Open App

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

The document RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-5 ) | RD Sharma Solutions for Class 8 Mathematics is a part of the Class 8 Course RD Sharma Solutions for Class 8 Mathematics.

All you need of Class 8 at this link: Class 8

	RD Sharma Solutions for Class 8 Mathematics 88 docs

RD Sharma Solutions for Class 8 Mathematics

88 docs

Join Course for Free

Top Courses for Class 8

View all

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.

Related Exams

Class 8

About this Document

	4.89/5 Rating
	Nov 22, 2024 Last updated

Document Description: RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-5 ) for Class 8 2024 is part of RD Sharma Solutions for Class 8 Mathematics preparation. The notes and questions for RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-5 ) have been prepared according to the Class 8 exam syllabus. Information about RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-5 ) covers topics like and RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-5 ) Example, for Class 8 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-5 ).

Introduction of RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-5 ) in English is available as part of our RD Sharma Solutions for Class 8 Mathematics for Class 8 & RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-5 ) in Hindi for RD Sharma Solutions for Class 8 Mathematics course. Download more important topics related with notes, lectures and mock test series for Class 8 Exam by signing up for free. Class 8: RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-5 ) | RD Sharma Solutions for Class 8 Mathematics

Description

Full syllabus notes, lecture & questions for RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-5 ) | RD Sharma Solutions for Class 8 Mathematics - Class 8 | Plus excerises question with solution to help you revise complete syllabus for RD Sharma Solutions for Class 8 Mathematics | Best notes, free PDF download

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

In this doc you can find the meaning of RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-5 ) defined & explained in the simplest way possible. Besides explaining types of RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-5 ) theory, EduRev gives you an ample number of questions to practice RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-5 ) tests, examples and also practice Class 8 tests

	RD Sharma Solutions for Class 8 Mathematics 88 docs

RD Sharma Solutions for Class 8 Mathematics

88 docs

Join Course for Free

Download as PDF

Explore Courses for Class 8 exam

Top Courses for Class 8

Explore Courses

Signup for Free!

Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.

Start learning for Free

10M+ students study on EduRev

mock tests for examination

video lectures

study material

Sample Paper

Extra Questions

practice quizzes

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

MCQs

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

past year papers

pdf

Free

Previous Year Questions with Solutions

Exam

Objective type Questions

Summary

shortcuts and tricks

Semester Notes

ppt

Important questions

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

Viva Questions

;

Additional Information about RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-5 ) for Class 8 Preparation

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

The RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-5 ) is an invaluable resource that delves deep into the core of the Class 8 exam. These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective. With the help of these notes, you can grasp complex subjects quickly, revise important points easily, and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner, allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material, or toppers' notes, this PDF has got you covered. Download the RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-5 ) now and kickstart your journey towards success in the Class 8 exam.

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

The importance of RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-5 ) cannot be overstated, especially for Class 8 aspirants. This document holds the key to success in the Class 8 exam. It offers a detailed understanding of the concept, providing invaluable insights into the topic. By knowing the concepts well in advance, students can plan their preparation effectively. Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.

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

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-5 ) Notes offer in-depth insights into the specific topic to help you master it with ease. This comprehensive document covers all aspects related to RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-5 ). It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation. Practice papers and question papers enable you to assess your progress effectively. Additionally, the paper analysis provides valuable tips for tackling the exam strategically. Access to Toppers' notes gives you an edge in understanding complex concepts. Whether you're a beginner or aiming for advanced proficiency, RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-5 ) Notes on EduRev are your ultimate resource for success.

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

The "RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-5 ) Class 8 Questions" guide is a valuable resource for all aspiring students preparing for the Class 8 exam. It focuses on providing a wide range of practice questions to help students gauge their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation. The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level. Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.

Study RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-5 ) on the App

Students of Class 8 can study RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-5 ) alongwith tests & analysis from the EduRev app, which will help them while preparing for their exam. Apart from the RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-5 ), students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc. The EduRev App will make your learning easier as you can access it from anywhere you want. The content of RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-5 ) is prepared as per the latest Class 8 syllabus.

Education Revolution

Signup to see your scores go up within 7 days!

Access 1000+ FREE Docs, Videos and Tests

Continue with Google

Takes less than 10 seconds to signup

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

RD Sharma Solutions for Class 8 Mathematics

Top Courses for Class 8

FAQs on 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 Mathematics

Top Courses for Class 8

mock tests for examination

video lectures

study material

Sample Paper

Extra Questions

practice quizzes

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

MCQs

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

past year papers

pdf

Free

Previous Year Questions with Solutions

Exam

Objective type Questions

Summary

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