Mathematics Assignment Class VIII

Algebraic Identities

Download or Print, free assignment with answer key for Class 8 on ALGEBRAIC IDENTITIES. Important and extra questions that cover all topics of ALGEBRAIC IDENTITIES and is useful and helpful for the students.

Algebraic Identities required to solve this assignment

Students need to learn these algebraic identities before start this assignment

(a + b)² = a² + b² + 2ab

(a - b)² = a² + b² - 2ab

a² - b² = (a + b)(a - b)

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

MATHEMATICS ASSIGNMENT

CLASS - 8 CHAPTER - 7

ALGEBRAIC IDENTITIES

Question 1

Expand: $\left(\sqrt{3}x-\sqrt{2}y-7\right)^{2}$

Answer : 3x² + 2y²+ 49 - 2 $\sqrt{6}$ xy + 14 $\sqrt{2}$ y - 14 $\sqrt{3}$ x

Question 2

Factorise: x² – 3x - 54

Answer : (x - 6) (x - 9)

Question 3

Factorise : p² – pq - 6q².

Answer : (p + 2q)(p - 3q)

Question 4

Factorise : (-3x + 4y – 5)².

Answer : 9x² + 16y²+ 25 - 24xy - 40y + 30x

Question 5

Simplify : (2x + 5y)² – (2x - 5y)²
Ans: 40xy

Question 6: Factorise: x² - 3x - 28

Ans: (x - 7)(x + 4)

Question 7

Factorise : 9x²y – 24xy² + 16y³

Answer : y(3x - 4y)²

Question 8

Factorise: (176)² - (24)²

Answer : 15200

Question 9

Evaluate : 211 ✖ 189 - 106 ✖ 94

Answer :29915

Question 10

If a + b + c = 12 and a² + b² + c² = 64, find the value of ab + bc + ca
Ans: 40

Question 11

Factorise : 4a² + b² + 25c² + 4ab - 10bc - 20ca

Answer :(2a + b - 5c)²

Question 12

Factorise: 25x² + y² + 9z² - 10xy – 6yz + 30xz

Answer : (- 5x + y + 3z)²

Question 13

Factorise: $a^{2}+4b^{2}+\frac{c^{2}}{9}+4ab-\frac{4bc}{3}-\frac{2ca}{3}$

Answer : $\left(a+2b-\frac{c}{3}\right)^{2}$

Question 14

Factorise: $a^{2}+\frac{1}{25}b^{2}+\frac{c^{2}}{4}-\frac{2}{5}ab-\frac{1}{5}bc+ca$

Answer : $\left(a-\frac{b}{5}+\frac{c}{2}\right)^{2}$

Question 15

Factorise : 9m² - 30mn + 25n²) - 64x²
Ans: (3m - 5n + 8x)(3m - 5n - 8x)

Question 16

Factorise: 4(x + y)² – 28(x² – y²) + 49(x – y)²

Answer : (-5x + 9y)² or (5x - 9y)²

Solution Hint :

4(x + y)² – 28(x² – y²) + 49(x – y)²
= [2(x + y)]² – 2 x 2(x + y) x 7(x + y) + [7(x - y)]²

Using: a² - 2ab + b² = (a - b)² we get

= [2(x + y) – 7(x - y)]².

= (2x + 2y - 7x + 7y)².

= (- 5x + 9y)² or (5x - 9y)²

Question 17

Simplify : $\frac{2.3\times 2.3-0.3\times 0.3}{2.3\times 2.3-2\times 2.3\times 0.3+0.3\times 0.3}$

Answer : 1.3

Solution Hint:

Above problem can be written as

$\frac{(2.3)^{2}-(0.3)^{2}}{(2.3)^{2}-2\times 2.3\times 0.3+(0.3)^{2}}$

$=\frac{(2.3+0.3)(2.3-0.3)}{(2.3-0.3)^{2}}$

$=\frac{2.6\times 2}{2^{2}}=\frac{2.6}{2}=1.3$

Question 18

Simplify : $\frac{8.63\times 8.63-1.37\times 1.37}{0.726}$

Answer : 100

Solution Hint:

Above problem can be written as

= $\frac{(8.63)^{2}-(1.37)^{2}}{0.726}$

$=\frac{(8.63+1.37)(8.63-1.37)}{0.726}$

$=\frac{10\times 7.26}{0.726}=\frac{72.6}{0.726}$

$=\frac{726}{726}\times\frac{1000}{10}=100$

Question 19

$Find\:\:\left(a^{4}+\frac{1}{a^{4}}\right)\:\:if\:\:\left(a+\frac{1}{a}\right)^{2}=6$

Answer : 14

Solution Hint:

$\left(a+\frac{1}{a}\right)^{2}=6$

$a^{2}+\frac{1}{a^{2}}+2\times a\times\frac{1}{a}=6$

$a^{2}+\frac{1}{a^{2}}=6-2=4$

Squaring on both side we get

$\left(a^{2}+\frac{1}{a^{2}}\right)^{2}=4^{2}$
$a^{4}+\frac{1}{a^{4}}+2\times a^{2}\times\frac{1}{a^{2}}=16$

$a^{4}+\frac{1}{a^{4}}+2=16$

$a^{4}+\frac{1}{a^{4}}=16-2=14$

$a^{4}+\frac{1}{a^{4}}=14$

Question 20

$Simplify:\;\:\frac{2.5\times 2.5-0.2\times 0.2}{2.5\times 2.5-2\times 2.5\times 0.2+0.2\times 0.2}$

Solution Hint

Above problem can be written as

$\frac{(2.5)^{2}-(0.2)^{2}}{(2.5)^{2}-2\times 2.5\times 0.2+(0.2)^{2}}$

$=\frac{(2.5-0.2)(2.5+0.2)}{(2.5-0.2)^{2}}$

$=\frac{(2.5+0.2)}{(2.5-0.2)}=\frac{2.7}{2.3}=\frac{27}{23}$

Question 21

Find the value of x if

x² – 91 = 2.3 x 2.3 + 2 x 0.7 x 2.3 + 0.7 x 0.7
Ans: 10

Solution Hint

x² – 91 = 2.3 ✕ 2.3 + 2 ✕ 0.7 ✕ 2.3 + 0.7 ✕ 0.7

= (2.3)² + 2 ✕ 2.3 ✕ 0.7 + (0.7)²
= (2.3 + 0.7)²= (3)²= 9

x² – 91 = 9

x² = 9 + 91 = 100

x = 10

Question 22

If $x^{4}+\frac{1}{x^{4}}=2,$ then find the value of $x+\frac{1}{x}$

Answer: 2

Solution Hint

$\left(x^{2}+\frac{1}{x^{2}}\right)^{2}=x^{4}+\frac{1}{x^{4}}+2$

$\left(x^{2}+\frac{1}{x^{2}}\right)^{2}=2+2=4$

Taking Square root on both side we get

$x^{2}+\frac{1}{x^{2}}=2$

$\left(x+\frac{1}{x}\right)^{2}=x^{2}+\frac{1}{x^{2}}+2$

$\left(x+\frac{1}{x}\right)^{2}=2+2=4$

Taking Square root on both side we get

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

Question 23
If $x+\frac{1}{x}=9$ and $x-\frac{1}{x}=6$ , then find the value of $x^{4}-\frac{1}{x^{4}}$

Answer: 4266

Solution Hint

$\left(x+\frac{1}{x}\right)^{2}=9^{2}$

$x^{2}+\frac{1}{x^{2}}+2\times x\times\frac{1}{x}=81$

$x^{2}+\frac{1}{x^{2}}=81-2=79$

$x^{2}-\frac{1}{x^{2}}=\left(x+\frac{1}{x}\right)\left(x-\frac{1}{x}\right)$

$x^{2}-\frac{1}{x^{2}}=9\times 6=54$

$x^{4}-\frac{1}{x^{4}}=\left(x^{2}+\frac{1}{x^{2}}\right)\left(x^{2}-\frac{1}{x^{2}}\right)$

$x^{4}-\frac{1}{x^{4}}=79\times 54=4266$

Question 24:  (5x - 3y + 1)² - (5x + 3y - 1)².
Answer: - 60xy + 20x
Solution Hint
(5x - 3y + 1)² - (5x + 3y - 1)².
Using a² – b² = (a + b)(a - b) we get
= (5x - 3y + 1+5x + 3y - 1)(5x - 3y + 1-5x - 3y + 1)
= (10x)(- 6y + 2) = - 60xy + 20x
Question 25:
If 3p - 4q = 8 and pq = 2, find the value of  (3p - 4q)².
Answer: 160
Solution Hint:
3p - 4q = 8
Squaring on both side we get
(3p – 4q)² = 8²
9p² + 16q² – 24pq = 64
9p² + 16q² – 24 🇽 2 = 64
9p² + 16q² – 48 = 64
9p² + 16q² = 64 + 48 = 112
(3p – 4q)² = 9p² + 16q² + 24pq
                = 112 + 24 🇽 2  = 160

Question 26:
Factorise : $5x^{2}+7y^{2}+z^{2}-2\sqrt{35}xy-2\sqrt{7}yz+2\sqrt{5}zx$
Answer $(\sqrt{5}x-\sqrt{7}y+z)^{2}$
Question 27: If 5x - 2y = 7 and xy = 2, find the value of (5x + 2y)².

Answer : 129

Question 28: Factorise: (64m² – 144mn + 81n²) – 25p²

Answer : (8m - 9n + 5p)(8m - 9n - 5p)
Question 29: Simplify : (a²- b²)(a²+ b²) - (a²- b²)².

Answer : 2b²(a + b)(a - b)

Question 30: Factorise: y² + y - 30

Answer : (y + 6)(y - 5)

Question 31

If $x-\frac{1}{x}=7$ , then find the value of $x-\frac{1}{x}=7$

Ans. 51

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