# NCERT Solutions Maths for Class 10 Chapter 4

NCERT Solutions Maths for Class 10 Chapter 4, titled ‘Quadratic Equations,’ serve as an indispensable resource for students preparing for their CBSE exams. These solutions meticulously cover all the problems presented in the Class 10 Maths NCERT textbook, ensuring comprehensive exam readiness. Expertly framed and accurately solved by subject specialists, these solutions provide step-by-step guidance for students’ queries. In this chapter, dealing with quadratic equations, precision and understanding are crucial for success. The solutions not only aid in problem-solving but also offer valuable tips and tricks to simplify complex mathematical concepts. Quadratic equations find applications in various real-world scenarios, making it essential for students to grasp these concepts thoroughly. By mastering this chapter, students can enhance their mathematical skills and boost their exam performance, knowing that in mathematics, there’s a clear distinction between right and wrong answers. Therefore, focused practice with these solutions is the key to achieving full marks in mathematics.

## Exercise 4.1

1. Check whether the following are quadratic equations:

(i) (x + 1)2 = 2(x – 3)

(ii) x2 – 2x = (–2) (3 – x)

(iii) (x – 2)(x + 1) = (x – 1)(x + 3)

(iv) (x – 3)(2x +1) = x(x + 5)

(v) (2x – 1)(x – 3) = (x + 5)(x – 1)

(vi) x2 + 3x + 1 = (x – 2)2

(vii) (x + 2)3 = 2x (x2 – 1)

(viii) x3 – 4x2 – x + 1 = (x – 2)3

Solutions:

(i) Given,(x + 1)2 = 2(x – 3)

⇒ x2 + 2x + 1 = 2x – 6

⇒ x2 + 7 = 0

The above equation is in the form of ax2 + bx + c = 0

Therefore, the given equation is a quadratic equation.

(ii) Given, x2 – 2x = (–2) (3 – x)

⇒ x 2x = -6 + 2x

⇒ x– 4x + 6 = 0

Therefore, the given equation is a quadratic equation.

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

⇒ x– x – 2 = x+ 2x – 3

⇒ 3x – 1 = 0

This equation is not in the form of ax2 + bx + c = 0

Therefore, the given equation is not a quadratic equation.

(iv) Given, (x – 3)(2x +1) = x(x + 5)

⇒ 2x– 5x – 3 = x+ 5x

⇒  x– 10x – 3 = 0

This equation is in the form of ax2 + bx + c = 0

Therefore, the given equation is a quadratic equation.

(v) Given, (2x – 1)(x – 3) = (x + 5)(x – 1)

⇒ 2x– 7x + 3 = x+ 4x – 5

⇒ x– 11x + 8 = 0

This equation is in the form of ax2 + bx + c = 0.

Therefore, the given equation is a quadratic equation.

(vi) Given, x2 + 3x + 1 = (x – 2)2

⇒ x2 + 3x + 1 = x2 + 4 – 4x

⇒ 7x – 3 = 0

This equation is not in the form of ax2 + bx + c = 0

Therefore, the given equation is not a quadratic equation.

(vii) Given, (x + 2)3 = 2x(x2 – 1)

Applying formula for (a+b)= a3+b3+3ab(a+b)

⇒ x3 + 8 + x2 + 12x = 2x3 – 2x

⇒ x3 + 14x – 6x2 – 8 = 0

This equation is not in the form of ax2 + bx + c = 0

Therefore, the given equation is not a quadratic equation.

(viii) Given, x3 – 4x2 – x + 1 = (x – 2)3

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

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

⇒ 2x2 – 13x + 9 = 0

This equation is in the form of ax2 + bx + c = 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) 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

Solutions:

(i) Let the breadth of the rectangular plot = x m

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

Area of rectangle = length × breadth = 528 m2

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

(2x + 1) × x = 528

⇒ 2x2 + x =528

⇒ 2x2 + x – 528 = 0

Therefore, the quadratic equation, 2x2 + x – 528 = 0, which is the required representation of the question mathematically.

(ii)  Let the first integer number = x

and the next consecutive positive integer will be = x + 1

Product of two consecutive integers = x × (x +1) = 306

⇒ x+ x = 306

⇒ x+ x – 306 = 0

Therefore, the quadratic equation, x+ x – 306 = 0, which is the required representation of the question mathematically.

(iii) Age of Rohan’s = x  years

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 =  360

(x + 3)(x + 29) = 360

⇒ x2 + 29x + 3x + 87 = 360

⇒ x2 + 32x + 87 – 360 = 0

⇒ x2 + 32x – 273 = 0

Therefore, the quadratic equation, x2 + 32x – 273 = 0, which is the required representation of the question mathematically.

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

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

From second condition, the speed of train = (x – 8) km/h

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

we know that,

Speed × Time = Distance

Therefore,

(x – 8)[(480/x )+ 3] = 480

⇒ 480 + 3x – 3840/x – 24 = 480

⇒ 3x – 3840/x = 24

⇒ x– 8x – 1280 = 0

Therefore, the quadratic equation, x– 8x – 1280 = 0, which is the required representation of the question mathematically.

## Exercise 4.2

1. Find the roots of the following quadratic equations by factorisation:

(i) x2 – 3x – 10 = 0
(ii) 2x2 + x – 6 = 0
(iii) √2 x2 + 7x + 5√2 = 0
(iv) 2x2 – x +1/8 = 0
(v) 100x2 – 20x + 1 = 0

Solutions:

(i) Given, x2 – 3x – 10 =0

⇒  x2 – 5x + 2x – 10 = 0

⇒  x(– 5) + 2(x – 5) = 0

⇒  (x – 5)(x + 2) = 0

⇒  (x – 5)(x + 2) = 0

Therefore, x – 5 = 0 or x + 2 = 0

⇒ x = 5 or x = -2

(ii) Given, 2x2 + x – 6 = 0

⇒  2x2 + 4x – 3x – 6 = 0

⇒  2x(x + 2) – 3(x + 2) = 0

⇒  (x + 2)(2x – 3) = 0

⇒ (x + 2)(2x – 3) = 0

Therefore, x + 2 = 0 or 2x – 3 = 0

⇒  x = -2 or x = 3/2

(iii) √2 x2 + 7x + 5√2=0

⇒√2 x+ 5x + 2x + 5√2 = 0

x (√2x + 5) + √2(√2x + 5)= (√2x + 5)(+ √2)

⇒ (√2x + 5)(+ √2) = 0

Therefore, √2x + 5 = 0 or x + √2 = 0

x = -5/√2 or x = -√2

(iv) 2x2 – x +1/8 = 0

⇒ 1/8 (16x2  – 8x + 1) = 0

⇒  1/8 (16x2 – 4x – 4x + 1)

⇒  1/8 (4x(4x  – 1) -1(4x – 1))

⇒  1/8 (4– 1)2

⇒  (4– 1)2= 0

Therefore, (4x – 1) = 0 or (4x – 1) = 0

⇒ x = 1/4 or x = 1/4

(v) Given, 100x2 – 20x + 1=0

⇒  100x2 – 10x – 10x + 1

⇒ 10x(10x – 1) -1(10x – 1)

⇒  (10x – 1)2

⇒  (10x – 1)2= 0

Therefore, (10x – 1) = 0 or (10x – 1) = 0

⇒x = 1/10 or x = 1/10

2. Solve the problems given in Example 1.

Represent the following situations mathematically:

(i) John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 124. We would like to find out how many marbles they had to start with.

(ii) A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production was Rs. 750. We would like to find out the number of toys produced on that day.

Solutions: (i) Let us say the number of marbles John has = x

Therefore, the number of marble Jivanti has = 45 – x

After losing 5 marbles each,

Number of marbles John has = x – 5

Number of marble Jivanti has = 45 – x – 5 = 40 – x

Given that the product of their marbles is 124.

∴ (– 5)(40 – x) = 124

⇒ x2 – 45x + 324 = 0

⇒ x2 – 36x – 9x + 324 = 0

⇒ x(x – 36) -9(x – 36) = 0

⇒ (x – 36)(x – 9) = 0

Thus, we can say,

x – 36 = 0 or x – 9 = 0

⇒ x = 36 or x = 9

Therefore,

If John’s marbles = 36

Then, Jivanti’s marbles = 45 – 36 = 9

And if John’s marbles = 9

Then, Jivanti’s marbles = 45 – 9 = 36

(ii) Let us say the number of toys produced in a day is x.

Therefore, cost of production of each toy = Rs(55 – x)

Given the total cost of production of the toys = Rs 750

∴ x(55 – x) = 750

⇒ x2 – 55x + 750 = 0

⇒ x2 – 25x – 30x + 750 = 0

x(x – 25) -30(x – 25) = 0

⇒ (x – 25)(x – 30) = 0

Thus, either x -25 = 0 or x – 30 = 0

⇒ x = 25 or x = 30

Hence, the number of toys produced in a day will be either 25 or 30.

3. Find two numbers whose sum is 27 and product is 182.

Solution:

Let the first number is x, and the second number is 27 – x.

Therefore, the product of two numbers

x(27 – x) = 182

⇒ x2 – 27x – 182 = 0

⇒ x2 – 13x – 14x + 182 = 0

⇒ x(x – 13) -14(x – 13) = 0

⇒ (x – 13)(x -14) = 0

Thus, x = -13 = 0 or x – 14 = 0

⇒ x = 13 or x = 14

Therefore, if first number = 13, then second number = 27 – 13 = 14

And if first number = 14, then second number = 27 – 14 = 13

Hence, the numbers are 13 and 14.

4. Find two consecutive positive integers, the sum of whose squares is 365.

Solution:

Let the two consecutive positive integers are x and x + 1.

Therefore, according to the questions,

x2 + (x + 1)2 = 365

⇒ xx+ 1 + 2x = 365

⇒ 2x2 + 2x – 364 = 0

⇒ x– 182 = 0

⇒ x+ 14x – 13x – 182 = 0

⇒ x(x + 14) -13(x + 14) = 0

⇒ (x + 14)(x – 13) = 0

Thus, either, x + 14 = 0 or x – 13 = 0,

⇒ x = – 14 or x = 13

Since the integers are positive, x can be 13 only.

∴ x + 1 = 13 + 1 = 14

Therefore, two consecutive positive integers will be 13 and 14.

5. The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.

Solution:

Let the base of the right triangle is x cm.

Given, the altitude of right triangle = (x – 7) cm

From Pythagoras’ theorem, we know,

Base2 + Altitude2  = Hypotenuse2

∴ x+ (x – 7)2 = 132

⇒ x+ x+ 49 – 14x = 169

⇒ 2x– 14x – 120 = 0

⇒ x– 7x – 60 = 0

⇒ x– 12x + 5x – 60 = 0

⇒ x(x – 12) + 5(x – 12) = 0

⇒ (x – 12)(x + 5) = 0

Thus, either x – 12 = 0 or x + 5 = 0,

⇒ x = 12 or x = – 5

Since sides cannot be negative, x can only be 12.

Therefore, the base of the given triangle is 12 cm, and the altitude of this triangle will be (12 – 7) cm = 5 cm.

6. A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was Rs.90, find the number of articles produced and the cost of each article.

Solution:

Let the number of articles produced is x.

Therefore, cost of production of each article = Rs (2x + 3)

Given the total cost of production is Rs.90

∴ x(2x + 3) = 90

⇒ 2x+ 3x – 90 = 0

⇒ 2x+ 15x -12x – 90 = 0

⇒ x(2x + 15) -6(2x + 15) = 0

⇒ (2x + 15)(x – 6) = 0

Thus, either 2x + 15 = 0 or x – 6 = 0

⇒ x = -15/2 or x = 6

As the number of articles produced can only be a positive integer, x can only be 6.

Hence, the number of articles produced = 6

Cost of each article = 2 × 6 + 3 = Rs 15

## Exercise 4.4

1. Find the nature of the roots of the following quadratic equations. If the real roots exist, find them.
(i) 2x2 – 3x + 5 = 0
(ii) 3x2 – 4√3x + 4 = 0
(iii) 2x2 – 6x + 3 = 0

Solutions:

(i) Given,

2x2 – 3x + 5 = 0

Comparing the equation with ax2 + bx c = 0, we get

a = 2, b = -3 and c = 5

Discriminant = b2 – 4ac

( – 3)2 – 4 (2) (5) = 9 – 40

= – 31

b2 – 4ac < 0

Therefore, no real root is possible for the given equation, 2x2 – 3x + 5 = 0

(ii) 3x2 – 4√3x + 4 = 0

Comparing the equation with ax2 + bx c = 0, we get

a = 3, b = -4√3 and c = 4

We know, Discriminant = b2 – 4ac

= (-4√3)– 4(3)(4)

= 48 – 48 = 0

b2 – 4ac = 0,

Real roots exist for the given equation, and they are equal to each other.

Hence, the roots will be –b/2a and –b/2a.

b/2= -(-4√3)/2×3 = 4√3/6 = 2√3/3 = 2/√3

Therefore, the roots are 2/√3 and 2/√3.

(iii) 2x2 – 6x + 3 = 0

Comparing the equation with ax2 + bx c = 0, we get

a = 2, b = -6, c = 3

As we know, Discriminant = b2 – 4ac

= (-6)2 – 4 (2) (3)

= 36 – 24 = 12

As b2 – 4ac > 0,

Therefore, there are distinct real roots that exist for this equation, 2x2 – 6x + 3 = 0 = (6±2√3 )/4

= (3±√3)/2

Therefore, the roots for the given equation are (3+√3)/2 and (3-√3)/2

2. Find the values of k for each of the following quadratic equations so that they have two equal roots.
(i) 2x2 + kx + 3 = 0
(ii) kx (x – 2) + 6 = 0

Solutions:

(i) 2x2 + kx + 3 = 0

Comparing the given equation with ax2 + bx c = 0, we get,

a = 2, b = k and c = 3

Discriminant = b2 – 4ac

= (k)2 – 4(2) (3)

k2 – 24

For equal roots, we know,

Discriminant = 0

k2 – 24 = 0

k2 = 24

k = ±√24 = ±2√6

(ii) kx(x – 2) + 6 = 0

or kx2 – 2kx + 6 = 0

Comparing the given equation with ax2 + bx c = 0, we get

a = kb = – 2k and c = 6

We know, Discriminant = b2 – 4ac

= ( – 2k)2 – 4 (k) (6)

= 4k2 – 24k

For equal roots, we know,

b2 – 4ac = 0

4k2 – 24k = 0

4k (k – 6) = 0

Either 4k = 0 or k = 6 = 0

k = 0 or k = 6

However, if k = 0, then the equation will not have the terms ‘x2′ and ‘x’

Therefore, if this equation has two equal roots, k should be 6 only.

3. Is it possible to design a rectangular mango grove whose length is twice its breadth and the area is 800 m2? If so, find its length and breadth.

Solution: Let the breadth of the mango grove b =  l.

The length of the mango grove  =  2l.

Area of the mango grove = (2l) (l)= 2l2

2l= 800

l= 800/2 = 400

l– 400 =0

Comparing the given equation with ax2 + bx c = 0, we get

a = 1, b = 0, c = 400

As we know, Discriminant = b2 – 4ac

⇒  (0)2 – 4 × (1) × ( – 400) = 1600

Here, b2 – 4ac > 0

Thus, the equation will have real roots. And hence, the desired rectangular mango grove can be designed.

= ±20

As we know, the value of length cannot be negative.

Therefore, the breadth of the mango grove = 20 m

Length of mango grove = 2 × 20 = 40 m

4. Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is 20 years. Four years ago, the product of their age in years was 48.

Solution:

Let’s the age of one friend =  x years.

Then, the age of the other friend =  (20 – x) years.

Four years ago,

Age of First friend = (x – 4) years

Age of Second friend = (20 – x – 4) = (16 – x) years

As per the given question, we can write,

(x – 4) (16 – x) = 48

16x – x2 – 64 + 4x = 48

– x2 + 20x – 112 = 0

x2 – 20x + 112 = 0

Comparing the equation with ax2 + bx c = 0, we get

a = 1b = -20 and c = 112

Discriminant = b2 – 4ac

= (-20)2 – 4 × 112

= 400 – 448 = -48

b2 – 4ac < 0

Therefore, there will be no real solution possible for the equations. Hence, the condition doesn’t exist.

5. Is it possible to design a rectangular park of perimeter 80 and an area of 400 m2? If so, find its length and breadth.

Solution:

Let the length and breadth of the park be and b.

Perimeter of the rectangular park = 2 (l + b) = 80

So, l + b = 40

Or, b = 40 – l

Area of the rectangular park = l×b = l(40 – l) = 40– l= 400

l2   40l + 400 = 0, which is a quadratic equation.

Comparing the equation with ax2 + bx c = 0, we get

a = 1, b = -40, c = 400

Since, Discriminant = b2 – 4ac

=(-40)2 – 4 × 400

= 1600 – 1600 = 0

Thus, b2 – 4ac = 0

Therefore, this equation has equal real roots. Hence, the situation is possible.

l2   40l + 400 = 0

⇒ l² – 20l – 20l – 400 = 0

⇒ l(l – 20 ) – 20(l – 20 ) = 0

⇒ (l – 20)(l – 20) = 0

⇒ l =20, 20

Therefore, the length of the rectangular park, = 20 m

And the breadth of the park, = 40 – = 40 – 20 = 20 m.

## Ncert solution Class 10

Chapter 1: Real Number

Ncert maths solution class 10 chapter 1.1

Ncert maths solution class 10 chapter 1.2

Chapter 2: Polynomials

Ncert maths solutions for class 10 chapter 2 Polynomials

NCERT Solutions Maths for Class 10 Chapter 4

Chapter 5: Arithmetic Progression

Ncert solutions class 10 maths chapter 5 A.P. 