myhelper: RD Sharma solution class 8 chapter 6 Algebraic Expressions and Identity Exercise 6.6

Exercise 6.6

Page-6.43

Question 1:

Write the following squares of binomials as trinomials:
(i) (x + 2)²
(ii) (8a + 3b)²
(iii) (2m + 1)²
(iv) ${(9 a + \frac{1}{6})}^{2}$
(v) ${(x + \frac{x^{2}}{2})}^{2}$
(vi) $(\frac{x}{4} - \frac{y}{3})$
(vii) ${(3 x - \frac{1}{3 x})}^{2}$
(viii) ${(\frac{x}{y} - \frac{y}{x})}^{2}$
(ix) ${(\frac{3 a}{2} - \frac{5 b}{4})}^{2}$
(x) (a²b − bc²)²
(xi) ${(\frac{2 a}{3 b} + \frac{2 b}{3 a})}^{2}$
(xii) (x² − ay)²

Answer 1:

We will use the identities ${(a + b)}^{2} = a^{2} + 2 a b + b^{2} and {(a - b)}^{2} = a^{2} - 2 a b + b^{2}$ to convert the squares of binomials as trinomials.

$(i) {(x + 2)}^{2} = x^{2} + 2 \times x \times 2 + b^{2} = x^{2} + 4 x + b^{2}$

$(ii) {(8 a + 3 b)}^{2} = {(8 a)}^{2} + 2 (8 a) (3 b) + {(6 b)}^{2} = 64 a^{2} + 48 a b + 36 b^{2}$

^{$(iii) {(2 m + 1)}^{2} = {(2 m)}^{2} + 2 (2 m) (1) + 1^{2} = 4 m^{2} + 4 m + 1$

$(iv) {(9 a + \frac{1}{6})}^{2} = {(9 a)}^{2} + 2 (9 a) (\frac{1}{6}) + {(\frac{1}{6})}^{2} = 81 a^{2} + 3 a + \frac{1}{36}$

$(v) {(x + \frac{x^{2}}{2})}^{2} = x^{2} + 2 x (\frac{x^{2}}{2}) + {(\frac{x^{2}}{2})}^{2} = x^{2} + x^{3} + \frac{x^{4}}{4}$

$(vi) {(\frac{x}{4} - \frac{y}{3})}^{2} = {(\frac{x}{4})}^{2} - 2 (\frac{x}{4}) (\frac{y}{3}) + {(\frac{y}{3})}^{2} = \frac{x^{2}}{16} - \frac{1}{6} x y + \frac{y^{2}}{9}$

$(vii) {(3 x - \frac{1}{3 x})}^{2} = {(3 x)}^{2} - 2 (3 x) (\frac{1}{3 x}) + {(\frac{1}{3 x})}^{2} = 9 x^{2} - 2 + \frac{1}{9 x^{2}}$

$(viii) {(\frac{x}{y} - \frac{y}{x})}^{2} = {(\frac{x}{y})}^{2} - 2 (\frac{x}{y}) (\frac{y}{x}) + {(\frac{y}{x})}^{2} = \frac{x^{2}}{y^{2}} - 2 + \frac{y^{2}}{x^{2}}$

$(ix) {(\frac{3 a}{2} - \frac{5 b}{4})}^{2} = {(\frac{3 a}{2})}^{2} - 2 (\frac{3 a}{2}) (\frac{5 b}{4}) + {(\frac{5 b}{4})}^{2} = \frac{9 a^{2}}{4} - \frac{15 a b}{4} + \frac{25 b^{2}}{16}$

$(x) {(a^{2} b - b c^{2})}^{2} = {(a^{2} b)}^{2} - 2 (a^{2} b) (b c^{2}) + {(b c^{2})}^{2} = a^{4} b^{2} - 2 a^{2} b^{2} c^{2} + b^{2} c^{4}$}

$(xi) {(\frac{2 a}{3 b} + \frac{2 b}{3 a})}^{2} = {(\frac{2 a}{3 b})}^{2} + 2 (\frac{2 a}{3 b}) (\frac{2 b}{3 a}) + {(\frac{2 b}{3 a})}^{2} = \frac{4 a^{2}}{9 b^{2}} + \frac{8}{9} + \frac{4 b^{2}}{9 a^{2}}$

$(xii) {(x^{2} - a y)}^{2} = {(x^{2})}^{2} - 2 x^{2} (a y) + {(a y)}^{2} = x^{4} - 2 x^{2} a y + a^{2} y^{2}$

Question 2:

Find the product of the following binomials:
(i) (2x + y)(2x + y)
(ii) (a + 2b)(a − 2b)
(iii) (a² + bc)(a²− bc)
(iv) $(\frac{4 x}{5} - \frac{3 y}{4}) (\frac{4 x}{5} + \frac{3 y}{4})$
(v) $(2 x + \frac{3}{y}) (2 x - \frac{3}{y})$
(vi) (2a³ + b³)(2a³ − b³)
(vii) $(x^{4} + \frac{2}{x^{2}}) (x^{4} - \frac{2}{x^{2}})$
(viii) $(x^{3} + \frac{1}{x^{3}}) (x^{3} - \frac{1}{x^{3}})$

Answer 2:

(i) We will use the identity ${(a + b)}^{2} = a^{2} + 2 a b + b^{2}$ in the given expression to find the product.
$(2 x + y) (2 x + y) = {(2 x + y)}^{2} = {(2 x)}^{2} + 2 (2 x) (y) + y^{2} = 4 x^{2} + 4 x y + y^{2}$

(ii) We will use the identity $(a + b) (a - b) = a^{2} - b^{2}$ in the given expression to find the product.
$(a + 2 b) (a - 2 b) = a^{2} - {(2 b)}^{2} = a^{2} - 4 b^{2}$

(iii) We will use the identity $(a + b) (a - b) = a^{2} - b^{2}$ in the given expression to find the product.
$(a^{2} + b c) (a^{2} - b c) = {(a^{2})}^{2} - {(b c)}^{2} = a^{4} - b^{2} c^{2}$

(iv)We will use the identity $(a + b) (a - b) = a^{2} - b^{2}$ in the given expression to find the product.
$(\frac{4 x}{5} - \frac{3 y}{4}) (\frac{4 x}{5} + \frac{3 y}{4}) = {(\frac{4 x}{5})}^{2} - {(\frac{3 y}{4})}^{2} = \frac{16 x^{2}}{25} - \frac{9 y^{2}}{16}$

(v) We will use the identity $(a + b) (a - b) = a^{2} - b^{2}$ in the given expression to find the product.
$(2 x + \frac{3}{y}) (2 x - \frac{3}{y}) = {(2 x)}^{2} - {(\frac{3}{y})}^{2} = 4 x^{2} - \frac{9}{y^{2}}$

(vi) We will use the identity $(a + b) (a - b) = a^{2} - b^{2}$ in the given expression to find the product.
$(2 a^{3} + b^{3}) (2 a^{3} - b^{3}) = {(2 a^{3})}^{2} - {(b^{3})}^{2} = 4 a^{6} - b^{6}$

(vii) We will use the identity $(a + b) (a - b) = a^{2} - b^{2}$ in the given expression to find the product.
$(x^{4} + \frac{2}{x^{2}}) (x^{4} - \frac{2}{x^{2}}) = {(x^{4})}^{2} - {(\frac{2}{x^{2}})}^{2} = x^{8} - \frac{4}{x^{4}}$

(viii) We will use the identity $(a + b) (a - b) = a^{2} - b^{2}$ in the given expression to find the product.

$(x^{3} + \frac{1}{x^{3}}) (x^{3} - \frac{1}{x^{3}}) = {(x^{3})}^{2} - {(\frac{1}{x^{3}})}^{2} = x^{6} - \frac{1}{x^{6}}$

Question 3:

Using the formula for squaring a binomial, evaluate the following:
(i) (102)²
(ii) (99)²
(iii) (1001)²
(iv) (999)²
(v) (703)²

Answer 3:

(i) Here, we will use the identity ${(a + b)}^{2} = a^{2} + 2 a b + b^{2}$
${(102)}^{2} = {(100 + 2)}^{2} = {(100)}^{2} + 2 \times 100 \times 2 + 2^{2} = 10000 + 400 + 4 = 10404$

(ii) Here, we will use the identity ${(a - b)}^{2} = a^{2} - 2 a b + b^{2}$
${(99)}^{2} = {(100 - 1)}^{2} = {(100)}^{2} - 2 \times 100 \times 1 + 1^{2} = 10000 - 200 + 1 = 9801$

(iii) Here, we will use the identity ${(a + b)}^{2} = a^{2} + 2 a b + b^{2}$
${(1001)}^{2} = {(1000 + 1)}^{2} = {(1000)}^{2} + 2 \times 1000 \times 1 + 1^{2} = 1000000 + 2000 + 1 = 1002001$

(iv) Here, we will use the identity ${(a - b)}^{2} = a^{2} - 2 a b + b^{2}$
${(999)}^{2} = {(1000 - 1)}^{2} = {(1000)}^{2} - 2 \times 1000 \times 1 + 1^{2} = 1000000 - 2000 + 1 = 998001$

(v) Here, we will use the identity ${(a + b)}^{2} = a^{2} + 2 a b + b^{2}$
${(703)}^{2} = {(700 + 3)}^{2} = {(700)}^{2} + 2 \times 700 \times 3 + 3^{2} = 490000 + 4200 + 9 = 494209$

Question 4:

Simplify the following using the formula: (a − b)(a + b) = a² − b²:
(i) (82)² − (18)²
(ii) (467)² − (33)²
(iii) (79)² − (69)²
(iv) 197 × 203
(v) 113 × 87
(vi) 95 × 105
(vii) 1.8 × 2.2
(viii) 9.8 × 10.2

Answer 4:

Here, we will use the identity $(a - b) (a + b) = a^{2} - b^{2}$

(i) Let us consider the following expression:

${(82)}^{2} - {(18)}^{2} = (82 + 18) (82 - 18) = 100 \times 64 = 6400$

(ii) Let us consider the following expression:

${(467)}^{2} - {(33)}^{2} = (467 + 33) (467 - 33) = 500 \times 434 = 217000$

(iii) Let us consider the following expression:

${(79)}^{2} - {(69)}^{2} = (79 + 69) (79 - 69) = 148 \times 10 = 1480$

(iv) Let us consider the following product:

$197 \times 203$

$∵ \frac{197 + 203}{2} = \frac{400}{2} = 200$ ; therefore, we will write the above product as:
$197 \times 203 = (200 - 3) (200 + 3) = {(200)}^{2} - {(3)}^{2} = 40000 - 9 = 39991$

Thus, the answer is $39991$ .

(v) Let us consider the following product:

$113 \times 87$

$∵ \frac{113 + 87}{2} = \frac{200}{2} = 100$ ; therefore, we will write the above product as:
$113 \times 87 = (100 + 13) (100 - 13) = {(100)}^{2} - {(13)}^{2} = 10000 - 169 = 9831$

Thus, the answer is 9831.

(vi) Let us consider the following product:

$95 \times 105$

$∵ \frac{95 + 105}{2} = \frac{200}{2} = 100$ ; therefore, we will write the above product as:
$95 \times 105 = (100 + 5) (100 - 5) = {(100)}^{2} - {(5)}^{2} = 10000 - 25 = 9975$

Thus, the answer is 9975.

(vii) Let us consider the following product:

$1.8 \times 2.2$

$∵ \frac{1.8 + 2.2}{2} = \frac{4}{2} = 2$ ; therefore, we will write the above product as:
$1.8 \times 2.2 = (2 - 0.2) (2 + 0.2) = {(2)}^{2} - {(0.2)}^{2} = 4 - 0.04 = 3.96$

Thus, the answer is 3.96.

(viii) Let us consider the following product:

$9.8 \times 10.2$

$∵ \frac{9.8 + 10.2}{2} = \frac{20}{2} = 10$ ; therefore, we will write the above product as:
$9.8 \times 10.2 = (10 - 0.2) (10 + 0.2) = {(10)}^{2} - {(0.2)}^{2} = 100 - 0.04 = 99.96$

Thus, the answer is 99.96.

Question 5:

Simplify the following using the identities:
(i) $\frac{58^{2} - 42^{2}}{16}$
(ii) 178 × 178 − 22 × 22
(iii) $\frac{198 \times 198 - 102 \times 102}{96}$
(iv) 1.73 × 1.73 − 0.27 × 0.27
(v) $\frac{8.63 \times 8.63 - 1.37 \times 1.37}{0.726}$

Answer 5:

(i) Let us consider the following expression:

$\frac{58^{2} - 42^{2}}{16}$
Using the identity $(a + b) (a - b) = a^{2} - b^{2}$ , we get:

$\frac{58^{2} - 42^{2}}{16} = \frac{(58 + 42) (58 - 42)}{16}$
$\Rightarrow \frac{58^{2} - 42^{2}}{16} = \frac{100 \times 16}{16} \Rightarrow \frac{58^{2} - 42^{2}}{16} = 100$

Thus, the answer is 100.

(ii) Let us consider the following expression:

$178 \times 178 - 22 \times 22$
Using the identity $(a + b) (a - b) = a^{2} - b^{2}$ , we get:

$178 \times 178 - 22 \times 22 = 178^{2} - 22^{2} = (178 + 22) (178 - 22) = 200 \times 156 = 31200$
Thus, the answer is 31200.

(iii) Let us consider the following expression:

$\frac{198 \times 198 - 102 \times 102}{96} = \frac{198^{2} - 102^{2}}{96}$
Using the identity $(a + b) (a - b) = a^{2} - b^{2}$ , we get:

$\frac{198 \times 198 - 102 \times 102}{96} = \frac{198^{2} - 102^{2}}{96} = \frac{(198 + 102) (198 - 102)}{96}$
$\Rightarrow \frac{198 \times 198 - 102 \times 102}{96} = \frac{(198 + 102) (198 - 102)}{96} \Rightarrow \frac{198 \times 198 - 102 \times 102}{96} = \frac{300 \times 96}{96} \Rightarrow \frac{198 \times 198 - 102 \times 102}{96} = 300$

Thus, the answer is 300.

(iv) Let us consider the following expression:

$1.73 \times 1.73 - 0.27 \times 0.27$
Using the identity $(a + b) (a - b) = a^{2} - b^{2}$ , we get:

$1.73 \times 1.73 - 0.27 \times 0.27 = 1 . 73^{2} - 0 . 27^{2} = (1.73 + 0.27) (1.73 - 0.27) = 2 \times 1.46 = 2.92$
Thus, the answer is 2.92.

(v) Let us consider the following expression:

$\frac{8.63 \times 8.63 - 1.37 \times 1.37}{0.726} = \frac{8 . 63^{2} - 1 . 37^{2}}{0.726}$
Using the identity $(a + b) (a - b) = a^{2} - b^{2}$ , we get:

$\frac{8.63 \times 8.63 - 1.37 \times 1.37}{0.726} = \frac{8 . 63^{2} - 1 . 37^{2}}{0.726} = \frac{(8.63 + 1.37) (8.63 - 1.37)}{0.726}$
$\Rightarrow \frac{8.63 \times 8.63 - 1.37 \times 1.37}{0.726} = \frac{(8.63 + 1.37) (8.63 - 1.37)}{0.726} \Rightarrow \frac{8.63 \times 8.63 - 1.37 \times 1.37}{0.726} = \frac{(8.63 + 1.37) (8.63 - 1.37)}{0.726} \Rightarrow \frac{8.63 \times 8.63 - 1.37 \times 1.37}{0.726} = \frac{10 \times 7.26}{0.726} \Rightarrow \frac{8.63 \times 8.63 - 1.37 \times 1.37}{0.726} = \frac{10 \times {7.26}^{10}}{0.726} \Rightarrow \frac{8.63 \times 8.63 - 1.37 \times 1.37}{0.726} = 100$

Thus, the answer is 100.

Question 6:

Find the value of x, if:
(i) 4x = (52)² − (48)²
(ii) 14x = (47)² − (33)²
(iii) 5x = (50)² − (40)²

Answer 6:

(i) Let us consider the following equation:

$4 x = {(52)}^{2} - {(48)}^{2}$
Using the identity $(a + b) (a - b) = a^{2} - b^{2}$ , we get:
$4 x = {(52)}^{2} - {(48)}^{2} 4 x = (52 + 48) (52 - 48) 4 x = 100 \times 4 = 400$
$\Rightarrow 4 x = 400$
$\Rightarrow x = 100$         (Dividing both sides by 4)

(ii) Let us consider the following equation:

$14 x = {(47)}^{2} - {(33)}^{2}$
Using the identity $(a + b) (a - b) = a^{2} - b^{2}$ , we get:
$14 x = {(47)}^{2} - {(33)}^{2} 14 x = (47 + 33) (47 - 33) 14 x = 80 \times 14 = 1120$
$\Rightarrow 14 x = 1120$
$\Rightarrow x = 80$         (Dividing both sides by 14)

(iii) Let us consider the following equation:

$5 x = {(50)}^{2} - {(40)}^{2}$
Using the identity $(a + b) (a - b) = a^{2} - b^{2}$ , we get:
$5 x = {(50)}^{2} - {(40)}^{2} 5 x = (50 + 40) (50 - 40) 5 x = 90 \times 10 = 900$
$\Rightarrow 5 x = 900$
$\Rightarrow x = 180$         (Dividing both sides by 5)

Question 7:

If $x + \frac{1}{x} = 20,$ find the value of $x^{2} + \frac{1}{x^{2}} .$

Answer 7:

Let us consider the following equation:

$x + \frac{1}{x} = 20$
Squaring both sides, we get:

${(x + \frac{1}{x})}^{2} = {(20)}^{2} = 400 \Rightarrow {(x + \frac{1}{x})}^{2} = 400 \Rightarrow x^{2} + 2 \times x \times \frac{1}{x} + {(\frac{1}{x})}^{2} = 400 [(a + b)^{2} = a^{2} + b^{2} + 2 a b] \Rightarrow x^{2} + 2 + \frac{1}{x^{2}} = 400$
$\Rightarrow x^{2} + \frac{1}{x^{2}} = 398$ (Subtracting 2 from both sides)

Thus, the answer is 398.

Question 8:

If $x - \frac{1}{x} = 3,$ find the values of $x^{2} + \frac{1}{x^{2}}$ and $x^{4} + \frac{1}{x^{4}} .$

Answer 8:

Let us consider the following equation:

$x - \frac{1}{x} = 3$

Squaring both sides, we get:

${(x - \frac{1}{x})}^{2} = {(3)}^{2} = 9 \Rightarrow {(x - \frac{1}{x})}^{2} = 9 \Rightarrow x^{2} - 2 \times x \times \frac{1}{x} + {(\frac{1}{x})}^{2} = 9 \Rightarrow x^{2} - 2 + \frac{1}{x^{2}} = 9$
$\Rightarrow x^{2} + \frac{1}{x^{2}} = 11$ (Adding 2 to both sides)

Squaring both sides again, we get:

${(x^{2} + \frac{1}{x^{2}})}^{2} = {(11)}^{2} = 121 \Rightarrow {(x^{2} + \frac{1}{x^{2}})}^{2} = 121 \Rightarrow {(x^{2})}^{2} + 2 (x^{2}) (\frac{1}{x^{2}}) + {(\frac{1}{x^{2}})}^{2} = 121 \Rightarrow x^{4} + 2 + \frac{1}{x^{4}} = 121$
$\Rightarrow x^{4} + \frac{1}{x^{4}} = 119$

Question 9:

If $x^{2} + \frac{1}{x^{2}} = 18,$ find the values of $x + \frac{1}{x} and x - \frac{1}{x} .$

Answer 9:

Let us consider the following expression:

$x + \frac{1}{x}$
Squaring the above expression, we get:

${(x + \frac{1}{x})}^{2} = x^{2} + 2 \times x \times \frac{1}{x} + {(\frac{1}{x})}^{2} = x^{2} - 2 + \frac{1}{x^{2}} [(a + b)^{2} = a^{2} + b^{2} + 2 a b] \Rightarrow {(x + \frac{1}{x})}^{2} = x^{2} + 2 + \frac{1}{x^{2}}$
$\Rightarrow {(x + \frac{1}{x})}^{2} = 20$                                        ( $∵$ $x^{2} + \frac{1}{x^{2}} = 18$ )
$\Rightarrow x + \frac{1}{x} = \pm \sqrt{20}$                                                     (Taking square root of both sides)

Now, let us consider the following expression:

$x - \frac{1}{x}$
Squaring the above expression, we get:

${(x - \frac{1}{x})}^{2} = x^{2} - 2 \times x \times \frac{1}{x} + {(\frac{1}{x})}^{2} = x^{2} - 2 + \frac{1}{x^{2}} [(a - b)^{2} = a^{2} + b^{2} - 2 a b] \Rightarrow {(x - \frac{1}{x})}^{2} = x^{2} - 2 + \frac{1}{x^{2}}$
$\Rightarrow {(x - \frac{1}{x})}^{2} = 16$                                  ( $∵$ $x^{2} + \frac{1}{x^{2}} = 18$ )
$\Rightarrow x - \frac{1}{x} = \pm 4$                                      (Taking square root of both sides)

Question 10:

If x + y = 4 and xy = 2, find the value of x² + y²

Answer 10:

We have:

${(x + y)}^{2} = x^{2} + 2 x y + y^{2} \Rightarrow x^{2} + y^{2} = {(x + y)}^{2} - 2 x y$
$\Rightarrow x^{2} + y^{2} = 4^{2} - 2 \times 2$ ( $∵$ $x + y = 4 and x y = 2$ )
$\Rightarrow x^{2} + y^{2} = 16 - 4 \Rightarrow x^{2} + y^{2} = 12$

Question 11:

If x − y = 7 and xy = 9, find the value of x² + y²

Answer 11:

We have:

${(x - y)}^{2} = x^{2} - 2 x y + y^{2} \Rightarrow x^{2} + y^{2} = {(x - y)}^{2} + 2 x y$
$\Rightarrow x^{2} + y^{2} = 7^{2} + 2 \times 9$ ( $∵$ $x - y = 7 and x y = 9$ )
$\Rightarrow x^{2} + y^{2} = 7^{2} + 2 \times 9 \Rightarrow x^{2} + y^{2} = 49 + 18 \Rightarrow x^{2} + y^{2} = 67$

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Question 12:

If 3x + 5y = 11 and xy = 2, find the value of 9x²+ 25y²

Answer 12:

We have:

${(3 x + 5 y)}^{2} = {(3 x)}^{2} + 2 (3 x) (5 y) + {(5 y)}^{2} \Rightarrow {(3 x + 5 y)}^{2} = 9 x^{2} + 30 x y + 25 y^{2} \Rightarrow 9 x^{2} + 25 y^{2} = {(3 x + 5 y)}^{2} - 30 x y$
$\Rightarrow 9 x^{2} + 25 y^{2} = 11^{2} - 30 \times 2$ ( $∵$ $3 x + 5 y = 11 and x y = 2$ )
$\Rightarrow 9 x^{2} + 25 y^{2} = 121 - 60 \Rightarrow 9 x^{2} + 25 y^{2} = 61$

Question 13:

Find the values of the following expressions:
(i) 16x² + 24x + 9, when $x = \frac{7}{4}$
(ii) 64x² + 81y² + 144xy, when x = 11 and $y = \frac{4}{3}$
(iii) 81x² + 16y² − 72xy, when $x = \frac{2}{3}$ and $y = \frac{3}{4}$

Answer 13:

(i) Let us consider the following expression:

$16 x^{2} + 24 x + 9$

Now

$16 x^{2} + 24 x + 9 = {(4 x + 3)}^{2}$                              (Using identity ${(a + b)}^{2} = a^{2} + 2 a b + b^{2}$ )
$\Rightarrow 16 x^{2} + 24 x + 9 = {(4 \times \frac{7}{4} + 3)}^{2} (Substituting x = \frac{7}{4}) \Rightarrow 16 x^{2} + 24 x + 9 = {(7 + 3)}^{2} \Rightarrow 16 x^{2} + 24 x + 9 = 10^{2} \Rightarrow 16 x^{2} + 24 x + 9 = 100$

(ii) Let us consider the following expression:

$64 x^{2} + 81 y^{2} + 144 x y$

Now

$64 x^{2} + 81 y^{2} + 144 x y = {(8 x + 9 y)}^{2}$                           (Using identity ${(a + b)}^{2} = a^{2} + 2 a b + b^{2}$ )
$\Rightarrow 64 x^{2} + 81 y^{2} + 144 x y = {[8 (11) + 9 (\frac{4}{3})]}^{2} (Substituting x = 11 and y = \frac{4}{3}) \Rightarrow 64 x^{2} + 81 y^{2} + 144 x y = {[88 + 12]}^{2} \Rightarrow 64 x^{2} + 81 y^{2} + 144 x y = 100^{2} \Rightarrow 64 x^{2} + 81 y^{2} + 144 x y = 10000$

(iii) Let us consider the following expression:

$81 x^{2} + 16 y^{2} - 72 x y$

Now

$81 x^{2} + 16 y^{2} - 72 x y = {(9 x - 4 y)}^{2}$                                   (Using identity ${(a + b)}^{2} = a^{2} - 2 a b + b^{2}$ )
$\Rightarrow 81 x^{2} + 16 y^{2} - 72 x y = {[9 (\frac{2}{3}) - 4 (\frac{3}{4})]}^{2} (Substituting x = \frac{2}{3} and y = \frac{3}{4}) \Rightarrow 81 x^{2} + 16 y^{2} - 72 x y = {[6 - 3]}^{2} \Rightarrow 81 x^{2} + 16 y^{2} - 72 x y = 3^{2} \Rightarrow 81 x^{2} + 16 y^{2} - 72 x y = 9$

Question 14:

If $x + \frac{1}{x} = 9,$ find the value of $x^{4} + \frac{1}{x^{4}} .$

Answer 14:

Let us consider the following equation:

$x + \frac{1}{x} = 9$

Squaring both sides, we get:

${(x + \frac{1}{x})}^{2} = {(9)}^{2} = 81 \Rightarrow {(x + \frac{1}{x})}^{2} = 81 \Rightarrow x^{2} + 2 \times x \times \frac{1}{x} + {(\frac{1}{x})}^{2} = 81 \Rightarrow x^{2} + 2 + \frac{1}{x^{2}} = 81$
$\Rightarrow x^{2} + \frac{1}{x^{2}} = 79$ (Subtracting 2 from both sides)

Now, squaring both sides again, we get:

${(x^{2} + \frac{1}{x^{2}})}^{2} = {(79)}^{2} = 6241 \Rightarrow {(x^{2} + \frac{1}{x^{2}})}^{2} = 6241 \Rightarrow {(x^{2})}^{2} + 2 (x^{2}) (\frac{1}{x^{2}}) + {(\frac{1}{x^{2}})}^{2} = 6241 \Rightarrow x^{4} + 2 + \frac{1}{x^{4}} = 6241$
$\Rightarrow x^{4} + \frac{1}{x^{4}} = 6239$

Question 15:

If $x + \frac{1}{x} = 12,$ find the value of $x - \frac{1}{x} .$

Answer 15:

Let us consider the following equation:

$x + \frac{1}{x} = 12$

Squaring both sides, we get:

${(x + \frac{1}{x})}^{2} = {(12)}^{2} = 144 \Rightarrow {(x + \frac{1}{x})}^{2} = 144 \Rightarrow x^{2} + 2 \times x \times \frac{1}{x} + {(\frac{1}{x})}^{2} = 144 [(a + b)^{2} = a^{2} + b^{2} + 2 a b] \Rightarrow x^{2} + 2 + \frac{1}{x^{2}} = 144$
$\Rightarrow x^{2} + \frac{1}{x^{2}} = 142$ (Subtracting 2 from both sides)

Now

${(x - \frac{1}{x})}^{2} = x^{2} - 2 \times x \times \frac{1}{x} + {(\frac{1}{x})}^{2} = x^{2} - 2 + \frac{1}{x^{2}} [(a - b)^{2} = a^{2} + b^{2} - 2 a b] \Rightarrow {(x - \frac{1}{x})}^{2} = x^{2} - 2 + \frac{1}{x^{2}} \Rightarrow {(x - \frac{1}{x})}^{2} = 142 - 2 (∵ x^{2} + \frac{1}{x^{2}} = 142) \Rightarrow {(x - \frac{1}{x})}^{2} = 140 \Rightarrow x - \frac{1}{x} = \pm \sqrt{140} (Taking square root)$

Question 16:

If 2x + 3y = 14 and 2x − 3y = 2, find the value of xy.
[Hint: Use (2x + 3y)² − (2x − 3y)² = 24xy]

Answer 16:

We will use the identity $(a + b) (a - b) = a^{2} - b^{2}$ to obtain the value of xy.
$Squaring (2 x + 3 y) and (2 x - 3 y) both and then subtracting them, we get :$
${(2 x + 3 y)}^{2} - {(2 x - 3 y)}^{2} = \{(2 x + 3 y) + (2 x - 3 y)\} \{(2 x + 3 y) - (2 x - 3 y)\} = 4 x \times 6 y = 24 x y \Rightarrow {(2 x + 3 y)}^{2} - {(2 x - 3 y)}^{2} = 24 x y$
$\Rightarrow 24 x y = {(2 x + 3 y)}^{2} - {(2 x - 3 y)}^{2} \Rightarrow 24 x y = {(14)}^{2} - {(2)}^{2} \Rightarrow 24 x y = (14 + 2) (14 - 2) (∵ (a + b) (a - b) = a^{2} - b^{2}) \Rightarrow 24 x y = 16 \times 12 \Rightarrow x y = \frac{16 \times 12}{24} (Dividing both sides by 24) \Rightarrow x y = 8$

Question 17:

If x² + y² = 29 and xy = 2, find the value of
(i) x + y
(ii) x − y
(iii) x⁴ + y⁴

Answer 17:

(i) We have:

${(x + y)}^{2} = x^{2} + 2 x y + y^{2} \Rightarrow (x + y) = \pm \sqrt{x^{2} + 2 x y + y^{2}} \Rightarrow (x + y) = \pm \sqrt{29 + 2 \times 2} (∵ x^{2} + y^{2} = 29 and x y = 2) \Rightarrow (x + y) = \pm \sqrt{29 + 4} \Rightarrow (x + y) = \pm \sqrt{33}$

(ii) We have:

${(x - y)}^{2} = x^{2} - 2 x y + y^{2} \Rightarrow (x - y) = \pm \sqrt{x^{2} - 2 x y + y^{2}} \Rightarrow (x + y) = \pm \sqrt{29 - 2 \times 2} (∵ x^{2} + y^{2} = 29 and x y = 2) \Rightarrow (x + y) = \pm \sqrt{29 - 4} \Rightarrow (x + y) = \pm \sqrt{25} \Rightarrow (x + y) = \pm 5$

(iii) We have:

${(x^{2} + y^{2})}^{2} = x^{4} + 2 x^{2} y^{2} + y^{4} \Rightarrow x^{4} + y^{4} = {(x^{2} + y^{2})}^{2} - 2 x^{2} y^{2} \Rightarrow x^{4} + y^{4} = {(x^{2} + y^{2})}^{2} - 2 {(x y)}^{2} \Rightarrow x^{4} + y^{4} = 29^{2} - 2 {(2)}^{2} (∵ x^{2} + y^{2} = 29 and x y = 2) \Rightarrow x^{4} + y^{4} = 841 - 8 \Rightarrow x^{4} + y^{4} = 833$

Question 18:

What must be added to each of the following expressions to make it a whole square?
(i) 4x² − 12x + 7
(ii) 4x²− 20x + 20

Answer 18:

(i) Let us consider the following expression:

$4 x^{2} - 12 x + 7$
The above expression can be written as:

$4 x^{2} - 12 x + 7 = {(2 x)}^{2} - 2 \times 2 x \times 3 + 7$
It is evident that if 2x is considered as the first term and 3 is considered as the second term, 2 is required to be added to the above expression to make it a perfect square. Therefore, 7 must become 9.

Therefore, adding and subtracting 2 in the above expression, we get:

$(4 x^{2} - 12 x + 7) + 2 - 2 = \{{(2 x)}^{2} - 2 \times 2 x \times 3 + 7\} + 2 - 2 = \{{(2 x)}^{2} - 2 \times 2 x \times 3 + 9\} - 2 = {(2 x + 3)}^{2} - 2$
Thus, the answer is 2.

(ii) Let's consider the following expression:

$4 x^{2} - 20 x + 20$
The above expression can be written as:

$4 x^{2} - 20 x + 20 = {(2 x)}^{2} - 2 \times 2 x \times 5 + 20$
It is evident that if 2x is considered as the first term and 5 is considered as the second term, 5 is required to be added to the above expression to make it a perfect square. Therefore, number 20 must become 25.

Therefore, adding and subtracting 5 in the above expression, we get:

$(4 x^{2} - 20 x + 20 + 5) - 5 = \{{(2 x)}^{2} - 2 \times 2 x \times 5 + 20\} + 5 - 5 = \{{(2 x)}^{2} - 2 \times 2 x \times 5 + 25\} - 5 = {(2 x + 5)}^{2} - 5$
Thus, the answer is 5.

Question 19:

Simplify:
(i) (x − y)(x + y) (x² + y²)(x⁴ + y²)
(ii) (2x − 1)(2x + 1)(4x² + 1)(16x⁴ + 1)
(iii) (4m − 8n)² + (7m + 8n)²
(iv) (2.5p − 1.5q)² − (1.5p − 2.5q)²
(v) (m² − n²m)² + 2m³n²

Answer 19:

To simplify, we will proceed as follows:
(i)
$(i) (x - y) (x + y) (x^{2} + y^{2}) (x^{4} + y^{4}) = (x^{2} - y^{2}) (x^{2} + y^{2}) (x^{4} + y^{4}) [∵ (a + b) (a - b) = a^{2} - b^{2}] = (x^{4} - y^{4}) (x^{4} + y^{4}) [∵ (a + b) (a - b) = a^{2} - b^{2}] = x^{8} - x^{8} [∵ (a + b) (a - b) = a^{2} - b^{2}]$

$(ii) (2 x - 1) (2 x + 1) (4 x^{2} + 1) (16 x^{4} + 1) = ({(2 x)}^{2} - 1^{2}) (4 x^{2} + 1) (16 x^{4} + 1) [∵ (a + b) (a - b) = a^{2} - b^{2}] = (4 x^{2} - 1) (4 x^{2} + 1) (16 x^{4} + 1) = \{{(4 x^{2})}^{2} - {(1^{2})}^{2}\} (16 x^{4} + 1) [∵ (a + b) (a - b) = a^{2} - b^{2}] = (16 x^{4} - 1) (16 x^{4} + 1) = {(16 x^{4})}^{2} - 1^{2} [∵ (a + b) (a - b) = a^{2} - b^{2}] = 256 x^{8} - 1$

$(iii) {(7 m - 8 n)}^{2} + {(7 m + 8 n)}^{2} = 2 {(7 m)}^{2} + 2 {(8 n)}^{2} [∵ {(a - b)}^{2} + {(a + b)}^{2} = 2 a^{2} + 2 b^{2}] = 98 m^{2} + 128 n^{2}$

$(iv) {(2.5 p - 1.5 q)}^{2} - {(1.5 p - 2.5 q)}^{2} = {(2.5 p)}^{2} + {(1.5 q)}^{2} - 2 (2.5 p) (1.5 q) - [{(1.5 p)}^{2} + {(2.5 q)}^{2} - 2 (1.5 p) (2.5 q)] = {(2.5 p)}^{2} + {(1.5 q)}^{2} - 2 (2.5 p) (1.5 q) - {(1.5 p)}^{2} - {(2.5 q)}^{2} + 2 (1.5 p) (2.5 q) = {(2.5 p)}^{2} - {(1.5 p)}^{2} + {(1.5 q)}^{2} - {(2.5 q)}^{2} = [(2.5 p + 1.5 p) (2.5 p - 1.5 p)] + [(1.5 q + 2.5) (1.5 q - 2.5 q)] [∵ (a + b) (a - b) = a^{2} - b^{2}] = 4 p \times p + 4 q \times (- q) = 4 p^{2} - 4 q^{2} = 4 (p^{2} - 4 q^{2})$

$v) {(m^{2} - n^{2} m)}^{2} + 2 m^{3} n^{2} = {(m^{2})}^{2} + {(n^{2} m)}^{2} [∵ {(a - b)}^{2} + 2 a b = a^{2} + b^{2}] = m^{4} + n^{4} m^{2}$

Question 20:

Show that:
(i) (3x + 7)² − 84x = (3x − 7)²
(ii) (9a − 5b)² + 180ab = (9a + 5b)²
(iii) ${(\frac{4 m}{3} - \frac{3 n}{4})}^{2} + 2 m n = \frac{16 m^{2}}{9} + \frac{9 n^{2}}{16}$
(iv) (4pq + 3q)² − (4pq − 3q)² = 48pq²
(v) (a − b)(a + b) + (b − c)(b + c) + (c − a)( c + a) = 0

Answer 20:

$(i) LHS = {(3 x + 7)}^{2} - 84 x = {(3 x + 7)}^{2} - 4 \times 3 x \times 7 = {(3 x - 7)}^{2} [∵ {(a + b)}^{2} - 4 a b = {(a - b)}^{2}] = RHS Because LHS is equal to RHS, the given equation is verified.$

$(ii) LHS = {(9 a - 5 b)}^{2} + 180 a b = {(9 a - 5 b)}^{2} + 4 \times 9 a \times 5 b = {(9 a + 5 b)}^{2} [∵ {(a - b)}^{2} + 4 a b = {(a + b)}^{2}] = RHS Because LHS is equal to RHS, the given equation is verified.$

$(iii) LHS = {(\frac{4 m}{3} - \frac{3 n}{4})}^{2} + 2 m n = {(\frac{4 m}{3} - \frac{3 n}{4})}^{2} + 2 \times \frac{4 m}{3} \times \frac{3 n}{4} = {(\frac{4 m}{3})}^{2} + {(\frac{3 n}{4})}^{2} [∵ {(a - b)}^{2} + 2 a b = a^{2} + b^{2}] = \frac{16 m^{2}}{9} + \frac{9 n^{2}}{16} = RHS Because LHS is equal to RHS, the given equation is verified.$

$(iv) LHS = {(4 p q + 3 q)}^{2} - {(4 p q - 3 q)}^{2} = 4 (4 p q) (3 q) [∵ {(a + b)}^{2} - {(a + b)}^{2} = 4 a b] = 48 p q^{2} = RHS Because LHS is equal to RHS, the given equation is verified.$

$(v) LHS = (a - b) (a + b) + (b - c) (b + c) + (c + a) (c - a) = a^{2} - b^{2} + b^{2} - c^{2} + c^{2} - a^{2} [∵ (a + b) (a - b) = a^{2} - b^{2}] = a^{2} - b^{2} + b^{2} - c^{2} + c^{2} - a^{2} = 0 = RHS Because LHS is equal to RHS, the given equation is verified.$

RD Sharma solution class 8 chapter 6 Algebraic Expressions and Identity Exercise 6.6