In this chapter, we provide KSEEB SSLC Class 8 Maths Chapter 2 Algebraic Expressions Ex 2.4 for English medium students, Which will very helpful for every student in their exams. Students can download the latest KSEEB SSLC Class 8 Maths Chapter 2 Algebraic Expressions Ex 2.4 pdf, free KSEEB SSLC Class 8 Maths Chapter 2 Algebraic Expressions Ex 2.4 pdf download. Now you will get step by step solution to each question.
Karnataka State Syllabus Class 8 Maths Chapter 2 Algebraic Expressions Ex 2.4
Karnataka Board Class 8 Maths Chapter 2 Algebraic Expressions Ex 2.4
Question 1.
Find the product:
i. (a + 3) ( a + 5)
ii. (3t + 1) (3t + 4)
iii. (a – 8) (a + 2)
iv. (a – 6) (a – 2)
Answer:
i. (a + 3) (a + 5) This is in the form.
(x + a)(x + b) = x² + (a + b)x + ab
Here x = a, a = 3, b = 5
(a + 3)(a + 5) = a² + (a + 5)a + 3.5
= a² + 8a + 15
(ii) (3t + 1)(3t + 4)This is in the form.
(x + a)(x + b) = x² +(a + b)x + ab
Here x =3t,a = 1, b = 4
(3t – 1)(3t + 4) = (3t)²+(1 + 4)3t + 1.4
= 9t² + 5(3t) + 4 = 9t² + 15t + 4
(iii) (a -8)(a + 2)
This is in the form (x + a)(x + b) = x² + (a + b)x + ab
Here x = a, a = -8, b = 2
(a – 8)(a + 2) = a² + (-8 + 2)a + (-8)(2)
= a² + (-6)a – 16
= a² – 6a – 16
(iv) (a – 6)(a – 2)
This is in the form (x + a)(x + b) = x² +(a + b)x + ab
Herex = a, a = -6,b = -2
(a – 6)(a – 2) = a² + (- 6 – 2) a + (-6)(-2)
=a² + (- 8)a + 12
= a² – 8a + 12
2. Evaluate using suitable identities
Question i.
53 × 55
Answer:
53 × 55 = (50 + 3) (50 + 5)
This is in the form (x + a)(x + b) = x² + (a + b)x + ab
Here x = 50, a = 3, b = 5
(50 + 3)(50 + 5) = 50² + (3 + 5) 50 + 3 × 5
= 2500 + (8)50 +15
= 2500 + 400 + 15
∴ 53 × 55 = 2915
Question ii.
102 × 106
Answer:
102 × 106 = (100 + 2) (100 + 6) This is in the form.
(x + a)(x + b) = x² +(a + b)x + ab
Here x = 100, a = 2, b = 6
(100 + 2)(100 + 6) = 100² +(2 + 6)100 + 2 × 6
= 10000 + (8)100 + 12
= 10000 + 800 + 12
∴102 × 106 = 10812
Question iii.
34 × 36
Answer:
34 × 36 = (30 + 4) (30 + 6) This is in the form.
(x + a)(x + b) = x²+ (a + b)x + ab
Here x =30, a = 4, b = 6
(30 + 4)(30 + 6) = 30² + (4 + 6)30 + 4 × 6
= 900 + 300 + 24
34 × 36 = 1224
Question iv.
103× 96
Answer: 103 × 96 = (100 + 3)(100-4)This is in the form.
(x + a)(x + b) = x² + (a + b)x + ab
Here x = 100,a = 3,b = -4 (100 + 3)(100-4) = 100² + (3 – 4)100 + 3 × -4
= 10000 + (- 1)100 – 12
= 10000 – 100 – 12
103 × 96 = 9888
Question 3.
Find the expression for the product (x+a) (x+b) (x+c) using the identity (x+a)(x+b) = x² + (a + b)x + ab
Answer:
(x + a) (x + b) (x + c)
Consider (x + a)(x + b)
= x² + (a + b)x + ab
= x² + ax + bx + ab (x + a)(x + b)(x + c)
= (x² + ax + bx + ab)(x + c)
= x(x² + ax + bx+ ab) + c
(x² + ax + bx + ab)
= x3 + ax² + bx² + abx + cx² + cax +cbx + abc
= x3 + ax² + bx² + cx² + abx + cax +cbx + abc [by rearranging]
= x3 + x²(a + b + c) + x(ab + ca +bc) + abc
∴ (x + a)(x + b)(x + c)
= x3 +x² (a + b + c) + x(ab + bc + ca) + abc
4. Using the identify (a+b)2 = a² + 2ab+b² simplify the following.
Question i.
(a + 6 )²
Answer:
Using the identity
(a + b)² = a² + 2ab + b²
(a + 6)² = a² +2.a.6 + 6²
= a² +12a + 36
Question ii.
(3x+2y)2
Answer:
Using the identity
(a + b)² = a² +2ab + b²
(3x + 2y)² =(3x)² + 2.3x.2y + (2y)²
= 9x² +12xy + 4y²
Question iii.
(2p + 3q)²
Answer:
Using the identity
(a + b)² =a² + 2ab + b²
(2p + 3q)² =(2p)² +2.2p.3q + (3q)²
= 4p² + 12pq + 9q²
Question iv.
(x² + 5)²
Answer:
Using the identity (a + b)² = a² +2ab + b²
(x² +5)² = (x²)² + 2.x².5 + 52 = x4 + 10x² + 25
5. Evaluate using the identity (a+b)² = a² + 2ab + b²
Question i.
(34)²
Answer:
34² = (30 + 4)²
(a + b)² = a² + 2ab+ b² Here a = 30, b = 4
(30 + 4)² = (30)² + 2 × 30 × 4 + (4)²
= 900 + 240 + 16
(34)² = 1156
Question ii.
(10.2)²
Answer:
(10.2)² = (10 + 0.2)²
(a + b)² = a² + 2ab + b² Here a = 10, b=0.2
[(10 + (0.2))]² = 10² +2.(10).(0.2) + (0.2)² ]
= 100 + 4 + 0.04
(10.2)² = 104.04
Question iii.
53²
Answer:
53² = (50 + 3)²
(a + b)² =a² +2ab + b²
Here a = 50, b = 3 (50 + 3)² =50² + 2 × 50 × 3 + (3)²
= 2500 + 300 + 9
53² = 2809
Question iv.
(41 )²
Answer:
41² = (40 + 1)²
(a + b)² = a² + 2ab + b²
Here a = 40,b = 1
(40 + 1)² = 40² + 2 × 40 × 1 + (1)²
= 1600 + 80 + 1
41² = 1681
6. Use the identify (a -b)² = a² – 2ab + b² to compute.
Question i.
(x – 6)²
Answer:
(a – b)² = a² – 2ab + b²
Herea = x, b=6
(x – 6)² = x² – (2)(x)(6) + 6²
= x² – 12x+ 36
Question ii.
(3x – 5y)²
Answer:
(a – b)² = a² – 2ab + b² Here a = 3x, b = 5y
(3x – 5y)² = (3x)² -(2)(3x)(5y) + (5y)² = 9x² – 30xy + 25y²
Question iii.
(5a – 4b)²
Answer:
(a – b)² = a² – 2ab + b²
Here a = 5a, b = 4b
(5a – 4b)² = (5a)² – (2)(5a)(4b) + (4b)²
= 25a² – 40ab + 16b²
Question iv.
(p² – q²)²
Answer:
(a – b)² = a² – 2ab + b² Here a = p² , b = q2
(p² – q²)² = (p²)² – (2)(p²)(q²) + (q²)²
= p4 – 2p²q² + q4
7. Evaluate using the identify (a – b)² = a² – 2ab + b².
Question i.
(49)²
Answer:
49² = (50 – 1)²
(a – b)² = a² – 2ab + b² Here a = 50, b = 1
(50 – 1)² = 50² – (2)(50)(1) + 1² = 2500 – 100 + 1
(49)² = 2401
Question ii.
(9.8)²
Answer:
(9.8)² = (10 – 0.2)²
(a – b)² =a² – 2ab + b² Here a = 10, b=0.2
(10 – 0.2)² = 10² – (2)(10)(0.2) + (0.2)²
= 100 – 4 + 0.04
(9.8)² =96.04
Question iii.
59²
Answer:
59² = (60 – 1)²
(a – b)² = a² – 2ab + b² Here a = 60, b = 1
(60 – 1)² = 60² – (2)(60)(1) +12 = 3600 – 120 + 1
59² =3481
Question iv.
(198)²
Answer:
(198)² = (200 – 2)²
(a – b)² = a² – 2ab + b² Here a = 200, b = 2
(200 – 2)² = 200² – (2)(200)(2) + 2²
= 40000 – 800 + 4
(198)² = 39204
8. Use the identify (a + b) (a – b) = a² – b² to find the product.
Question i.
(x + 6) (x – 6)
Answer:
(a + b) (a – b) = a2 – b2
Here a = x,b = 6
(x + 6)(x – 6) = x² – 6² = x² – 36
Question ii.
(3x + 5) (3x – 5)
Answer:
(a + b) (a – b) = a² – b²
Here a = 3x, b = 5
(3x + 5)(3x – 5) = (3x)² – (5)²
= 9x² -25
Question iii.
(2a + 4b) (2a – 4b)
Answer:
(a + b) (a – b) = a² – b²
Here a = 2a,b = 4b
(2a + 4b)(2a-4b) = (2a)² – (4b)²
= 4a² – 16b²
Question iv.
(2×3+1)(2×3−1)
Answer:
(a + b) (a – b) = a² – b²
Here a = 2×3 ,b = 1
(2×3+1)(2×3−1) = (2×3)2−12
= 4×29−1
9. Evaluate these using identity :
Question i.
55 × 45
Answer:
55 × 45 = (50 + 5) (50 – 5)
Identity (a + b)(a – b) = a² – b²
Here a = 50,b=5
(50 + 5)(50 – 5) = (50)² – 5²
= 2500 – 25
55 × 45 = 2475
Question ii.
33 × 27
Answer:
33 × 27 = (30 + 3) (30 – 3)
identity (a + b)(a – b) = a² – b²
Here a = 30,b = 3
(30 + 3)(30 – 3) = 30² – 3² = 900 – 9
33 × 27 = 891
Question iii.
8.5 × 9.5
Answer:
8.5 × 9.5 = (9 – 0.5) (9 + 0.5)
Here a = 9, b = 0.5
(9 – 0.5)(9 + 0.5) = (9²) – (0.5)² = 81 – 0.25
= 80.75
Question iv.
102 × 98
Answer:
102 × 98 = (100+ 2) (100 – 2)
Identity (a + b)(a – b) = a2 – b2
Here a = 100, b= 2
(100 + 2)(100 – 2) = 100² – 2² =10000 – 4
102 × 98 = 9996
10. Find the product.
Question i.
(x – 3)(x + 3) (x2 + 9)
Answer:
(x – 3)(x + 3) (x2 + 9)
Identity(a + b)(a-b) = a² – b²
= (x² – 3²)(x² + 9)
= (x² – 9)(x² + 9)
= (x²)² – 9²
= x4 – 81
Question ii.
(2a + 3)(2a – 3) (4a² + 9)
Answer:
(2a + 3)(2a – 3) (4a² + 9)
Identity (a + b)(a-b) = a² – b²
= [((2a)² – 3²)(4a²+9)²
= (4a² – 9)(4a² + 9)
= (4a²)² – 9² = 16a4 – 81
Question iii.
(P + 2) (P – 2) (P² + 4)
Answer:
(P + 2) (P – 2) (P² + 4)
Identity (a + b)(a – b) = a² – b²
= (P² – 2²)(P²+ 4)
= (P² – 4)(P² + 4)
= (P²)² – 4² = P4 – 16
Question iv.
(12m−13)(12m+13)(14m2+19)
Answer:
Question v.
(2x – y)(2x + y) (4x² + y²)
Answer:
(2x – y)(2x + y) (4x² + y²)
Identity (a + b)(a – b) = a² – b²
[(2x)² – y²](4x² + y²)
= (4x² – y²)(4x² + y²)
= I6x4 – y4
Question vi.
(2x – 3y)(2x + 3y) (4x² + 9y²)
Answer:
(2x – 3y)(2x + 3y) (4x² + 9y²)
Identity (a + b)(a-b) = a² -b²
= [(2x)² – (3y)²](4x²+ 9y²)
= (4x² – 9y²)(4x² + 9y²)
= (4x²)² – (9y²)²
= 16x4 – 81y4
All Chapter KSEEB Solutions For Class 8 maths
—————————————————————————–
All Subject KSEEB Solutions For Class 9
*************************************************
I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment on the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.
If these solutions have helped you, you can also share kseebsolutionsfor.com to your friends.
Best of Luck!!