﻿ Class 10 NCERT Math Solution
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Exercise - 4.1

Question-1 :- Check whether the following are quadratic equations :
(i) (x + 1)² = 2(x – 3)
(ii) x² – 2x = (–2) (3 – x)
(iii) (x – 2)(x + 1) = (x – 1)(x + 3)
(iv) (x – 3)(2x +1) = x(x + 5)
(v) (2x – 1)(x – 3) = (x + 5)(x – 1)
(vi) x² + 3x + 1 = (x – 2)²
(vii) (x + 2)³ = 2x (x² – 1)
(viii) x³ – 4x² – x + 1 = (x – 2)³

Solution :-
```(i) (x + 1)² = 2(x – 3)
x² + 1 + 2x = 2x - 6
x² + 1 + 2x - 2x + 6 = 0
x² + 7 = 0
It is the form of ax² + bx + c = 0
So, it is a quadratic equation.
```
```(ii) x² – 2x = (–2) (3 – x)
x² - 2x = -6 + 2x
x² - 2x + 6 - 2x = 0
x² - 4x + 6 = 0
It is the form of ax² + bx + c = 0
So, it is a quadratic equation.
```
```(iii) (x – 2)(x + 1) = (x – 1)(x + 3)
x² + x - 2x - 2 = x² + 3x - x - 3
x² - x - 2 - x² - 2x + 3 = 0
-3x + 1 = 0
It is not in the form of ax² + bx + c = 0
So, it is not a quadratic equation.
```
```(iv) (x – 3)(2x + 1) = x(x + 5)
2x² + x - 6x - 3 = x² + 5x
2x² - 5x - 3 - x² - 5x = 0
x² - 10x - 3 = 0
It is the form of ax² + bx + c = 0
So, it is a quadratic equation.
```
```(v) (2x – 1)(x – 3) = (x + 5)(x – 1)
2x² - 6x - x + 3 = x² - x + 5x - 5
2x² - 7x + 3 - x² - 4x + 5 = 0
x² - 11x + 8 = 0
It is the form of ax² + bx + c = 0
So, it is a quadratic equation.
```
```(vi) x² + 3x + 1 = (x – 2)²
x² + 3x + 1 = x² + 4 - 4x
x² + 3x + 1 - x² - 4 + 4x = 0
7x - 3 = 0
It is not in the form of ax² + bx + c = 0
So, it is not a quadratic equation.
```
```(vii) (x + 2)³ = 2x (x² – 1)
x² + 4 + 4x = 2x³ - 2x
x² + 4 + 4x - 2x³ + 2x = 0
-2x³ + x² + 6x + 4 = 0
It is not the form of ax² + bx + c = 0
So, it is not a quadratic equation.
```
```(viii) x³ – 4x² – x + 1 = (x – 2)³
x³ - 4x² - x + 1 = x³ - 8 - 6x² + 12x
x³ - 4x² - x + 1 - x³ + 8 + 6x² - 12x = 0
2x² - 13x + 9 = 0
It is the form of ax² + bx + c = 0
So, it is a quadratic equation.
```

Question-2 :-  Represent the following situations in the form of quadratic equations :
(i) The area of a rectangular plot is 528 m². The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.
(ii) The product of two consecutive positive integers is 306. We need to find the integers.
(iii) Rohan’s mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohan’s present age.
(iv) A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.

Solution :-
```(i) Let the breadth of the plot = x m.
The length of the plot = (2x + 1) m.
Area of a rectangle = 528 m²

According to Question :
Area of a rectangle = Length × Breadth
528 = x(2x + 1)
2x² + x - 528 = 0
```
```(ii) Let the consecutive integers be x and x + 1.
It is given that their product is 306.

According to Question :
x(x + 1) = 306
x² + x - 306 = 0
```
```(iii) Let Rohan’s age = x.
His mother’s age = x + 26
3 years Now,
Rohan’s age = x + 3
Mother’s age = x + 26 + 3 = x + 29
It is given that the product of their ages after 3 years is 360.

According to Question :
(x + 3)(x + 29) = 360
x² + 29x + 3x + 87 - 360 = 0
x² + 32x - 273 = 0
```
```(iv) Let the speed of train be x km/h.
Time taken to travel 480 km = 480/x hrs
In second condition, let the speed of train =	(x -8) km/h

It is also given that the train will take 3 hours to cover the same distance.
Therefore, time taken to travel 480 km = (480/x + 3)hrs

According to Question :
Speed × Time = Distance
(x - 8)(480/x + 3) = 480
480 + 3x - 3840/x - 24 = 480
3x - 3840/x = 24
3x² - 24x + 3840 = 0
x² - 8x - 1280 = 0
```
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