RBSE Solutions for Class 8 Maths Chapter 10 Factorization In Text Exercise is part of RBSE Solutions for Class 8 Maths. Here we have given Rajasthan Board RBSE Class 8 Maths Chapter 10 Factorization In Text Exercise.
| Board | RBSE |
| Textbook | SIERT, Rajasthan |
| Class | Class 8 |
| Subject | Maths |
| Chapter | Chapter 10 |
| Chapter Name | Factorization |
| Exercise | In Text Exercise |
| Number of Questions | 4 |
| Category | RBSE Solutions |
Rajasthan Board RBSE Class 8 Maths Chapter 10 Factorization In Text Exercise
Page No: 119
Question 1.
Find the two integers a and b such that
| a+b | ab | a | b |
| 8 | 15 | 5 | 3 |
| 13 | 12 | ||
| -1 | -20 | -5 | 4 |
| -5 | 4 | ||
| 10 | 21 | ||
| -1 | -12 | ||
| -11 | 10 | ||
| -7 | 10 |
Solution:
(a + b)² = a² + 2ab + b²,
(a – b)² = a² – 2ab + b²
(a + b)² – (a – b)² – 4ab
⇒ (a – b)² = (a + b)² = 4ab
⇒ a – b = \(\sqrt { { (a+b) }^{ 2 }-4ab } \)
Now (i) a + b = 13, and ab = 12
then a – b = \(\sqrt { { (13) }^{ 2 }-4\times 12 } \)
= \(=\sqrt { 169-48 } =\sqrt { 121 } \) = 11
then a + b + a – b = 13 + 11 = 24
⇒ 2a = 24
⇒ a = 12
and b = 13 – a = 13 – 12 = 1
⇒ b = 1
(ii) When a + b = – 5 and ab = 4
then a – b = \(\sqrt { { (13) }^{ 2 }-4\times 12 } \)
Filling the(RBSESolutions.com)above values in table given below
| a+b | ab | a | b |
| 8 | 15 | 5 | 3 |
| 13 | 12 | 12 | 1 |
| -1 | -20 | -5 | 4 |
| -5 | 4 | -4 | -1 |
| 10 | 21 | 7 | 3 |
| -1 | -12 | -4 | 3 |
| -11 | 10 | -10 | -1 |
| -7 | 10 | -5 | -2 |
Page No: 123
Question 2.
3x + x + 4x = 56
7x = 56
x = \(\frac { 56 }{ 7 }\)
= 8
Find the error.
Solution:
3x + x + 4x = 56
=> 3x + 1x + 4x = 56
=> (3 + 1 + 4) x = 56
=> 8x = 56
=> x = \(\frac { 56 }{ 8 }\) = 7 (correct value)
Question 3.
Find the value of 5x at x = – 2 = 5 – 2 = 3
Find the(RBSESolutions.com)error and also find the correct value.
Solution:
Value of 5x at x = – 2
= 5 x (- 2) | 5x = 5 × x
= – 10 (correct value)
Question 4.
Solutions of the expression is given in the column A and B. Find which of the solution is correct.
| Expression | A | B |
| 3(x-4) | 3x-4 | 3x-12 |
| (2x)² | 2x² | 4x² |
| (x+4)² | x²+16 | x²+8x+16 |
| (x-3)² | x²-9 | x²-6x+9 |
| \(\frac { y+1 }{ 5 }\) | y+1 | \(\frac { y }{ 5 }+1\) |
Solution:
(i) 3(x – 4)
= 3 × x – 3 × 4 =
3x – 12
Hence B is correct.
(ii) (2x)²
= (2x) × (2x)
= (2 × x) × (2 × x)
= 2 × 2 × x × x
= 4 × x²
= 4x²
Hence B is correct.
(iii) (x + 4)² = x² + 2(x) (4) + (4)²
= x² + 8x + 16
Hence B is correct.
(iv) (x – 3)² = x² – 2 (x) (3) + (3)²
= x² – 6x + 9
Hence B is correct.
(v)
Hence B is correct.
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