# NCERT Solutions for Class 10 Maths Chapter 4 - Quadratic Equations Exercise 4.1

NCERT Solutions are the best guide for students. NCERT Solutions Class 10 Maths Chapter 4 – Quadratic Equations Exercise 4.1 have all the solutions to the questions provided in the exercise on page no. 73. The subject experts have provided accurate stepwise NCERT Class 10 Maths Solutions to the questions.

NCERT Solutions Class 10 Maths provides solutions to all the exercises mentioned in the textbook. They also contain miscellaneous exercises for practice. The students are advised to practise all the questions repeatedly to excel in the board examinations. NCERT Solutions Class 10 Maths Chapter 4 – Quadratic Equations should be dealt with thoroughly. It is an important chapter from the examination perspective.

The more students practise, the better they become. In subjects like Maths, using reference materials to clarify the concepts is a must. NCERT Solutions are the best way to prepare for an examination. The students should solve all the questions provided here for better practice. The students can analyse their shortcomings and work on further improvement.

## NCERT Solutions for Class 10 Maths Chapter 4 – Quadratic Equations Exercise 4.1

### Access other exercise Solutions of Class 10 Maths Chapter 4 – Quadratic Equations

Exercise 4.2 Solutions– 6 Questions

Exercise 4.3 Solutions– 11 Questions

Exercise 4.4 Solutions– 5 Questions

### Access answers of Maths NCERT Class 10 Chapter 4 – Quadratic Equations Exercise 4.1

1. Check whether the following are quadratic equations:

(i) (x + 1)2 = 2(x â€“ 3)

(ii) x2 â€“ 2x = (â€“2) (3 â€“ x)

(iii) (x â€“ 2)(x + 1) = (x â€“ 1)(x + 3)

(iv) (x â€“ 3)(2x +1) = x(x + 5)

(v) (2x â€“ 1)(x â€“ 3) = (x + 5)(x â€“ 1)

(vi) x2 + 3x + 1 = (x â€“ 2)2

(vii) (x + 2)3 = 2x (x2 â€“ 1)

(viii) x3 â€“ 4x2 â€“ x + 1 = (x â€“ 2)3

Solutions:

(i) Given,

(x + 1)2 = 2(x â€“ 3)

By using the formula for (a+b)2 = a2+2ab+b2

â‡’Â x2Â + 2xÂ + 1 = 2xÂ â€“ 6

â‡’Â x2Â + 7 = 0

The above equation is in the form of ax2Â +Â bxÂ +Â cÂ = 0.

Therefore, the given equation is a quadratic equation.

(ii) Given, x2 â€“ 2x = (â€“2) (3 â€“ x)

â‡’Â x2Â â€“Â 2x = -6Â + 2x

â‡’Â x2Â â€“ 4xÂ + 6 = 0

The above equation is in the form of ax2Â +Â bxÂ +Â cÂ = 0.

Therefore, the given equation is a quadratic equation.

(iii) Given, (x â€“ 2)(x + 1) = (x â€“ 1)(x + 3)

By multiplication

â‡’Â x2Â â€“Â xÂ â€“ 2 =Â x2Â + 2xÂ â€“ 3

â‡’ 3xÂ â€“ 1 = 0

The above equation is not in the form of ax2Â +Â bxÂ +Â cÂ = 0.

Therefore, the given equation isÂ not a quadratic equation.

(iv) Given, (x â€“ 3)(2x +1) = x(x + 5)

By multiplication

â‡’ 2x2Â â€“ 5xÂ â€“ 3 =Â x2Â + 5x

â‡’ Â x2Â â€“ 10xÂ â€“ 3 = 0

The above equation is in the form of ax2Â +Â bxÂ +Â cÂ = 0.

Therefore, the given equation is a quadratic equation.

(v) Given, (2xÂ â€“ 1)(xÂ â€“ 3) = (xÂ + 5)(xÂ â€“ 1)

By multiplication

â‡’ 2x2Â â€“ 7xÂ +Â 3 =Â x2Â + 4xÂ â€“ 5

â‡’Â x2Â â€“ 11xÂ +Â 8 = 0

The above equation is in the form of ax2Â +Â bxÂ +Â cÂ = 0.

Therefore, the given equation is a quadratic equation.

(vi) Given, x2Â + 3xÂ + 1 = (xÂ â€“ 2)2

By using the formula for (a-b)2=a2-2ab+b2

â‡’Â x2Â + 3xÂ + 1Â =Â x2Â + 4Â â€“ 4x

â‡’ 7xÂ â€“ 3 = 0

The above equation is not in the form of ax2Â +Â bxÂ +Â cÂ = 0.

Therefore, the given equation isÂ not a quadratic equation.

(vii) Given, (xÂ + 2)3Â = 2x(x2Â â€“ 1)

By using the formula for (a+b)3 = a3+b3+3ab(a+b)

â‡’Â x3Â + 8Â +Â x2Â + 12xÂ = 2x3Â â€“ 2x

â‡’Â x3Â + 14xÂ â€“ 6x2Â â€“ 8 = 0

The above equation is not in the form of ax2Â +Â bxÂ +Â cÂ = 0.

Therefore, the given equation isÂ not a quadratic equation.

(viii) Given,Â x3Â – 4x2Â –Â xÂ + 1 = (xÂ – 2)3

By using the formula for (a-b)3 = a3-b3-3ab(a-b)

â‡’ Â x3Â – 4x2Â –Â xÂ + 1Â =Â x3Â – 8 – 6x2Â Â + 12x

â‡’ 2x2Â – 13xÂ + 9 = 0

The above equation is in the form of ax2Â +Â bxÂ +Â cÂ = 0.

Therefore, the given equation 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 positive integers is 306. We need to find the integers.

(iii) Rohanâ€™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 Rohanâ€™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

Solutions:

(i) Let us consider,

The breadth of the rectangular plot = x m

Thus, the length of the plot = (2x + 1) m

As we know,

Area of rectangle = lengthÂ Ã—Â breadth = 528 m2

Putting the value of the length and breadth of the plot in the formula, we get

(2xÂ + 1) Ã—Â xÂ = 528

â‡’ 2x2Â +Â xÂ =528

â‡’ 2x2Â +Â xÂ – 528 = 0

Therefore, the length and breadth of the plot satisfy the quadratic equation, 2x2Â +Â xÂ – 528 = 0, which is the required representation of the problem mathematically.

(ii) Let us consider,

The first integer number =Â x

Thus, the next consecutive positive integer will be =Â xÂ + 1

Product of two consecutive integers =Â xÂ Ã—Â (xÂ +1) = 306

â‡’Â x2Â +Â xÂ = 306

â‡’Â x2Â +Â xÂ â€“ 306 = 0

Therefore, the two integers, x and x+1, satisfy the quadratic equation, x2Â +Â xÂ â€“ 306 = 0, which is the required representation of the problem mathematically.

(iii) Let us consider,

Age of Rohanâ€™s =Â xÂ  years

Therefore, as per the given question,

Rohanâ€™s motherâ€™s age =Â xÂ + 26

After 3 years,

Age of Rohanâ€™s =Â xÂ + 3

Age of Rohanâ€™s mother will be =Â xÂ + 26 + 3 =Â xÂ + 29

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

(xÂ + 3)(xÂ + 29) = 360

â‡’Â x2Â + 29xÂ + 3xÂ + 87 = 360

â‡’Â x2Â + 32xÂ + 87 â€“ 360 = 0

â‡’Â x2Â + 32xÂ â€“ 273 = 0

Therefore, the age of Rohan and his mother satisfies the quadratic equation, x2Â + 32xÂ â€“ 273 = 0, which is the required representation of the problem mathematically.

(iv) Let us consider,

The speed of the train = xÂ  km/h

And

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

As per second condition, the speed of train = (xÂ – 8) km/h

Also given, the train will take 3 hours to cover the same distance.

Therefore, time taken to travel 480 km = (480/x)+3 km/h

As we know,

Speed Ã— Time = Distance

Therefore,

(xÂ – 8)(480/x)Â + 3 = 480

â‡’ 480Â + 3xÂ â€“ 3840/xÂ – 24 = 480

â‡’ 3xÂ â€“ 3840/xÂ = 24

â‡’ x2Â – 8xÂ – 1280 = 0

Therefore, the speed of the train satisfies the quadratic equation, x2Â – 8xÂ – 1280 = 0, which is the required representation of the problem mathematically.

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