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Class 10 NCERT Solutions- Chapter 4 Quadratic Equations – Exercise 4.1

  • Last Updated : 16 Sep, 2021

Question 1. Check whether the following are quadratic equations :

(i) (x + 1)2 = 2(x – 3) 

Solution:

Here,

LHS = (x + 1)2

= x2 + 2x + 1        (Using identity (a+b)2 = a2 + 2ab + b2)

and, RHS = 2(x–3)



= 2x – 6

As, LHS = RHS

x2 + 2x + 1 = 2x – 6

x2 + 7 = 0 ………….(I)

As, the eqn. (I) is in the form of ax2 + bx + c = 0 where (a≠0).

Hence, the equation is QUADRATIC equation because highest power of x is 2.

(ii) x2 – 2x = (–2) (3 – x)

Solution:

Here,

LHS = x2 – 2x

and, RHS = (–2) (3 – x)

= 2x–6

As, LHS = RHS

x2 – 2x = 2x – 6

x2 – 4x + 6 = 0 ………….(I)

As, the eqn. (I) is in the form of ax2 + bx + c = 0 where (a≠0).

Hence, the equation is QUADRATIC equation because highest power of x is 2.

(iii) (x – 2)(x + 1) = (x – 1)(x + 3) 

Solution:

Here,



LHS = (x – 2)(x + 1)

= x2 + (–2+1)x + (–2)(1)    (Using identity (x+a) (x+b) = x2 + (a+b)x + ab)

= x2 – x – 2

and, RHS = (x – 1)(x + 3) 

= x2 + (–1+3)x + (–1)(3)   (Using identity (x+a) (x+b) = x2 + (a+b)x + ab)

= x2 + 2x – 3

As, LHS = RHS

x2 – x – 2 = x2 + 2x – 3

3x – 1 = 0 ………….(I)

As, the eqn. (I) is not in the form of ax2 + bx + c = 0 because (a=0).

Hence, the equation is NOT QUADRATIC equation because highest power of x is 1.

(iv) (x – 3)(2x +1) = x(x + 5)

Solution:

Here,

LHS = (x – 3)(2x +1)

= 2x2 + x +(–3)(2x) + (–3)(1)

= 2x2 – 5x – 3

and, RHS = x(x + 5)

= x2 + 5x

As, LHS = RHS

2x2 – 5x – 3 = x2 + 5x



x2 – 10x – 3 = 0 ………….(I)

As, the eqn. (I) is in the form of ax2 + bx + c = 0 where (a≠0).

Hence, the equation is QUADRATIC equation because highest power of x is 2.

(v) (2x – 1)(x – 3) = (x + 5)(x – 1) 

Solution:

Here,

LHS = (2x – 1)(x – 3)

= 2x2 + (2x)(–3) +(–1)(x) + (–1)(–3)

= 2x2 – 7x + 3

and, RHS = (x + 5)(x – 1) 

= x2 + 5(x) + (–1)(x) + (5)(–1)     (Using identity (x+a) (x+b) = x2 + (a+b)x + ab)

= x2 + 4(x) – 5

As, LHS = RHS

2x2 – 7x + 3 = x2 + 4(x) – 5

x2 – 11x + 8 = 0 ………….(I)

As, the eqn. (I) is in the form of ax2 + bx + c = 0 where (a≠0).

Hence, the equation is QUADRATIC equation because highest power of x is 2.

(vi) x2 + 3x + 1 = (x – 2)2

Solution:

Here,

LHS = x2 + 3x + 1

and, RHS = (x – 2)2



= x2 – 4x + 4      (Using identity (ab)2 = a2 2ab + b2)

As, LHS = RHS

x2 + 3x + 1 = x2 – 4x + 4

7x – 3 = 0 ………….(I)

As, the eqn. (I) is not in the form of ax2 + bx + c = 0 because (a=0).

Hence, the equation is NOT QUADRATIC equation because highest power of x is 1.

(vii) (x + 2)3 = 2x (x2 – 1) 

Solution:

Here,

LHS = (x + 2)3

= x3 + 23 + 3x(2)(x+2)   (Using identity (x+y)3 = x3 + y3 + 3xy(x+y))

= x3 + 8 + 6x2 +12x

and, RHS = 2x (x2 – 1) 

= 2x3  – 2x      

As, LHS = RHS

x3 + 8 + 6x2 +12x = 2x3  – 2x 

x3 – 6x2 – 14x – 8 = 0 ………….(I)

As, the eqn. (I) is not in the form of ax2 + bx + c = 0 where (a≠0).

Hence, the equation is NOT QUADRATIC equation because highest power of x is 3.

(viii) x3 – 4x2 – x + 1 = (x – 2)3

Solution:

Here,

LHS = x3 – 4x2 – x + 1

and, RHS = (x – 2)3

= x3 – 23 – 3x(2)(x–2)     (Using identity (xy)3 = x3 y3 3xy(x-y))

= x3 – 8 – 6x2 +12x

As, LHS = RHS

x3 – 4x2 – x + 1 = x3 – 8 – 6x2 +12x

2x2 – 13x + 9 = 0 ………….(I)

As, the eqn. (I) is in the form of ax2 + bx + c = 0 where (a≠0).

Hence, the equation is QUADRATIC equation because highest power of x is 2.

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 meters) is one more than twice its breadth. We need to find the length and breadth of the plot.

Solution:



Let’s consider,

Breadth of the rectangular plot = b m

Then, length of the plot = (2b + 1) m.

As, Area of rectangle = length × breadth 

528 m2 = (2x + 1) × x

2x2 + x =528

2x2 + x – 528 = 0 ……………….(I)

As, the eqn. (I) is in the form of ax2 + bx + c = 0 where (a≠0).

Hence, the equation is QUADRATIC equation because highest power of x is 2.

Therefore, the length and breadth of plot, satisfies the quadratic equation, 2x2 + x – 528 = 0, which is the required representation of the problem mathematically.

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

Solution:

Let’s consider,

The first integer number = x

Next consecutive positive integer will be = x + 1

Product of two consecutive integers = x × (x +1)

x2 + x = 306

x2 + x – 306 = 0 ……………….(I)

As, the eqn. (I) is in the form of ax2 + bx + c = 0 where (a≠0).

Hence, the equation is QUADRATIC equation because highest power of x is 2.

Therefore, the two integers x and x+1, satisfies the quadratic equation, x2 + x – 306 = 0, which is the required representation of the problem mathematically.

(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’s consider,

Age of Rohan’s = x  years

So, Rohan’s mother’s age = x + 26

After 3 years,

Age of Rohan’s = x + 3

Age of Rohan’s mother will be = x + 26 + 3 = x + 29

The product of their ages after 3 years will be equal to 360, such that

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

x2 + 29x + 3x + 87 = 360

x2 + 32x + 87 – 360 = 0

x2 + 32x – 273 = 0 ……………….(I)

As, the eqn. (I) is in the form of ax2 + bx + c = 0 where (a≠0).

Hence, the equation is QUADRATIC equation because highest power of x is 2.

Therefore, the age of Rohan and his mother, satisfies the quadratic equation, x2 + 32x – 273 = 0, which is the required representation of the problem mathematically.

(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’s consider,

The speed of train = x  km/hr

Time taken to travel 480 km = (480/x) hr

Time = Distance / Speed



Here, According to the given condition,

The speed of train = (x – 8) km/hr

As, the train will take 3 hours more to cover the same distance.

Hence, Time taken to travel 480 km = 480/(x+3) km/hr

As,

Speed × Time = Distance

(x – 8)(480/(x + 3) = 480

480 + 3x – 3840/x – 24 = 480

3x – 3840/x = 24

3x2 – 8x – 1280 = 0 ……………….(I)

As, the eqn. (I) is in the form of ax2 + bx + c = 0 where (a≠0).

Hence, the equation is QUADRATIC equation because highest power of x is 2.

Therefore, the speed of the train, satisfies the quadratic equation, 3x2 – 8x – 1280 = 0, which is the required representation of the problem mathematically.

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