow ab = \dfrac{6^2 - 4^2}{4} \\[1em] = \dfrac{36 - 16}{4} \\[1em] =\dfrac{20}{4} \\[1em] = 5.$

Hence, ab = 5.

If $a + \dfrac{1}{a} = 6$ and a ≠ 0; find :

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

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

Answer

(i) By formula,

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

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

(ii) Solving,

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

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

If $a - \dfrac{1}{a} = 8$ and a ≠ 0; find :

(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 :

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

Hence, $a + \dfrac{1}{a} = \pm 2\sqrt{17}.$

(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 :

$\Rightarrow a^2 - \dfrac{1}{a^2} = \pm 2\sqrt{17} \times 8 \\[1em] = \pm 16\sqrt{17}.$

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

If a² - 3a + 1 = 0 and a ≠ 0; find :

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

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

Answer

(i) Given,

⇒ a² - 3a + 1 = 0

⇒ a² + 1 = 3a

Dividing above equation by a, we get :

$\Rightarrow \dfrac{a^2 + 1}{a} = \dfrac{3a}{a}$

$\Rightarrow a + \dfrac{1}{a} = 3$

Hence, $a + \dfrac{1}{a} = 3$.

(ii) By formula,

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

Hence, $a^2 + \dfrac{1}{a^2} = 7$.

If a² - 5a - 1 = 0 and a ≠ 0, find :

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

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

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

Answer

(i) Given,

⇒ a² - 5a - 1 = 0

⇒ a² - 1 = 5a

Dividing above equation by a, we get :

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

$\Rightarrow a - \dfrac{1}{a} = 5$

Hence, $a - \dfrac{1}{a} = 5$.

(ii) By formula,

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

Hence, $a + \dfrac{1}{a} = \pm \sqrt{29}$.

(iii) By formula,

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

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

If 3x + 4y = 16 and xy = 4; find the value of 9x² + 16y².

Answer

Solving,

⇒ (3x + 4y)² = (3x)² + (4y)² + 2 × 3x × 4y

⇒ (3x + 4y)² = 9x² + 16y² + 24xy

Substituting values we get :

⇒ 16² = 9x² + 16y² + 24 × 4

⇒ 256 = 9x² + 16y² + 96

⇒ 9x² + 16y² = 256 - 96 = 160.

Hence, 9x² + 16y² = 160.

The difference between two positive numbers is 5 and the sum of their squares is 73. Find the product of these numbers.

Answer

Let two positive numbers be x and y.

Given,

Difference between two positive numbers is 5.

∴ x - y = 5

⇒ x = y + 5 ...........(1)

Given,

Sum of squares of numbers is 73.

∴ x² + y² = 73

Substituting value of x from equation (1) in above equation, we get :

⇒ (y + 5)² + y² = 73

⇒ y² + 5² + 2 × y × 5 + y² = 73

⇒ 2y² + 25 + 10y = 73

⇒ 2y² + 10y + 25 - 73 = 0

⇒ 2y² + 10y - 48 = 0

⇒ 2(y² + 5y - 24) = 0

⇒ y² + 5y - 24 = 0

⇒ y² + 8y - 3y - 24 = 0

⇒ y(y + 8) - 3(y + 8) = 0

⇒ (y - 3)(y + 8) = 0

⇒ y - 3 = 0 or y + 8 = 0

⇒ y = 3 or y = -8.

If y = 3, x = y + 5 = 3 + 5 = 8, xy = 24,

If y = -8, x = y + 5 = -8 + 5 = -3, xy = 24.

Hence, product of numbers = 24.