- RD Chapter 8- Quadratic Equations Ex-8.1
- RD Chapter 8- Quadratic Equations Ex-8.2
- RD Chapter 8- Quadratic Equations Ex-8.3
- RD Chapter 8- Quadratic Equations Ex-8.4
- RD Chapter 8- Quadratic Equations Ex-8.5
- RD Chapter 8- Quadratic Equations Ex-8.6
- RD Chapter 8- Quadratic Equations Ex-8.8
- RD Chapter 8- Quadratic Equations Ex-8.9
- RD Chapter 8- Quadratic Equations Ex-8.10
- RD Chapter 8- Quadratic Equations Ex-8.11
- RD Chapter 8- Quadratic Equations Ex-8.12
- RD Chapter 8- Quadratic Equations Ex-8.13

RD Chapter 8- Quadratic Equations Ex-8.1 |
RD Chapter 8- Quadratic Equations Ex-8.2 |
RD Chapter 8- Quadratic Equations Ex-8.3 |
RD Chapter 8- Quadratic Equations Ex-8.4 |
RD Chapter 8- Quadratic Equations Ex-8.5 |
RD Chapter 8- Quadratic Equations Ex-8.6 |
RD Chapter 8- Quadratic Equations Ex-8.8 |
RD Chapter 8- Quadratic Equations Ex-8.9 |
RD Chapter 8- Quadratic Equations Ex-8.10 |
RD Chapter 8- Quadratic Equations Ex-8.11 |
RD Chapter 8- Quadratic Equations Ex-8.12 |
RD Chapter 8- Quadratic Equations Ex-8.13 |

**Answer
1** :

Let first number = x

Then second number = x + 1

According to the condition

x² + (x + 1)2 = 85

=> x² + x² + 2x + 1 = 85

=> 2x² + 2x + 1 – 85 = 0

=> 2x² + 2x – 84 = 0

=> x² + x – 42 = 0

=> x² + 7x – 6x – 42 = 0

=> x (x + 7) – 6 (x + 7) = 0

=> (x + 7) (x – 6) = 0

Either x + 7 = 0, then x = -7 or x – 6 = 0, then x = 6

(i) If x = -7, then the first number = -7 and second number = -7 + 1 = -6

(ii) If x = 6, then the first number = 6 and second number = 6 + 1 = 7

Hence numbers are -7, -6 or 6, 7

**Answer
2** :

Total = 29

Let first part = x

Then second part = 29 – x

According to the condition

x² + (29 – x)2 = 425

=> x² + 841 + x² – 58x = 425

=> 2x² – 58x + 841 – 425 = 0

=> 2x² – 58x + 416 = 0

=> x² – 29x + 208 = 0 (Dividing by 2)

=> x² – 13x – 16x + 208 = 0

=> x(x – 13) – 16 (x – 13) = 0

=> (x – 13) (x – 16) = 0

Either x – 13 = 0, then x = 13 or x – 16 = 0, then x = 16

(i) If x = 13, then First part =13 and second part = 29 – 13 = 16

(ii) If x = 16, then First part =16 and second part = 29 – 16 = 13

Parts are 13, 16

**Answer
3** :

Side of the first square = x cm

Its area = (side)2 = x² cm2

Side of the second square = (x + 4) cm

Its area = (x + 4)2 cm2

According to the condition,

x² + (x + 4)2 = 656

=> x² + x² + 8x + 16 = 656

=> 2x² + 8x + 16 – 656 = 0

=> 2x² + 8x – 640 = 0

=> x² + 4x – 320 = 0 (Dividing by 2)

=> x² + 20x – 16x – 320 = 0

=> x (x + 20) – 16 (x + 20) = 0

=> (x + 20) (x – 16) = 0

Either x + 20 = 0, then x = -20 Which is not possible being negative

or x – 16 = 0, then x = 16

Side of the first square = 16 cm

and side of the second square = 16 + 4 = 20 cm

**Answer
4** :

Sum of two numbers = 48

Let first number = x

The second number = 48 – x

According to the condition,

x (48 – x) = 432

=> 48x – x² = 432

=> – x² + 48x – 432 = 0

=> x² – 48x + 432 = 0

=> x² – 12x – 36x + 432 = 0

=> x (x – 12) – 36 (x – 12) = 0

=> (x – 12) (x – 36) = 0

Either x – 12 = 0, then x = 12 or x – 36 = 0, then x = 36

(i) If x = 12, then First number = 12 and second number = 48 – 12 = 36

(ii) If x = 36, then First number = 36 and second number = 48 – 36 = 12

Numbers are 12, 36

**Answer
5** :

Let the given integer be = x

According to the condition

x² + x = 90

=> x² + x – 90 = 0

=> x² + 10x – 9x – 90 = 0

=> x (x + 10) – 9 (x + 10) = 0

=> (x + 10) (x – 9) = 0

Either x + 10 = 0, then x = -10 or x – 9 = 0, then x = 9.

The integer will be -10 or 9

**Answer
6** :

Let the given whole number = x

Then its reciprocal = 1/x

According to the condition,

x – 20 = 69 x (1/x)

=> x – 20 = 69/x

=> x² – 20x = 69

=> x² – 20x – 69 = 0

=> x² – 23x + 3x – 69 = 0

=> x (x – 23) + 3 (x – 23) = 0

=> (x – 23) (x + 3) = 0

Either x – 23 = 0, then x = 23

or x + 3 = 0, then x = -3, but it is not a whole number

Required whole number = 23

**Answer
7** :

Let first natural number = x

Then second number = x + 1

According to the condition,

x (x + 1) = 20

=> x² + x – 20 = 0

=> x² + 5x – 4x – 20 = 0

=> x (x + 5) – 4 (x + 5) = 0

=> (x + 5) (x – 4) = 0

Either x + 5 = 0, then x = -5 which is not a natural number

or x – 4 = 0, then x = 4

First natural number = 4 and second number = 4 + 1=5

**Answer
8** :

Let first odd number = 2x + 1

Then second odd number = 2x + 3

According to the condition

(2x + 1)2 + (2x + 3)2 = 394

=> 4x² + 4x + 1 + 4x² + 12x + 9 = 394

=> 8x² + 16x + 10 = 394

=> 8x² + 16x + 10 – 394 = 0

=> 8x² + 16x – 384 = 0

=> x² + 2x – 48 = 0 (Dividingby8)

=> x² + 8x – 6x – 48 = 0

=> x(x + 8) – 6(x + 8) = 0

=> (x + 8) (x – 6) = 0

Either x + 8 = 0, then x = 8 but it is not possible as it is negative

or x – 6 = 0, then x = 6

First odd number = 2x + 1 = 2 x 6 + 1 = 13

and second odd number = 13 + 2 = 15

**Answer
9** :

Sum of two numbers = 8

Let first number = x

Then second number = 8 – x

According to the condition,

(ii) If x = 5, then First number = 5 and second number = 8 – 5 = 3

Numbers are 3, 5

**Answer
10** :

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