# NCERT Solutions Exercise 4.1 Class 10 Maths

Q.1. Check whether the following are quadratic equations ÷

(i) (x + 1)2 = 2 (x – 3)

Solution:

(x + 1)2 = 2 (x – 3)

[By Applying ( a + b )² = a²+2ab+b2]

⇒ x²+2x+1= 2x – 6

⇒ x²+2x+1 – 2x + 6 = 0

⇒ x² + 7  = 0

By comparing ax²+bx+c= 0

a=1 , b=0 , c=7.

yes, it is a quadratic equation.

(ii) (x-2) (x+1) = (x-1) (x+3)

Solution:

⇒ x(x+1) -2(x+1) = x(x+3)-1(x+3)

⇒ x² + x – 2x – 2 = x² + 3x – x – 3

⇒ x² – x – 2-x² – 3x + x + 3 = 0

⇒ – 2 – 3x + 3 = 0

⇒ – 3x + 1 = 0

by comparing  ax² +bx+c=0 ; a≠0

a=0 , b=-3 , c=1

• No it is not a quadratic equation
•  It is a linear equation.

(iii) x² – 2x = (-2) (3-x)

⇒ x² – 2x = -6 +2x

⇒ x² – 2x + 6 – 2x = 0

⇒ x² + 6 = 0

⇒ x² + 0 x + 6 = 0

By comparing standard form

ax²+bx+c=0

a=1 , b=0 , c=6

Yes, it is a quadratic equations.

(iv) ( x – 3) (2x + 1) = x(x+5)

⇒ x(2x + 1) – 3(2x + 1) = x² + 5x

⇒ 2x² + x – 6x – 3 =  x² + 5x

⇒ 2x² – 5x – 3 – x²- 5x = 0

⇒ x²-10x-3 = 0

By Comparing.

a = 1

b = -10

c = -3

Yes it is a quadratic equation

(v) (2x-1) (x-3) = (x+5) (x-1)

Solution:

(2x-1) (x-3) = (x+5) (x-1)

⇒ 2x( x-3) -1(x-3) = x(x -1) + 5(x – 1)

⇒ 2x²- 6x – x + 3 = x²- x + 5x – 5

⇒ 2x² – 7x + 3 = x² + 4x – 5

⇒ 2x² – 7x + 3 – x² – 4x + 5 = 0

⇒ x²-11x + 8 = 0

By comparing ax² + bx + c = 0; a≠0

a = 1

b = -11

c = 8

Yes it is a quadratic equation

(vii) (x + 2)³ = 2x(x² – 1)

[By Applying identity (a + b)³ = a³ + b³ + 3a²b + 3ab²

⇒ x³+ (2)³+3(x)²(2)+ 3(x)(2)²=2x³-1

⇒ x³ + 8 + 6x² + 3x × 4 = 2x³-1

⇒ x³ + 8 + 6x² + 12x – 2x³ + 1 = 0

⇒ – x³ + 6x² + 12x + 9 = 0

• No it is not a quadratic equation
•  It is a quibic equation

(viii) x³- 4x²- x + 1 = (x – 2)³

Solution:

x³- 4x²- x + 1 = (x – 2)³

⇒ x³ – 4x² – x + 1 = x³ + 8 – 3(x)²(2) + 3(x)(2)²

⇒ x³ – 4x² – x + 1 = x³ + 8 – 6x² + 3(x)4

⇒ x³- 4x² – x + 1 = x³ + 8 – 6x² + 12x

⇒ x³ – 4x² – x + 1 = x³ – 6x² + 12x + 8

⇒ x³- 4x²- x + 1 – x³ + 6x² – 12x – 8 = 0

⇒ 2x² – 13x – 7 = 0

Yes it is a quadratic equation

2. Represent the following situation in the form of quadratic equations:

(i) The area of a rectangular plot is 528m². The length of the plot (in meters) is one more than twice its breadth.

we need to find the length and breadth of the plot.

Solution:

Let breadth of rectangular plot be = x m

Length = one more than twice its breadth

⇒ L = 2( Breadth) + 2

length(L) = 2x + 1

length(L) × breadth(B)    =   Area of rectangular plot

⇒ (2x+1) x = 528

⇒ 2x²+x = 528

⇒ 2x² + x – 528 = 0

by middle term splitting

⇒ 2x² + (33 – 32)x – 528 = 0

⇒ 2x²+33x-32x-528 = 0

⇒ x(2x + 33) – 16(2x + 33) = 0

⇒ (2x + 33)(x – 16) = 0

⇒ 2x + 33 = 0  ;  x – 16 = 0

⇒ 2x = – 33    ;  x = 16

⇒ x = -33/2 (not possible)

Hence

• length   =  2x+1= 2×16+1= 32+1=33metres

(ii) The product of two consecutive positive integers is 306. We need to find the integers .

Solution:

Let the two consecutive integers be = x , x+1

Product of two consecutive integers = 306

⇒ x(x + 1) = 306

⇒ x² + x – 306 = 0

⇒ x² + (18 – 17)x – 306 = 0

⇒ x² + 18x – 17x – 306 = 0

⇒ x(x + 18) – 17(x + 18) = 0

⇒ (x + 18)(x – 17) = 0

⇒ x + 18 = 0   ;  x – 17 = 0

⇒ x = – 18     ;      x = 17

Hence two consecutive  positive integers are  [17, (17+1)] = 17,18

(iii) Rohan’s mothers is 26 years older then him the product of their ages( in years) 3 years from now will be 360. we would like to find Rohan’s present age.

Solution:

Let Rohan’s present age be = x years.

His mother’s Present age be = (x+26) years .

3 years from now (present)

Rohan’s age will be =(x+3) years

His mother’s age will be = x+26+3 = (x+ 29) years.

Product of their ages will be = 360

(x+3) (x+29) = 360

⇒ x(x + 29) + 3(x + 29) = 360

⇒ x² + 29x + 3x + 87 = 360

⇒ x² + 32x +  87 – 360 = 0

⇒ x² + 32x – 273 = 0

⇒ x² + (39 – 7)x – 273 = 0

⇒ x² + 39x – 7x – 273 = 0

⇒ x(x + 39) -7(x + 39) = 0

⇒ (x + 39) (x – 7) = 0

⇒ x + 39 = 0   ;    x – 7 = 0

⇒ x = – 39     ;   x = 7

Hence Rohan’s Present age =7 years .

(iv) A train travels a distance of 480 km at a uniform speed. If the speed had been 8km/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:

Let the uniform speed of train be = x km/h

Case 1st

t1= 480/ x

case 2nd

When speed had been 8km/h less than actual speed of train = ( x – 8) km/h

t2 =   480/ (x – 8)

According to condition:

T1 + 3 = T2

480/x + 3 = 480/x – 8

⇒ (480 + 3x) (x – 8) = 480x

⇒ x(480 + 3x) – 8(480 + 3x) = 480x

⇒ 480x + 3x² – 3840 – 24x – 480x = 0

⇒ 3x² – 3840 – 24x = 0

⇒ 3x²- 24x – 3840 = 0

⇒3(x² – 8x – 1280) = 0

⇒ x² – 8x – 1280 = 0

⇒ x² – (40 – 32)x – 1280 = 0

⇒ x² – 40x – 32x – 1280 = 0

⇒ x(x – 40) + 32(x – 40) = 0

⇒ (x – 40) (x + 32) = 0

⇒ x – 40 = 0 ;  x + 32 = 0

⇒ x = 40  ;   x = – 32

Hence speed of train = 40km/h