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Class 8 Exam > Class 8 Notes > RD Sharma Solutions for Class 8 Mathematics > RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-5 )

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-5 ) | RD Sharma Solutions for Class 8 Mathematics PDF Download

Question 1:Write the following squares of binomials as trinomials:
(i) (x + 2)²
(ii) (8a + 3b)²
(iii) (2m + 1)²

^{(x) (a2b − bc2)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=x2+2×x×2+b2=x2+4x+b2

(ii) (8a+3b)2=(8a)2+2(8a)(3b)+(6b)2=64a2+48ab+36b2(iii) (2m+1)2=(2m)2+2(2m)(1)+12=4m2+4m+1

(iv) (9a+1/6)2=(9a)2+2(9a)(1/6)+(1/6)2=81a2+3a+1/36

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-5 ) | RD Sharma Solutions for Class 8 Mathematics

(x) (a2b−bc2)2=(a2b)2−2(a2b)(bc2)+(bc2)2=a4b2−2a2b2c2+b2c4 RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-5 ) | RD Sharma Solutions for Class 8 Mathematics

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-5 ) | RD Sharma Solutions for Class 8 Mathematics

(xii) (x2−ay)2=(x2)2−2x2(ay)+(ay)2=x4−2x2ay+a2y2

Question 2: Find the product of the following binomials:
(i) (2x + y)(2x + y)
(ii) (a + 2b)(a − 2b)
(iii) (a² + bc)(a²− bc)

(vi) (2a³ + b³)(2a³ − b³)

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.

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-5 ) | RD Sharma Solutions for Class 8 Mathematics

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

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-5 ) | RD Sharma Solutions for Class 8 Mathematics

(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.

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-5 ) | RD Sharma Solutions for Class 8 Mathematics

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

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-5 ) | RD Sharma Solutions for Class 8 Mathematics

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

${(79)}^{2} - {(69)}^{2} = (79 + 69) (79 - 69) = 148 \times 10 = 1480$ RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-5 ) | RD Sharma Solutions for Class 8 Mathematics $∵ \frac{197 + 203}{2} = \frac{400}{2} = 200$ ; therefore, we will write the above product as:

197×203=(200−3)(200+3)=(200)2−(3)2=40000−9=39991 Thus, the answer is $39991$ .
(v) Let us consider the following product:

$113 \times 87$
RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-5 ) | RD Sharma Solutions for Class 8 Mathematics $∵ \frac{113 + 87}{2} = \frac{200}{2} = 100$ ; therefore, we will write the above product as:

113×87=(100+13)(100−13)=(100)2−(13)2=10000−169=9831 Thus, the answer is 9831.

(vi) Let us consider the following product:
$95 \times 105$

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-5 ) | RD Sharma Solutions for Class 8 Mathematics $∵ \frac{95 + 105}{2} = \frac{200}{2} = 100$ ; therefore, we will write the above product as:

95×105=(100+5)(100−5)=(100)2−(5)2=10000−25=9975 Thus, the answer is 9975.
(vii) Let us consider the following product:
$1.8 \times 2.2$
RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-5 ) | RD Sharma Solutions for Class 8 Mathematics $∵ \frac{1.8 + 2.2}{2} = \frac{4}{2} = 2$ ; therefore, we will write the above product as:

1.8×2.2=(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:
$9.8 \times 10.2$
RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-5 ) | RD Sharma Solutions for Class 8 Mathematics $∵ \frac{9.8 + 10.2}{2} = \frac{20}{2} = 10$ ; therefore, we will write the above product as:

9.8×10.2=(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:

(ii) 178 × 178 − 22 × 22

(iv) 1.73 × 1.73 − 0.27 × 0.27

Answer 5:(i) Let us consider the following expression:

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

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:

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

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:

1.73×1.73−0.27×0.27=1.73²−0.27²=(1.73+0.27)(1.73−0.27)=2×1.46=2.92
Thus, the answer is 2.92.

(v) Let us consider the following expression:

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

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:

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-5 ) | RD Sharma Solutions for Class 8 Mathematics

[(a+b)2=a2 +b2 +2ab]

⇒x2+2+1/x² =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:

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

Squaring both sides again, we get:

(x2+1/x² )2=(11)2=121

⇒(x2+1/x²)2=121

⇒x4+2+1/x⁴ =121

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+1/x )2=x2+2×x×1/x +(1/x)2=x2−2+ 1/x²

[(a+b)2=a2+b2+2ab]

$\Rightarrow {(x + \frac{1}{x})}^{2} = 20$ ( $∵$ $x^{2} + \frac{1}{x^{2}} = 18$ )

(Taking square root of both sides)

Now, let us consider the following expression:

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

$x - \frac{1}{x}$ (∵ x²+ 1/x² =18)

⇒x−1/x =±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=x2+2xy+y2⇒x2+y2=(x+y)2−2xy $\Rightarrow x^{2} + y^{2} = 4^{2} - 2 \times 2$ ( $∵$ $x + y = 4 and x y = 2$ )
$\Rightarrow x^{2} + y^{2} = 16 - 4 \Rightarrow x^{2} + y^{2} = 12$

Question 11: If x − y = 7 and xy = 9, find the value of x² + y²

Answer 11: We have:

(x−y)2=x2−2xy+y2⇒x2+y2=(x−y)2+2xy $\Rightarrow x^{2} + y^{2} = 7^{2} + 2 \times 9$ ( $∵$ $x - y = 7 and x y = 9$ )
$\Rightarrow x^{2} + y^{2} = 7^{2} + 2 \times 9 \Rightarrow x^{2} + y^{2} = 49 + 18 \Rightarrow x^{2} + y^{2} = 67$

Question 12: If 3x + 5y = 11 and xy = 2, find the value of 9x²+ 25y²

Answer 12: We have:

(3x+5y)2=(3x)2+2(3x)(5y)+(5y)2⇒(3x+5y)2=9x2+30xy+25y2⇒9x2+25y2=(3x+5y)2−30xy $\Rightarrow 9 x^{2} + 25 y^{2} = 11^{2} - 30 \times 2$ ( $∵$ $3 x + 5 y = 11 and x y = 2$ )
$\Rightarrow 9 x^{2} + 25 y^{2} = 121 - 60 \Rightarrow 9 x^{2} + 25 y^{2} = 61$

Question 13: Find the values of the following expressions:

(i) 16x² + 24x + 9, when $x = \frac{7}{4}$ 7/4
(ii) 64x² + 81y² + 144xy, when x = 11 and $y = \frac{4}{3}$ 4/3
(iii) 81x² + 16y² − 72xy, when $x = \frac{2}{3}$ and $y = \frac{3}{4}$ 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 ${(a + b)}^{2} = a^{2} + 2 a b + b^{2}$ )
⇒64x²+81y²+144xy=[8(11)+9(4/3)]² (Substituting x=11 and y=4/3) $\Rightarrow 64 x^{2} + 81 y^{2} + 144 x y = {[8 (11) + 9 (\frac{4}{3})]}^{2} (Substituting x = 11 and y = \frac{4}{3}) \Rightarrow 64 x^{2} + 81 y^{2} + 144 x y = {[88 + 12]}^{2} \Rightarrow 64 x^{2} + 81 y^{2} + 144 x y = 100^{2} \Rightarrow 64 x^{2} + 81 y^{2} + 144 x y = 10000$ (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:

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

⇒(x+1/x )2=81

⇒x2+2×x×1/x +(1/x )2=81

⇒x2+2+1/x² =81

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

Now, squaring both sides again, we get:

(x2+1/x² )2=(79)2=6241

⇒(x2+1/x² )2=6241

⇒(x2)2+2(x2)(1/x²)+(1/x²)2=6241

⇒x4+2+1/x⁴=6241

⇒x4+1/x⁴ =6239⇒x4+1x4=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:

⇒x2+2×x×1/x +(1/x)2=144 [ (a+b)2=a2+b2+2ab]

⇒x2+2+1/x² =144

$\Rightarrow x^{2} + \frac{1}{x^{2}} = 142$ (Subtracting 2 from both sides)

Now

(x−1/x)²=x²−2×x×1/x+(1/x)²=x²−2+1/x² [(a−b)²=a²+b²−2ab]

⇒(x−1/x )2=x2−2+1/x²

^{⇒(x−1/x )2=142−2}(∵ x²+1/x² =142)

⇒(x−1/x)2=140

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-5 ) | RD Sharma Solutions for Class 8 Mathematics (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.
$Squaring (2 x + 3 y) and (2 x - 3 y) both and then subtracting them, we get :$
${(2 x + 3 y)}^{2} - {(2 x - 3 y)}^{2} = \{(2 x + 3 y) + (2 x - 3 y)\} \{(2 x + 3 y) - (2 x - 3 y)\} = 4 x \times 6 y = 24 x y \Rightarrow {(2 x + 3 y)}^{2} - {(2 x - 3 y)}^{2} = 24 x y$ 6y=24xy ${(2 x + 3 y)}^{2} - {(2 x - 3 y)}^{2} = \{(2 x + 3 y) + (2 x - 3 y)\} \{(2 x + 3 y) - (2 x - 3 y)\} = 4 x \times 6 y = 24 x y \Rightarrow {(2 x + 3 y)}^{2} - {(2 x - 3 y)}^{2} = 24 x y$
$\Rightarrow 24 x y = {(2 x + 3 y)}^{2} - {(2 x - 3 y)}^{2} \Rightarrow 24 x y = {(14)}^{2} - {(2)}^{2} \Rightarrow 24 x y = (14 + 2) (14 - 2) (∵ (a + b) (a - b) = a^{2} - b^{2}) \Rightarrow 24 x y = 16 \times 12 \Rightarrow x y = \frac{16 \times 12}{24} (Dividing both sides by 24) \Rightarrow x y = 8$

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=x2+2xy+y2

⇒(x+y)=±

⇒(x+y)=± ( ∵ x2+y2=29 and xy=2)

⇒(x+y)=±

(ii) We have:

(x−y)2=x2−2xy+y2

⇒(x−y)=±

⇒(x−y)=± (∵ x2+y2=29 and xy=2)

⇒(x−y)=±

⇒(x−y)=± √25

⇒(x−y)=± 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:
$(4 x^{2} - 12 x + 7) + 2 - 2 = \{{(2 x)}^{2} - 2 \times 2 x \times 3 + 7\} + 2 - 2 = \{{(2 x)}^{2} - 2 \times 2 x \times 3 + 9\} - 2 = {(2 x + 3)}^{2} - 2$ 3+9}−2=(2x+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:
$(4 x^{2} - 20 x + 20 + 5) - 5 = \{{(2 x)}^{2} - 2 \times 2 x \times 5 + 20\} + 5 - 5 = \{{(2 x)}^{2} - 2 \times 2 x \times 5 + 25\} - 5 = {(2 x + 5)}^{2} - 5$ 5+25}−5=(2x+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)
$(i) (x - y) (x + y) (x^{2} + y^{2}) (x^{4} + y^{4}) = (x^{2} - y^{2}) (x^{2} + y^{2}) (x^{4} + y^{4}) [∵ (a + b) (a - b) = a^{2} - b^{2}] = (x^{4} - y^{4}) (x^{4} + y^{4}) [∵ (a + b) (a - b) = a^{2} - b^{2}] = x^{8} - x^{8} [∵ (a + b) (a - b) = a^{2} - b^{2}]$ $(i) (x - y) (x + y) (x^{2} + y^{2}) (x^{4} + y^{4}) = (x^{2} - y^{2}) (x^{2} + y^{2}) (x^{4} + y^{4}) [∵ (a + b) (a - b) = a^{2} - b^{2}] = (x^{4} - y^{4}) (x^{4} + y^{4}) [∵ (a + b) (a - b) = a^{2} - b^{2}] = x^{8} - x^{8} [∵ (a + b) (a - b) = a^{2} - b^{2}]$ (iii) (7m−8n)2+(7m+8n)2=2(7m)2+2(8n)2 [∵ (a−b)2+(a+b)2=2a2+2b2]=98m2+128n2

(iv) (2.5p−1.5q)2−(1.5p−2.5q)2=(2.5p)2+(1.5q)2−2(2.5p)(1.5q)−[(1.5p)2+(2.5q)2−2(1.5p)(2.5q)] =(2.5p)2+(1.5q)2−2(2.5p)(1.5q)−(1.5p)2−(2.5q)2+2(1.5p)(2.5q)=(2.5p)2−(1.5p)2+(1.5q)2−(2.5q)2 =[(2.5p+1.5p)(2.5p−1.5p)]+[(1.5q+2.5)(1.5q−2.5q)] [∵(a+b)(a−b)=a2−b2] =4p×p+4q×(−q)=4p2−4q2 =4(p2−4q2 )v) (m2−n2m)2+2m3n2=(m2)2+(n2m)2 [∵ (a−b)2+2ab=a2+b2]=m4+n4m2

Question 20: Show that:
(i) (3x + 7)² − 84x = (3x − 7)²
(ii) (9a − 5b)² + 180ab = (9a + 5b)²

^{(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=(3x+7)2−84x=(3x+7)2−4×3x×7=(3x−7)2 [∵ (a+b)2−4ab=(a−b)2]=RHSBecause LHS is equal to RHS, the given equation is verified. (ii) LHS=(9a−5b)2+180ab=(9a−5b)2+4×9a×5b=(9a+5b)2 [∵ (a−b)2+4ab=(a+b)2]=RHSBecause LHS is equal to RHS, the given equation is verified. (iii) LHS=(4m/3 −3n/4)2+2mn=(4m/3 - 3n/4)2+2×4m/3 × 3n/4=(4m/3)2+(3n/4)2 [∵ (a−b)2+2ab=a2+b2]=16m²/9 +9n²/16 =RHSBecause LHS is equal to RHS, the given equation is verified.

(iv) LHS=(4pq+3q)2−(4pq−3q)2=4(4pq)(3q) [∵(a+b)2−(a+b)2=4ab]=48pq2 =RHSBecause LHS is equal to RHS, the given equation is verified.(v) LHS=(a−b)(a+b)+(b−c)(b+c)+(c+a)(c−a)=a2−b2+b2−c2+c2−a2 [∵ (a+b)(a−b)=a2−b2]=a2−b2+b2−c2+c2−a2 =0=RHSBecause LHS is equal to RHS, the given equation is verified.

Question 21: Find the following products:
(i) (x + 4) (x + 7)
(ii) (x − 11) (x + 4)
(iii) (x + 7) (x − 5)
(iv) (x − 3) ( x − 2)
(v) (y² − 4) (y² − 3)

(vi) $(x + \frac{4}{3}) (x + \frac{3}{4})$

(vii) (3x + 5) (3x + 11)
(viii) (2x² − 3) (2x² + 5)
(ix) (z² + 2) (z² − 3)
(x) (3x − 4y) (2x − 4y)
(xi) (3x² − 4xy) (3x² − 3xy)

(xii) $(x + \frac{1}{5}) (x + 5)$
(xiii) $(z + \frac{3}{4}) (z + \frac{4}{3})$
(xiv) (x² + 4) (x² + 9)
(xv) (y²+ 12) (y²+ 6)
(xvi) $(y^{2} + \frac{5}{7}) (y^{2} - \frac{14}{5})$
(xvii) (p² + 16) $(p^{2} - \frac{1}{4})$

Answer 21: (i) Here, we will use the identity $(x + a) (x + b) = x^{2} + (a + b) x + a b .$
(x+4)(x+7)

=x2+(4+7)x+4×7

=x2+11x+28

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

(x−11)(x+4)

=x2+(4−11)x−11×4

=x2−7x−44

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

(x+7)(x−5)

=x2+(7−5)x−7×5

=x2+2x−35

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

(x−3)(x−2)

=x2−(3+2)x+3×2

=x2−5x+6

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

(y2−4)(y2−3)

=(y2)2−(4+3)(y2)+4×3

=y4−7y2+12

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

(x+4/3)(x+3/4)

=x2+25/12 x+1

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

(3x+5)(3x+11)

=(3x)2+(5+11)(3x)+5×11

=9x2+48x+55

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

(2x2−3)(2x2+5)

=(2x2)2+(5−3)(2x2)−3×5

=4x4+4x2−15

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

(z2+2)(z2−3)

=(z2)2+(2−3)(z2)−2×3

=z4−z2−6

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

(3x−4y)(2x−4y)

=(4y−3x)(4y−2x)
(Taking common −1 from both parentheses)

=(4y)2−(3x+2x)(4y)+3x×2x

=16y2−(12xy+8xy)+6x2

=16y2−20xy+6x2

(xi) Here, we will use the identity $(x - a) (x - b) = x^{2} - (a + b) x + a b$
(3x2−4xy)(3x2−3xy)

=(3x2)2−(4xy+3xy)(3x2)+4xy×3xy

=9x4−(12x3y+9x3y)+12x2y2

=9x4−21x3y+12x2y2

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

(x+1/5)(x+5)

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-5 ) | RD Sharma Solutions for Class 8 Mathematics

=x2+26/5 x+1

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

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-5 ) | RD Sharma Solutions for Class 8 Mathematics

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

(x2+4)(x2+9)

=(x2)2+(4+9)(x2)+4×9

=x4+13x2+36

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

(y2+12)(y2+6)

=(y2)2+(12+6)(y2)+12×6

=y4+18y2+72

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

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-5 ) | RD Sharma Solutions for Class 8 Mathematics

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

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-5 ) | RD Sharma Solutions for Class 8 Mathematics

Question 22: Evaluate the following:
(i) 102 × 106
(ii) 109 × 107
(iii) 35 × 37
(iv) 53 × 55
(v) 103 × 96
(vi) 34 × 36
(vii) 994 × 1006

Answer 22: (i) Here, we will use the identity (x+a)(x+b)=x2+(a+b)x+abx+ax+b=x2+a+bx+ab

102×106

=(100+2)(100+6)

=1002+(2+6)100+2×6

=10000+800+12

=10812

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

109 × 107

=(100+9)(100+7)

=1002+(9+7)100+9×7

=10000+1600+63

=11663

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

35 × 37

=(30+5)(30+7)

=302+(5+7)30+5×7

=900+360+35

=1295

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

53 × 55

=(50+3)(50+5)

=502+(3+5)50+3×5

=2500+400+15

=2915

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

103 × 96

=(100+3)(100−4)

=1002+(3−4)100−3×4

=10000−100−12

=9888

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

34 × 36

=(30+4)(30+6)

=302+(4+6)30+4×6

=900+300+24

=1224

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

994 × 1006

=(1000−6)×(1000+6)

=10002+(6−6)×1000−6×6

=1000000−36

=999964

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FAQs on RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-5 ) - RD Sharma Solutions for Class 8 Mathematics

1. What are algebraic expressions?

Ans. Algebraic expressions are mathematical expressions that contain variables, constants, and mathematical operations like addition, subtraction, multiplication, and division. These expressions can be used to represent real-life situations and solve problems in algebra.

2. How do you simplify algebraic expressions?

Ans. To simplify algebraic expressions, you need to combine like terms and perform the indicated operations. Start by combining the constants and then the variables with the same exponent. Use the distributive property to simplify expressions with parentheses. Finally, simplify any remaining terms by performing the required operations.

3. What are identities in algebraic expressions?

Ans. Identities in algebraic expressions are equations that are true for all values of the variables involved. These equations depict the equality of two algebraic expressions, regardless of the specific values of the variables. For example, (a + b)^2 = a^2 + 2ab + b^2 is an identity as it holds true for any values of a and b.

4. How can algebraic expressions be used in real-life situations?

Ans. Algebraic expressions can be used in various real-life situations, such as calculating the cost of items based on their quantity and price, determining the distance traveled based on speed and time, and predicting future outcomes based on patterns and trends. They are widely used in fields like finance, engineering, and physics to model and solve real-world problems.

5. What is the importance of studying algebraic expressions and identities?

Ans. Studying algebraic expressions and identities helps in developing critical thinking and problem-solving skills. It provides a foundation for higher-level mathematics and scientific studies. Understanding these concepts enables individuals to analyze and manipulate mathematical relationships, make predictions, and solve complex equations. Additionally, algebraic expressions and identities are used extensively in various professions and everyday life situations.

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RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-5 ) | RD Sharma Solutions for Class 8 Mathematics PDF Download

Question 1:Write the following squares of binomials as trinomials:(i) (x + 2)2(ii) (8a + 3b)2(iii) (2m + 1)2

Answer 1: We will use the identities (a+b)2=a2+2ab+b2 and (a−b)2=a2−2ab+b2a+b2=a2+2ab+b2 and a-b2=a2-2ab+b2 to convert the squares of binomials as trinomials.

Question 2: Find the product of the following binomials:(i) (2x + y)(2x + y)(ii) (a + 2b)(a − 2b)(iii) (a2 + bc)(a2 − bc)

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

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)=a2 −b2(a-b)(a+b)=a2 -b2(i) Let us consider the following expression:(82)2−(18)2=(82+18)(82−18)=100×64=6400(ii) Let us consider the following expression:(467)2−(33)2=(467+33)(467−33)=500×434=217000

Question 5: Simplify the following using the identities:

Answer 5:(i) Let us consider the following expression:

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:

Question 7: If x+1/x =20,x+1x=20, find the value of x2+ 1/x2 .x2+1x2.

Answer 7: Let us consider the following equation:

Question 8: If x−1/x =3,x-1x=3, find the values of x2+1/x2x2+1x2 and x4+1/x4 .x4+1x4.

Answer 8: Let us consider the following equation:

Question 9: If x2+1/x2 =18,x2+1x2=18, find the values of x+1/x and x−1/x .

Answer 9: Let us consider the following expression:

Question 10: If x + y = 4 and xy = 2, find the value of x2 + y2

Answer 10: We have:

Question 11: If x − y = 7 and xy = 9, find the value of x2 + y2

Answer 11: We have:

Question 12: If 3x + 5y = 11 and xy = 2, find the value of 9x2 + 25y2

Answer 12: We have:

Question 13: Find the values of the following expressions:

Answer 13: (i) Let us consider the following expression:16x2+24x+916x2+24x+9Now16x2+24x+9=(4x+3)216x2+24x+9=4x+32 (Using identity (a+b)2=a2+2ab+b2a+b2=a2+2ab+b2)⇒16x2+24x+9=(4×7/4 +3)2 (Substituting x=7/4)⇒16x2+24x+9=(7+3)2⇒16x2+24x+9=102⇒16x2+24x+9=100

Question 14: If x+1/x =9,x+1x=9, find the value of x4+1/x4.x4+1x4.

Answer 14: Let us consider the following equation:

Question 15: If x+1/x =12,x+1x=12, find the value of x−1/x.

Answer 15: Let us consider the following equation:

Question 16: If 2x + 3y = 14 and 2x − 3y = 2, find the value of xy.[Hint: Use (2x + 3y)2 − (2x − 3y)2 = 24xy]

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:

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

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

Question 20: Show that:(i) (3x + 7)2 − 84x = (3x − 7)2(ii) (9a − 5b)2 + 180ab = (9a + 5b)2

Answer 20:

Question 21: Find the following products:(i) (x + 4) (x + 7)(ii) (x − 11) (x + 4)(iii) (x + 7) (x − 5)(iv) (x − 3) ( x − 2)(v) (y2 − 4) (y2 − 3)

Answer 21: (i) Here, we will use the identity (x+a)(x+b)=x2+(a+b)x+ab.x+ax+b=x2+a+bx+ab.(x+4)(x+7)

=x2+(4+7)x+4×7

=x2+11x+28

(ii) Here, we will use the identity (x−a)(x+b)=x2+(b−a)x−abx-ax+b=x2+b-ax-ab.

(x−11)(x+4)

=x2+(4−11)x−11×4

=x2−7x−44

(iii) Here, we will use the identity (x+a)(x−b)=x2+(a−b)x−abx+ax-b=x2+a-bx-ab.

(x+7)(x−5)

=x2+(7−5)x−7×5

=x2+2x−35

(iv) Here, we will use the identity (x−a)(x−b)=x2−(a+b)x+abx-ax-b=x2-a+bx+ab.

(x−3)(x−2)

=x2−(3+2)x+3×2

=x2−5x+6

(v) Here, we will use the identity (x−a)(x−b)=x2−(a+b)x+abx-ax-b=x2-a+bx+ab.

(y2−4)(y2−3)

=(y2)2−(4+3)(y2)+4×3

=y4−7y2+12

(vi) Here, we will use the identity (x+a)(x+b)=x2+(a+b)x+abx+ax+b=x2+a+bx+ab.

(x+4/3)(x+3/4)

Question 22: Evaluate the following:(i) 102 × 106(ii) 109 × 107(iii) 35 × 37(iv) 53 × 55(v) 103 × 96(vi) 34 × 36(vii) 994 × 1006

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RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-5 ) | RD Sharma Solutions for Class 8 Mathematics

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Question 1:Write the following squares of binomials as trinomials:
(i) (x + 2)²
(ii) (8a + 3b)²
(iii) (2m + 1)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.

Question 2: Find the product of the following binomials:
(i) (2x + y)(2x + y)
(ii) (a + 2b)(a − 2b)
(iii) (a² + bc)(a²− bc)

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

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

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

Question 7: If $x + \frac{1}{x} = 20,$ find the value of $x^{2} + \frac{1}{x^{2}} .$

Question 8: If $x - \frac{1}{x} = 3,$ find the values of $x^{2} + \frac{1}{x^{2}}$ and $x^{4} + \frac{1}{x^{4}} .$

Question 9: If $x^{2} + \frac{1}{x^{2}} = 18,$ find the values of $x + \frac{1}{x} and x - \frac{1}{x} .$

Question 10: If x + y = 4 and xy = 2, find the value of x² + y²

Question 11: If x − y = 7 and xy = 9, find the value of x² + y²

Question 12: If 3x + 5y = 11 and xy = 2, find the value of 9x²+ 25y²

Question 14: If $x + \frac{1}{x} = 9,$ find the value of $x^{4} + \frac{1}{x^{4}} .$

Question 15: If $x + \frac{1}{x} = 12,$ find the value of $x - \frac{1}{x} .$

Question 16: If 2x + 3y = 14 and 2x − 3y = 2, find the value of xy.
[Hint: Use (2x + 3y)² − (2x − 3y)² = 24xy]

Question 17: If x² + y² = 29 and xy = 2, find the value of
(i) x + y
(ii) x − y
(iii) x⁴ + y⁴

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

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

Question 20: Show that:
(i) (3x + 7)² − 84x = (3x − 7)²
(ii) (9a − 5b)² + 180ab = (9a + 5b)²

Question 21: Find the following products:
(i) (x + 4) (x + 7)
(ii) (x − 11) (x + 4)
(iii) (x + 7) (x − 5)
(iv) (x − 3) ( x − 2)
(v) (y² − 4) (y² − 3)

Answer 21: (i) Here, we will use the identity $(x + a) (x + b) = x^{2} + (a + b) x + a b .$
(x+4)(x+7)

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

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

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

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

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

Question 22: Evaluate the following:
(i) 102 × 106
(ii) 109 × 107
(iii) 35 × 37
(iv) 53 × 55
(v) 103 × 96
(vi) 34 × 36
(vii) 994 × 1006