# NCERT Solution Exercise 4.1 Class10

Q.1 Check whether the Following are quadratic equations:

(i) (x+1)² = 2(x-3)

x² + 2x + 1 = 2x – 6

⇒ x² + 2x + 1 – 2x + 6 = 0

⇒ x² + 7 = 0

by comparing a x² + b x + c = 0

a = 1, b = 0 c = 7

Yes  it is a Quadratic Equations

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

⇒ x(x+1)-2(x+1) = x(x+3) – 1(x+3)

⇒ x² + x – 2x – 2 = x² + 3x – x – 3

⇒ x² + x – 2x – 2 – x² – 3x + x + 3 = 0

⇒ – 3x + 1 = 0

By comparing a x² + b x + c = 0 ; a ≠ 0

a = 0, b = – 3, and 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 + 0x + 6 = 0

By comparing standard form : a x² + b x + c = 0

We get

a = 1, b = 0 and c = 6

(iv) ( x – 3 )( 2x + 1 ) = x ( x + 5 )

⇒ x(2x + 1) – 3(2x + 1) = x² + 5x

⇒ 2x² + x – 6x -3 – x² – 5x = 0

⇒ x² – 10x – 3 = 0

By comparing standard of the Quadratic Equation  a x² + b x + c = 0

a = 1 , b = -10 and c = -3

Yes it is a quadratic equations

(v) (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 Standard form of the Quadratic Equation: a x² + b x + c = 0

We get

a = 1, b = – 11 and c = 8

Yes it is Quadratic a Equations

(vi)(x + 2)³ = 2x (x² – 1)

[by using identity (a + b)³ = a³ + b³ + 3a²b + 3ab²]

⇒ x³ + (2)³ + 3(x)²×2 + 3 x 2² = 2x³ – 2x

⇒ x³ + 8 + 6x² + 12x = 2x³ – 2x

⇒ x³ + 8 + 6x² + 12x – 2x³ + 2x = 0

⇒ – x³ + 6x² + 14x + 8 = 0

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

(vii) x³ – 4x² – x + 1 = (x – 2)³

⇒ x³ – 4x² – x + 1 = x³ – 2³ – 3x²×2 + 3x×2²

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

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

⇒ 2x² – 13x + 9 = 0

By Comparing Standard Form of the  Quadratic Equation: a x² + b x + c = 0

We get

a = 2 , b = -13 and c = 9

Yes it is a quadratic equations

Q.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 meters) is one more than twice its breadth we need to find the length and breadth if the plot.

Solution:

Let breadth of rectangular plot be = x m

Length  = one more than twice its breadth

⇒ Length = 2x + 1

Length × Breadth = 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 Breadth = 16 m and

Length = 2x+1

⇒ Length = 2× 16 + 1 = 33 m

(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

⇒ x(x+1) – 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, 18

(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.

Solution:

Let Rohan’s Present age be = x years

His mothers age will 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

⇒ x² + 32x + 87 – 360 = 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 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:

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

Case I :

T1 = 480/x

Case II :

When speed had been 8 km/hrs. less then actual speed of train = (x – 8)km/hrs.

T2 = 480/(x-8)

According to Condition:

T1 + 3 = T2

480/x + 3 = 480/(x – 8)

⇒ (480 + 3x) / x = 480 /(x -8)

⇒ (480 + 3x)(x -8) = 480x

⇒ 480(x – 8) + 3x(x – 8) = 480x

⇒ 480x +  3840 + 3x² – 24x – 480x = 0

⇒ 3x² – 24x – 3840 = 0

⇒ 3( x² – 8x – 1280 ) = 0

⇒  x² – 8x – 1280 =0/3

⇒  x² – 8x – 1280 =0

By middle term splitting

⇒ 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 = 40 km/h