Expansions — Exercise 4(C)

The expansion of (x - 3y)(x + 5y) is :

1. x² + 2xy + 15y²

2. x² + 2xy - 15y²

3. x² - 2xy + 15y²

4. x² - 2xy - 15y²

Answer

Expansion of (x - a)(x + b) = x² - (a - b)x - ab

∴ Expansion of (x - 3y)(x + 5y) = x² - (3y - 5y)x - 3y × 5y

= x² - (-2y)x - 15y²

= x² + 2xy - 15y².

Hence, Option 2 is the correct option.

x² - (a + b)x + ab is the expansion of :

1. (x - b)(x - a)

2. (x - b)(x + a)

3. (x + b)(x - a)

4. (x + a)(x + b)

Answer

Given,

⇒ x² - (a + b)x + ab

⇒ x² - ax - bx + ab

⇒ x(x - a) - b(x - a)

⇒ (x - a)(x - b).

Hence, Option 1 is the correct option.

If a + b - c = 4 and a² + b² + c² = 14, the value of ab - bc - ca is :

1. 2

2. 1

3. 0.5

4. -0.5

Answer

By formula,

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

Substituting values we get :

⇒ 4² = 14 + 2(ab - bc - ca)

⇒ 16 = 14 + 2(ab - bc - ca)

⇒ 2(ab - bc - ca) = 16 - 14

⇒ 2(ab - bc - ca) = 2

⇒ ab - bc - ca = 1

Hence, Option 2 is the correct option.

$\Big(a - \dfrac{1}{2a}\Big)^2$ is equal to :

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

2. $\Big(a^2 + \dfrac{1}{4a^2} + 2\Big)$

3. $\Big(a^2 + \dfrac{1}{4a^2} - 1\Big)$

4. $\Big(a^2 - \dfrac{1}{4a^2} - 2\Big)$

Answer

Given,

⇒ $\Big(a - \dfrac{1}{2a}\Big)^2$

Expanding,

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

Hence, Option 3 is the correct option.

Expand (x + 8)(x + 10)

Answer

Given,

⇒ (x + 8)(x + 10)

Expanding,

⇒ x² + 10x + 8x + 80

⇒ x² + 18x + 80.

Hence, expansion of (x + 8)(x + 10) = x² + 18x + 80.

Expand (x + 8)(x - 10)

Answer

Given,

⇒ (x + 8)(x - 10)

Expanding,

⇒ x² - 10x + 8x - 80

⇒ x² - 2x - 80.

Hence, expansion of (x + 8)(x - 10) = x² - 2x - 80.

Expand (x - 8)(x + 10)

Answer

Given,

⇒ (x - 8)(x + 10)

Expanding,

⇒ x² + 10x - 8x - 80

⇒ x² + 2x - 80.

Hence, expansion of (x - 8)(x + 10) = x² + 2x - 80.

Expand (x - 8)(x - 10)

Answer

Given,

⇒ (x - 8)(x - 10)

Expanding,

⇒ x² - 10x - 8x + 80

⇒ x² - 18x + 80.

Hence, expansion of (x - 8)(x - 10) = x² - 18x + 80.

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

Answer

Given,

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

Expanding,

$\Rightarrow 2x \times 3x + 2x \times \dfrac{2}{x} - \dfrac{1}{x} \times 3x - \dfrac{1}{x} \times \dfrac{2}{x} \\[1em] \Rightarrow 6x^2 + 4 - 3 - \dfrac{2}{x^2} \\[1em] \Rightarrow 6x^2 + 1 - \dfrac{2}{x^2}.$

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

Expand $\Big(3a + \dfrac{2}{b}\Big)\Big(2a - \dfrac{3}{b}\Big)$

Answer

Given,

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

Expanding,

$\Rightarrow 3a \times 2a + 3a \times -\dfrac{3}{b} + \dfrac{2}{b} \times 2a + \dfrac{2}{b} \times -\dfrac{3}{b} \\[1em] \Rightarrow 6a^2 - \dfrac{9a}{b} + \dfrac{4a}{b} - \dfrac{6}{b^2} \\[1em] \Rightarrow 6a^2 - \dfrac{5a}{b} - \dfrac{6}{b^2}.$

Hence, $\Big(3a + \dfrac{2}{b}\Big)\Big(2a - \dfrac{3}{b}\Big) = 6a^2 - \dfrac{5a}{b} - \dfrac{6}{b^2}.$

Expand (x + y - z)²

Answer

Given,

⇒ (x + y - z)²

Expanding,

⇒ x² + y² + z² + 2xy - 2yz - 2zx

⇒ x² + y² + z² + 2(xy - yz - zx).

Hence, (x + y - z)² = x² + y² + z² + 2(xy - yz - zx).

Expand (x - 2y + 2)²

Answer

Given,

⇒ (x - 2y + 2)²

Expanding,

⇒ (x - 2y + 2)(x - 2y + 2)

⇒ x² - 2xy + 2x - 2xy + 4y² - 4y + 2x - 4y + 4

⇒ x² + 4y² + 4 - 4xy - 8y + 4x.

Hence, (x - 2y + 2)² = x² + 4y² + 4 - 4xy - 8y + 4x.

Expand (5a - 3b + c)²

Answer

Given,

⇒ (5a - 3b + c)²

Expanding,

⇒ (5a - 3b + c)(5a - 3b + c)

⇒ (5a)² - 15ab + 5ac - 15ab + 9b² - 3bc + 5ac - 3bc + c²

⇒ 25a² + 9b² + c² - 30ab - 6bc + 10ac.

Hence, (5a - 3b + c)² = 25a² + 9b² + c² - 30ab - 6bc + 10ac.

Expand (5x - 3y - 2)²

Answer

Given,

⇒ (5x - 3y - 2)²

Expanding,

⇒ (5x - 3y - 2)(5x - 3y - 2)

⇒ 25x² - 15xy - 10x - 15xy + 9y² + 6y - 10x + 6y + 4

⇒ 25x² + 9y² - 30xy - 20x + 12y + 4.

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

Expand $\Big(x - \dfrac{1}{x} + 5\Big)^2$

Answer

Given,

⇒ $\Big(x - \dfrac{1}{x} + 5\Big)^2$

Expanding,

$\Rightarrow \Big(x - \dfrac{1}{x} + 5\Big)\Big(x - \dfrac{1}{x} + 5\Big) \\[1em] \Rightarrow x^2 - 1 + 5x - 1 + \dfrac{1}{x^2} - \dfrac{5}{x} + 5x - \dfrac{5}{x} + 25 \\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} - \dfrac{10}{x} + 23 + 10x.$

Hence, $\Big(x - \dfrac{1}{x} + 5\Big)^2 = x^2 + \dfrac{1}{x^2} - \dfrac{10}{x} + 23 + 10x.$

If a + b + c = 12 and a² + b² + c² = 50; find ab + bc + ca.

Answer

By formula,

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

Substituting values we get :

⇒ 12² = 50 + 2(ab + bc + ca)

⇒ 144 = 50 + 2(ab + bc + ca)

⇒ 2(ab + bc + ca) = 144 - 50

⇒ 2(ab + bc + ca) = 94

⇒ ab + bc + ca = 47.

Hence, ab + bc + ca = 47.

If a² + b² + c² = 35 and ab + bc + ca = 23; find a + b + c.

Answer

By formula,

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

Substituting values we get :

⇒ (a + b + c)² = 35 + 2 × 23

⇒ (a + b + c)² = 35 + 46

⇒ (a + b + c)² = 81

⇒ (a + b + c) = $\sqrt{81} = \pm 9$.

Hence, (a + b + c) = $\pm 9$.

If a + b + c = p and ab + bc + ca = q; find a² + b² + c².

Answer

By formula,

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

Substituting values we get :

⇒ p² = a² + b² + c² + 2q

⇒ a² + b² + c² = p² - 2q.

Hence, a² + b² + c² = p² - 2q.

If a² + b² + c² = 50 and ab + bc + ca = 47, find a + b + c.

Answer

By formula,

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

Substituting values we get :

⇒ (a + b + c)² = 50 + 2 × 47

⇒ (a + b + c)² = 50 + 94

⇒ (a + b + c)² = 144

⇒ (a + b + c) = $\sqrt{144} = \pm 12$.

Hence, (a + b + c) = $\pm 12$.

If x + y - z = 4 and x² + y² + z² = 30, then find the value of xy - yz - zx.

Answer

By formula,

(x + y - z)² = x² + y² + z² + 2(xy - yz - zx)

Substituting values we get :

⇒ 4² = 30 + 2(xy - yz - zx)

⇒ 16 = 30 + 2(xy - yz - zx)

⇒ 2(xy - yz - zx) = 16 - 30

⇒ 2(xy - yz - zx) = -14

⇒ xy - yz - zx = -7.

Hence, xy - yz - zx = -7.

The longest road that can be placed in a rectangular box = 20 cm and the sum of its length breadth and height is 30 cm. Find the total surface area of the box.

Answer

Given, the longest road that can be placed in a rectangular box, d = 20 cm

The longest rod which can be kept inside a rectangular box will be equal to the diagonal of the box.

By formula,

⇒ Diagonal² = l² + b² + h²

⇒ 20² = l² + b² + h²

⇒ 400 = l² + b² + h² ..................(1)

Given,

The sum of its length breadth and height equals to 30 cm.

⇒ l + b + h = 30 cm.

Squaring both sides, we get :

⇒ (l + b + h)² = 30²

⇒ l² + b² + h² + 2(lb + bh + hl) = 900

From equation (1), we get

⇒ 400 + 2(lb + bh + hl) = 900

⇒ 2(lb + bh + hl) = 900 - 400

⇒ 2(lb + bh + hl) = 500

Hence, the total surface area of the box = 500 cm².