NCERT Solutions Maths Class 10 Chapter 4

# NCERT Solutions Maths Class 10 Chapter 4

## NCERT Solutions Maths Class 10 Exercise 4.1

1. Check whether the following are Quadratic Equations.

(i) (x + 1)2 = 2(− 3)

(ii) x2 − 2x = (−2)(3 − x)

(iii) (− 2)(+ 1) = (− 1)(+ 3)

(iv) (− 3)(2+ 1) = x(+ 5)

(v) (2− 1)(− 3) = (+ 5)(− 1)

(vi) x2 + 3x + 1 = (x − 2)2

(vii) (x + 2)3 = 2x(x2 – 1)

(viii) x3 – 4x2x + 1 = (x – 2)3

Solution: (i) (x + 1)2 = 2(− 3)

x2 + 2x + 1 = 2– 6                           [Since (a + b)2 = a2 + 2ab + b2]

x2 + 7 = 0

Here, the degree of the equation is 2.

Therefore, it is a quadratic equation.

(ii) x2 − 2x = (−2)(3 − x)

x2 – 2x = −6 + 2x

x2 – 2x − 2+ 6 = 0

x2 – 4x + 6 = 0

Here, the degree of the equation is 2.

Therefore, it is a quadratic equation.

(iii) (− 2)(+ 1) = (− 1)(+ 3)

x2 + x – 2x – 2 x2 + 3x x – 3

x2 + x – 2x – 2 – x2 – 3x + x + 3 = 0

x − 2x – 2 − 3x + x + 3 = 0

3+ 1 = 0

Here, the degree of the equation is 1.

Therefore, it is not a quadratic equation.

(iv) (− 3)(2+ 1) = x(+ 5)

2x2 + x – 6x – 3 = x2 + 5x

2x2 + x – 6x – 3 – x2 – 5x = 0

x2 – 10x – 3 = 0

Here, the degree of the equation is 2.

Therefore, it is a quadratic equation.

(v) (2− 1)(− 3) = (+ 5)(− 1)

2x2 – 6xx + 3 = x2 x + 5x – 5

2x2 – 7x + 3 – x2 + x – 5x + 5 = 0

x− 11+ 8 = 0

Here, the degree of the equation is 2.

Therefore, it is a quadratic equation.

(vi) x2 + 3x + 1 = (x − 2)2

x2 + 3x + 1 = x2 – 4x + 4              [ Since (a b)2 = a2 2ab + b2]

x2 + 3x + 1 x2 + 4x – 4 = 0

7– 3 = 0

Here, the degree of the equation is 1.

Therefore, it is not a quadratic equation.

(vii) (x + 2)3 = 2x(x2 – 1)

x3 + 23 + 3(x)(2)(x + 2) = 2x(x2 – 1)         [Since (a + b)3 = a3 + b3 + 3ab(a + b)]

x3 + 8 + 6x(x + 2) = 2x3 – 2x

x3 + 8 + 6x2 + 12x = 2x3 – 2x

2x3 – 2xx3 – 8 – 6x2 – 12x = 0

x3 – 6x2 – 14x – 8 = 0

Here, the degree of the equation is 3.

Therefore, it is not a quadratic equation.

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(viii) x3 – 4x2x + 1 = (x – 2)3

x3 – 4x2 x + 1 = x3 – 23 – 3(x)(2)(− 2)           [Since (ab)3 = a3b3 – 3ab(ab)]

–4x2 x + 1 = –8 – 6x2 + 12x

2x2 – 13x + 9 = 0

Here, the degree of the equation is 2.

Therefore, it is a quadratic equation.

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 numbers 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) It is given that the area of a rectangular plot is 528 m2.

Let the breadth of the rectangular plot be x metres.

Then, according to the given condition, length = (2+ 1) metres

Area of a rectangle = length × breadth

528 = (2+ 1)x

528 = 2x2 + x

2x2 + x – 528 = 0

This is the required quadratic equation.

(ii) Let the two consecutive positive integers be x and (x + 1).

According to the given condition, we have

x(+ 1) = 306

x2 + x = 306

x2 + x – 306 = 0

This is the required quadratic equation.

(iii) Let the present age of Rohan be x years.

Then, the present age of Rohan’s mother = (x + 26) years

Age of Rohan after 3 years = (x + 3) years

Age of Rohan’s mother after 3 years = x + 26 + 3 = (x + 29) years

According to the given condition, we have

(+ 3)(+ 29) = 360

x2 + 29x + 3x + 87 = 360

x2 + 32x – 273 = 0

This is the required quadratic equation.

(iv) Let the speed of the train be x km/h.

Time taken by the train to cover 480 km = 480/x hours

If the speed had been 8 km/h less, then the time taken would be 480/(x − 8) hours

According to the given condition, if the speed had been 8 km/h less, then the time taken to cover the same distance is 3 hours more.

Therefore, 480/(x − 8) = 480/+ 3

480/(x − 8) − 480/= 3

480[x − (x − 8)] = 3x(x – 8)

480(– + 8) = 3x(x – 8)

480 × 8 = 3x(x – 8)

3840 = 3x2 – 24x

3x2 – 24x – 3840 = 0

Dividing the above equation by 3, we get

x2 – 8x – 1280 = 0

This is the required quadratic equation.