NCERT Solutions for Class 10 Maths Chapter 4 : Quadratic Equations

Q: Q1. Why Should I Practice NCERT Solutions Class 10 Maths Chapter 4?

Answer: Quadratic equation is a topic that is not only essential in mathematics but also plays a vital part in many real-life events. The quadratic equation can be used to calculate the length and width of a garden. You can plan the quantity of grass carpet needed for the garden based on this information. Quadratic equations are frequently employed in astronomy, science, and architecture. Because of its wide range of applications, students should thoroughly practise the NCERT Solutions Class 10 Maths Chapter 4.

Q: Q2. How many exercises are there in Chapter 4 of Class 10 Maths?

Answer: There are four exercises in the fourth chapter of NCERT Solutions for Class 10 Maths. Class 10 Maths Chapter 4 Quadratic Equations contains a total of 24 questions, 15 of which are simple, 5 of which are intermediate, and 4 of which are challenging. These questions are answered step by step. Students can answer all quadratic equation-based questions by completing these activities. In addition, the problems are answered in more than one way to help students learn basic quadratic equation ideas.

Q: Q3. What major topics are addressed in NCERT Solutions Class 10 Maths Chapter 4?

Answer: Quadratic Equations are the foundation of Chapter 4 of Class 10 Maths. The important topics covered in NCERT Solutions Class 10 Maths Chapter 4 are how to mathematically represent the given problem statements, what is the standard form of a quadratic equation, and how to solve quadratic equations by factoring and completing the squares, which is an essential topic that requires regular practice.

Q: Q4. In Class 10, how do you solve Quadratic Equations?

Answer: If you want to learn how to solve quadratic equations in Class 10, you can use the Toppr website or app to access NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations. All solutions are written in simple language by specialists. The equations are easily understood by students. Students must use the quadratic formula to discover the roots. They can compute the sum and product of both roots. The procedure is straightforward and well-explained for clarity.

Question 1

Check whether the following are quadratic equations :
(i)

(x + 1)^{2} = 2 (x - 3)

(ii)

x^{2} - 2 x = (- 2) (3 - x)

(iii)

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

(iv)

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

(v)

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

(vi)

x^{2} + 3 x + 1 = (x - 2)^{2}

(vii)

(x + 2)^{3} = 2 x (x^{2} - 1)

(viii)

x^{3} - 4 x^{2} - x + 1 = (x - 2)^{3}

Medium

Solution

Verified by Toppr

Answer

i)

(x + 1)^{2} = 2 (x - 3)

x^{2} + 2 x + 1 = 2 x - 6

x^{2} + 7 = 0 \to

Quadratic

ii)

x^{2} - 2 x = - 2 (3 - x)

x^{2} - 2 x = - 6 + 2 x

x^{2} + 4 x + 6 = 0 \to

Quadratic

iii)

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

x^{2} - x - 2 = x^{2} + 2 x - 3

3 x = 1 \to

Non Quadratic

iv)

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

2 x^{2} - 5 x - 3 = x^{2} + 5 x

x^{2} + 10 x - 3 = 0 \to

Quadratic

v)

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

2 x^{2} - 7 x + 3 = x^{2} + 4 x - 5

x^{2} - 11 x + 8 = 0 \to

Quadratic

vi)

x^{2} + 3 x + 1 = (x - 2)^{2}

x^{2} + 3 x + 1 = x^{2} - 4 x + 4

7 x = 3 \to

Non Quadratic

vii)

(x + 2)^{3} = 2 x (x^{2} - 1)

x^{3} + 8 + 6 x (x + 2) = 2 x^{3} - 2 x

x^{3} + 8 + 6 x^{2} + 12 x = 2 x^{3} - 2 x

- x^{3} + 6 x^{2} + 14 x + 8 = 0 \to

Non Quadratic

viii)

x^{3} - 4 x^{2} - x + 1

= x^{3} + 3 x^{2} (- 2) + 3 x (- 2)^{2} + (- 2)^{3}

x^{3} - 4 x^{2} - x + 1 = x^{3} - 6 x^{2} + 12 x - 8

x^{3} - 4 x^{2} - x + 1 - x^{3} + 6 x^{2} - 12 x + 8 = 0

2 x^{2} - 13 x + 9 = 0 \to

Non Quadratic

Answer

i)

(x + 1)^{2} = 2 (x - 3)

x^{2} + 2 x + 1 = 2 x - 6

x^{2} + 7 = 0 \to

Quadratic

ii)

x^{2} - 2 x = - 2 (3 - x)

x^{2} - 2 x = - 6 + 2 x

x^{2} + 4 x + 6 = 0 \to

Quadratic

iii)

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

x^{2} - x - 2 = x^{2} + 2 x - 3

3 x = 1 \to

Non Quadratic

iv)

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

2 x^{2} - 5 x - 3 = x^{2} + 5 x

x^{2} + 10 x - 3 = 0 \to

Quadratic

v)

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

2 x^{2} - 7 x + 3 = x^{2} + 4 x - 5

x^{2} - 11 x + 8 = 0 \to

Quadratic

vi)

x^{2} + 3 x + 1 = (x - 2)^{2}

x^{2} + 3 x + 1 = x^{2} - 4 x + 4

7 x = 3 \to

Non Quadratic

vii)

(x + 2)^{3} = 2 x (x^{2} - 1)

x^{3} + 8 + 6 x (x + 2) = 2 x^{3} - 2 x

x^{3} + 8 + 6 x^{2} + 12 x = 2 x^{3} - 2 x

- x^{3} + 6 x^{2} + 14 x + 8 = 0 \to

Non Quadratic

viii)

x^{3} - 4 x^{2} - x + 1

= x^{3} + 3 x^{2} (- 2) + 3 x (- 2)^{2} + (- 2)^{3}

x^{3} - 4 x^{2} - x + 1 = x^{3} - 6 x^{2} + 12 x - 8

x^{3} - 4 x^{2} - x + 1 - x^{3} + 6 x^{2} - 12 x + 8 = 0

2 x^{2} - 13 x + 9 = 0 \to

Non Quadratic

Question 2

Represent the following situations in the form of quadratic equations :
(i) The area of a rectangular plot is

528 m^{2}

. 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) Rohans 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 Rohans 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 3 hours more to cover the same distance. We need to find the speed of the train.

Hard

Solution

Verified by Toppr

Answer

i)

Let the breadth be

x

m and the length will be

2 x + 1

m.

Area

= l \times b

Area

= x (2 x + 1) = 528

2 x^{2} + x - 528 = 0 2 x^{2} + 33 x - 32 x - 528 = 0

\Rightarrow 2 x (x - 16) + 33 (x - 16) = 0

\Rightarrow (2 x + 33) (x - 16) = 0

\Rightarrow x = 16, \frac{3 3}{2}

Breadth

= 16

m and length

= 33

m

ii)

Let one number be

x

then the next number will be

x + 1

x (x + 1) = 306

\Rightarrow x^{2} + x - 306 = 0

\Rightarrow x^{2} + 18 x - 17 x - 306 = 0

\Rightarrow x (x + 18) - 17 (x + 18) = 0

\Rightarrow (x + 18) (x - 17) = 0

\Rightarrow x = 17, - 18

The numbers are

17

and

18 .

iii)

Let Rohan's present age

= x

yrs.

Then his mother's present age

= x + 26

yrs

After

3

yrs

Rohan's age

= x + 3

yrs

His mother's age

= x + 29

yrs

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

\Rightarrow x^{2} + 32 x + 87 - 360 = 0

\Rightarrow x^{2} + 32 x - 273 = 0

\Rightarrow x^{2} + 39 x - 7 x - 273 = 0

\Rightarrow x (x + 39) - 7 (x + 39) = 0

\Rightarrow (x + 39) (x - 7) = 0

\Rightarrow x = 7, - 39

So, Rohan's present age

= 7

yrs.

iv)

Let the speed of the train

= x

km/hr

\frac{4 8 0}{( x - 8 )} - \frac{4 8 0}{x} = 3

\Rightarrow 480 x - 480 x + 3840 = 3 x (x - 8)

\Rightarrow 3 x^{2} - 24 x - 3840 = 0

\Rightarrow x^{2} - 8 x - 1280 = 0

\Rightarrow x^{2} - 40 x + 32 x - 1280 = 0

\Rightarrow x (x - 40) + 32 (x - 40) = 0

\Rightarrow (x + 32) (x - 40) = 0

\Rightarrow x = 40, - 32

The speed of the train is

40

km/hr.

Answer

i)

Let the breadth be

x

m and the length will be

2 x + 1

m.

Area

= l \times b

Area

= x (2 x + 1) = 528

2 x^{2} + x - 528 = 0 2 x^{2} + 33 x - 32 x - 528 = 0

\Rightarrow 2 x (x - 16) + 33 (x - 16) = 0

\Rightarrow (2 x + 33) (x - 16) = 0

\Rightarrow x = 16, \frac{3 3}{2}

Breadth

= 16

m and length

= 33

m

ii)

Let one number be

x

then the next number will be

x + 1

x (x + 1) = 306

\Rightarrow x^{2} + x - 306 = 0

\Rightarrow x^{2} + 18 x - 17 x - 306 = 0

\Rightarrow x (x + 18) - 17 (x + 18) = 0

\Rightarrow (x + 18) (x - 17) = 0

\Rightarrow x = 17, - 18

The numbers are

17

and

18 .

iii)

Let Rohan's present age

= x

yrs.

Then his mother's present age

= x + 26

yrs

After

3

yrs

Rohan's age

= x + 3

yrs

His mother's age

= x + 29

yrs

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

\Rightarrow x^{2} + 32 x + 87 - 360 = 0

\Rightarrow x^{2} + 32 x - 273 = 0

\Rightarrow x^{2} + 39 x - 7 x - 273 = 0

\Rightarrow x (x + 39) - 7 (x + 39) = 0

\Rightarrow (x + 39) (x - 7) = 0

\Rightarrow x = 7, - 39

So, Rohan's present age

= 7

yrs.

iv)

Let the speed of the train

= x

km/hr

\frac{4 8 0}{( x - 8 )} - \frac{4 8 0}{x} = 3

\Rightarrow 480 x - 480 x + 3840 = 3 x (x - 8)

\Rightarrow 3 x^{2} - 24 x - 3840 = 0

\Rightarrow x^{2} - 8 x - 1280 = 0

\Rightarrow x^{2} - 40 x + 32 x - 1280 = 0

\Rightarrow x (x - 40) + 32 (x - 40) = 0

\Rightarrow (x + 32) (x - 40) = 0

\Rightarrow x = 40, - 32

The speed of the train is

40

km/hr.

Question 3

Find the roots of the following quadratic equations by factorisation:
(i)

x^{2} - 3 x - 10 = 0

(ii)

2 x^{2} + x - 6 = 0

(iii)

2 x^{2} + 7 x + 52 = 0

(iv)

2 x^{2} - x + \frac{1}{8} = 0

(v)

100 x^{2} - 20 x + 1 = 0

Medium

Solution

Verified by Toppr

Answer

i)

x^{2} - 3 x - 10 = 0

\Rightarrow x^{2} - 5 x + 2 x - 10 = 0

\Rightarrow x (x - 5) + 2 (x - 5) = 0

\Rightarrow (x + 2) (x - 5) = 0

\Rightarrow x = 5, - 2

ii)

2 x^{2} + x - 6 = 0

\Rightarrow 2 x^{2} + 4 x - 3 x - 6 = 0

\Rightarrow 2 x (x + 2) - 3 (x + 2) = 0

\Rightarrow (x + 2) (2 x - 3) = 0

\Rightarrow x = \frac{3}{2}, - 2

iii)

2 x^{2} + 7 x + 52 = 0

\Rightarrow 2 x^{2} + 2 x + 5 x + 52 = 0

\Rightarrow 2 x (x + 2) + 5 (x + 2) = 0

\Rightarrow (x + 2) (2 x + 5) = 0

\Rightarrow x = - 2, \frac{- 5}{2}

iv)

2 x^{2} - x + \frac{1}{8} = 0

\Rightarrow 16 x^{2} - 8 x + 1 = 0

\Rightarrow 16 x^{2} - 4 x - 4 x + 1 = 0

\Rightarrow 16 x (x - \frac{1}{4}) - 4 (x - \frac{1}{4}) = 0

\Rightarrow (x - \frac{1}{4}) (16 x - 4) = 0

\Rightarrow x = \frac{1}{4}, \frac{1}{4}

v)

100 x^{2} - 20 x + 1 = 0

\Rightarrow 100 x^{2} - 10 x - 10 x + 1 = 0

\Rightarrow 10 x (10 x - 1) - 1 (10 x - 1) = 0

\Rightarrow (10 x - 1) (10 x - 1) = 0

\Rightarrow x = \frac{1}{1 0}, \frac{1}{1 0}

Answer

i)

x^{2} - 3 x - 10 = 0

\Rightarrow x^{2} - 5 x + 2 x - 10 = 0

\Rightarrow x (x - 5) + 2 (x - 5) = 0

\Rightarrow (x + 2) (x - 5) = 0

\Rightarrow x = 5, - 2

ii)

2 x^{2} + x - 6 = 0

\Rightarrow 2 x^{2} + 4 x - 3 x - 6 = 0

\Rightarrow 2 x (x + 2) - 3 (x + 2) = 0

\Rightarrow (x + 2) (2 x - 3) = 0

\Rightarrow x = \frac{3}{2}, - 2

iii)

2 x^{2} + 7 x + 52 = 0

\Rightarrow 2 x^{2} + 2 x + 5 x + 52 = 0

\Rightarrow 2 x (x + 2) + 5 (x + 2) = 0

\Rightarrow (x + 2) (2 x + 5) = 0

\Rightarrow x = - 2, \frac{- 5}{2}

iv)

2 x^{2} - x + \frac{1}{8} = 0

\Rightarrow 16 x^{2} - 8 x + 1 = 0

\Rightarrow 16 x^{2} - 4 x - 4 x + 1 = 0

\Rightarrow 16 x (x - \frac{1}{4}) - 4 (x - \frac{1}{4}) = 0

\Rightarrow (x - \frac{1}{4}) (16 x - 4) = 0

\Rightarrow x = \frac{1}{4}, \frac{1}{4}

v)

100 x^{2} - 20 x + 1 = 0

\Rightarrow 100 x^{2} - 10 x - 10 x + 1 = 0

\Rightarrow 10 x (10 x - 1) - 1 (10 x - 1) = 0

\Rightarrow (10 x - 1) (10 x - 1) = 0

\Rightarrow x = \frac{1}{1 0}, \frac{1}{1 0}

Question 4

Solve:

(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

R s . 750

. We would like to find out the number of toys produced on that day.

Hard

Solution

Verified by Toppr

Answer

(i)

Let John have

x

marbles, so Jayanti will have

(45 - x)

marbles.

After losing

5

marbles,

John will have

(x - 5)

marbles

Jivanti will have

(45 - x - 5) = (40 - x)

marbles

As per the given condition,

(x - 5) (40 - x) = 124

\Rightarrow - (x - 5) (x - 40) = 124

\Rightarrow - x^{2} + 45 x - 200 = 124

\Rightarrow x^{2} - 45 x + 324 = 0

\Rightarrow x^{2} - 36 x - 9 x - 324 = 0

\Rightarrow x (x - 36) - 9 (x - 36) = 0

\Rightarrow (x - 36) (x - 9) = 0

\Rightarrow x - 36 = 0

or

x - 9 = 0

\Rightarrow x = 36

or

x = 9

So, if John has

36

marbles, Jivanti will have

9

marbles.

If John has

9

marbles, Jivanti will have

36

marbles.

(i i)

Let the total number of toys produced in a day be

x

.

Cost of production of one toy

= R s . (55 - x)

∴ x (55 - x) = 750

\Rightarrow x^{2} - 55 x + 750 = 0

\Rightarrow x^{2} - 25 x - 30 x + 750 = 0

\Rightarrow x (x - 25) - 30 (x - 25) = 0

\Rightarrow (x - 25) (x - 30) = 0

\Rightarrow x - 25 = 0

or

x - 30 = 0

\Rightarrow x = 25

or

x = 30

So, total number of toys produced in a day are

25

or

30

.

Answer

(i)

Let John have

x

marbles, so Jayanti will have

(45 - x)

marbles.

After losing

5

marbles,

John will have

(x - 5)

marbles

Jivanti will have

(45 - x - 5) = (40 - x)

marbles

As per the given condition,

(x - 5) (40 - x) = 124

\Rightarrow - (x - 5) (x - 40) = 124

\Rightarrow - x^{2} + 45 x - 200 = 124

\Rightarrow x^{2} - 45 x + 324 = 0

\Rightarrow x^{2} - 36 x - 9 x - 324 = 0

\Rightarrow x (x - 36) - 9 (x - 36) = 0

\Rightarrow (x - 36) (x - 9) = 0

\Rightarrow x - 36 = 0

or

x - 9 = 0

\Rightarrow x = 36

or

x = 9

So, if John has

36

marbles, Jivanti will have

9

marbles.

If John has

9

marbles, Jivanti will have

36

marbles.

(i i)

Let the total number of toys produced in a day be

x

.

Cost of production of one toy

= R s . (55 - x)

∴ x (55 - x) = 750

\Rightarrow x^{2} - 55 x + 750 = 0

\Rightarrow x^{2} - 25 x - 30 x + 750 = 0

\Rightarrow x (x - 25) - 30 (x - 25) = 0

\Rightarrow (x - 25) (x - 30) = 0

\Rightarrow x - 25 = 0

or

x - 30 = 0

\Rightarrow x = 25

or

x = 30

So, total number of toys produced in a day are

25

or

30

.

Question 5

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

Medium

Solution

Verified by Toppr

Answer

Let the first number be

x

then the second number will be

27 - x .

$x(27-x)=182$$

\Rightarrow 27 x - x^{2} = 182

\Rightarrow x^{2} - 27 x + 182 = 0

\Rightarrow x^{2} - 13 x - 14 x + 182 = 0

\Rightarrow x (x - 13) - 14 (x - 13) = 0

\Rightarrow (x - 14) (x - 13) = 0

\Rightarrow x = 14, 13

If the first number is

14,

then the second number is

13

and if the first number is

13,

then the second number is

14

.

Answer

Let the first number be

x

then the second number will be

27 - x .

$x(27-x)=182$$

\Rightarrow 27 x - x^{2} = 182

\Rightarrow x^{2} - 27 x + 182 = 0

\Rightarrow x^{2} - 13 x - 14 x + 182 = 0

\Rightarrow x (x - 13) - 14 (x - 13) = 0

\Rightarrow (x - 14) (x - 13) = 0

\Rightarrow x = 14, 13

If the first number is

14,

then the second number is

13

and if the first number is

13,

then the second number is

14

.

Question 6

Two consecutive positive integers, sum of whose squares is

365

are

A

13, 14

B

14, 15

C

12, 13

D

11, 12

Medium

Solution

Verified by Toppr

Answer

Let the two consecutive positive integers be

x

and

x + 1

Then,

x^{2} + (x + 1)^{2} = 365

\Rightarrow x^{2} + x^{2} + 2 x + 1 = 365

\Rightarrow 2 x^{2} + 2 x - 364 = 0

\Rightarrow x^{2} + x - 182 = 0

Using the quadratic formula, we get

x = \frac{- 1 \pm 1 + 7 2 8}{2}

\Rightarrow \frac{- 1 \pm 2 7}{2}

\Rightarrow x = 13

and

x = - 14

But

x

is given to be a positive integer.

∴ x \neq = - 14

Hence, the two consecutive positive integers are

13

and

14

.

Answer

Let the two consecutive positive integers be

x

and

x + 1

Then,

x^{2} + (x + 1)^{2} = 365

\Rightarrow x^{2} + x^{2} + 2 x + 1 = 365

\Rightarrow 2 x^{2} + 2 x - 364 = 0

\Rightarrow x^{2} + x - 182 = 0

Using the quadratic formula, we get

x = \frac{- 1 \pm 1 + 7 2 8}{2}

\Rightarrow \frac{- 1 \pm 2 7}{2}

\Rightarrow x = 13

and

x = - 14

But

x

is given to be a positive integer.

∴ x \neq = - 14

Hence, the two consecutive positive integers are

13

and

14

.

Question 7

The altitude of a right triangle is

7 c m

less then its base. If the hypotenuse is

13 c m

, find the other two sides.

Hard

Solution

Verified by Toppr

Answer

Let $x$ be the base of the triangle, then the altitude will be $(x - 7)$ .

By Pythagoras theorem,

$x^{2} + (x - 7)^{2} = (13)^{2}$

$2 x^{2} - 14 x + 49 - 169 = 0$

$2 x^{2} - 14 x - 120 = 0$

$x^{2} - 7 x - 60 = 0$

$x^{2} - 12 x + 5 x - 60 = 0$

$(x - 12) (x + 5) = 0$

$x = 12, x = - 5$

Since the side of the triangle cannot be negative, so the base of the triangle is $12 c m$ and the altitude of the triangle will be $12 - 7 = 5 c m$ .

Answer

Let $x$ be the base of the triangle, then the altitude will be $(x - 7)$ .

By Pythagoras theorem,

$x^{2} + (x - 7)^{2} = (13)^{2}$

$2 x^{2} - 14 x + 49 - 169 = 0$

$2 x^{2} - 14 x - 120 = 0$

$x^{2} - 7 x - 60 = 0$

$x^{2} - 12 x + 5 x - 60 = 0$

$(x - 12) (x + 5) = 0$

$x = 12, x = - 5$

Since the side of the triangle cannot be negative, so the base of the triangle is $12 c m$ and the altitude of the triangle will be $12 - 7 = 5 c m$ .

Question 8

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.

Medium

Solution

Verified by Toppr

Answer

Let there be

x

articles.

Then the production cost of each article

= 3 + 2 x

Total production cost

= x (3 + 2 x) = 90

3 x + 2 x^{2} = 90 2 x^{2} + 3 x - 90 = 0 2 x^{2} + 15 x - 12 x - 90 = 0 2 x (x - 6) + 15 (x - 6) = 0 (2 x + 15) (x - 6) = 0 x = 6, - \frac{1 5}{2}

Number cannot be negative.

So, number of articles

= 6

and the cost of each article

= R s . 15

Answer

Let there be

x

articles.

Then the production cost of each article

= 3 + 2 x

Total production cost

= x (3 + 2 x) = 90

3 x + 2 x^{2} = 90 2 x^{2} + 3 x - 90 = 0 2 x^{2} + 15 x - 12 x - 90 = 0 2 x (x - 6) + 15 (x - 6) = 0 (2 x + 15) (x - 6) = 0 x = 6, - \frac{1 5}{2}

Number cannot be negative.

So, number of articles

= 6

and the cost of each article

= R s . 15

Question 9

Find the roots of the following quadratic equations, if they exist, by the method of completing the square :
(i)

2 x^{2} - 7 x + 3 = 0

(ii)

2 x^{2} + x - 4 = 0

(iii)

4 x^{2} + 43 x + 3 = 0

(iv)

2 x^{2} + x + 4 = 0

Medium

Solution

Verified by Toppr

Answer

a) .

2 x^{2} - 7 x + 3 = 0

\Rightarrow x^{2} - \frac{7}{2} x = - \frac{3}{2}

Adding

(\frac{7}{4})^{2}

on both sides

\Rightarrow x^{2} - \frac{7}{2} x + (\frac{7}{4})^{2} = \frac{- 3}{2} + (\frac{7}{4})^{2}

\Rightarrow (x - \frac{7}{4})^{2} = \frac{- 3}{2} + \frac{4 9}{1 6}

\Rightarrow (x - \frac{7}{4})^{2} = \frac{2 5}{1 6}

\Rightarrow (x - \frac{7}{4})^{2} = (\frac{5}{4})^{2}

Taking square root on both sides

\Rightarrow (x - \frac{7}{4}) = \pm \frac{5}{4}

\Rightarrow x - \frac{7}{4} = \frac{5}{4},

x - \frac{7}{4} = \frac{- 5}{4}

x = \frac{5}{4} + \frac{7}{4}

x = \frac{- 5}{4} + \frac{7}{4}

x = 3

x = \frac{1}{2}

b) .

2 x^{2} + x - 4 = 0

x^{2} + \frac{x}{2} = 2

Adding

(\frac{1}{4})^{2}

on both sides

\Rightarrow x^{2} + \frac{x}{2} + (\frac{1}{4})^{2} = 2 + (\frac{1}{4})^{2}

(x + \frac{1}{4})^{2} = 2 + \frac{1}{1 6}

(x + \frac{1}{4})^{2} = \frac{3 3}{1 6}

Taking square root on both sides

\Rightarrow x + \frac{1}{4} = \pm \frac{3 3}{4}

\Rightarrow x = \pm \frac{3 3}{4} - \frac{1}{4},

x = \frac{- 3 3}{4} - \frac{1}{4}

\Rightarrow x = \pm \frac{3 3 - 1}{4},

x = \frac{- 3 3 - 1}{4}

c) .

4 x^{2} + 432 + 3 = 0

\Rightarrow x^{2} + 3 x + \frac{3}{4} = 0

x^{2} + 3 x = \frac{- 3}{4}

Adding

(\frac{3}{2}) 62

on both sides

\Rightarrow x^{2} + 3 x + (\frac{3}{2})^{2} = \frac{- 3}{4} + (\frac{3}{2})^{2}

\Rightarrow (x + \frac{3}{2})^{2} = \frac{- 3}{4} + \frac{3}{4}

\Rightarrow (x + \frac{3}{2})^{2} = 0

x = \frac{- 3}{2}, \frac{- 3}{2}

same roots.

d) .

2 x^{2} + x + 4 = 0

\Rightarrow x^{2} + \frac{x}{2} + 2 = 0

x^{2} + \frac{x}{2} = - 2

Adding

(\frac{1}{4})^{2}

on both sides

\Rightarrow x^{2} + \frac{x}{2} + (\frac{1}{4})^{2} = - 2 + (\frac{1}{4}) 62

\Rightarrow (x + \frac{1}{4})^{2} = - 2 + \frac{1}{1 6}

\Rightarrow (x + \frac{1}{4})^{2} = - \frac{- 3 1}{1 6}

Hence, solved.

Answer

a) .

2 x^{2} - 7 x + 3 = 0

\Rightarrow x^{2} - \frac{7}{2} x = - \frac{3}{2}

Adding

(\frac{7}{4})^{2}

on both sides

\Rightarrow x^{2} - \frac{7}{2} x + (\frac{7}{4})^{2} = \frac{- 3}{2} + (\frac{7}{4})^{2}

\Rightarrow (x - \frac{7}{4})^{2} = \frac{- 3}{2} + \frac{4 9}{1 6}

\Rightarrow (x - \frac{7}{4})^{2} = \frac{2 5}{1 6}

\Rightarrow (x - \frac{7}{4})^{2} = (\frac{5}{4})^{2}

Taking square root on both sides

\Rightarrow (x - \frac{7}{4}) = \pm \frac{5}{4}

\Rightarrow x - \frac{7}{4} = \frac{5}{4},

x - \frac{7}{4} = \frac{- 5}{4}

x = \frac{5}{4} + \frac{7}{4}

x = \frac{- 5}{4} + \frac{7}{4}

x = 3

x = \frac{1}{2}

b) .

2 x^{2} + x - 4 = 0

x^{2} + \frac{x}{2} = 2

Adding

(\frac{1}{4})^{2}

on both sides

\Rightarrow x^{2} + \frac{x}{2} + (\frac{1}{4})^{2} = 2 + (\frac{1}{4})^{2}

(x + \frac{1}{4})^{2} = 2 + \frac{1}{1 6}

(x + \frac{1}{4})^{2} = \frac{3 3}{1 6}

Taking square root on both sides

\Rightarrow x + \frac{1}{4} = \pm \frac{3 3}{4}

\Rightarrow x = \pm \frac{3 3}{4} - \frac{1}{4},

x = \frac{- 3 3}{4} - \frac{1}{4}

\Rightarrow x = \pm \frac{3 3 - 1}{4},

x = \frac{- 3 3 - 1}{4}

c) .

4 x^{2} + 432 + 3 = 0

\Rightarrow x^{2} + 3 x + \frac{3}{4} = 0

x^{2} + 3 x = \frac{- 3}{4}

Adding

(\frac{3}{2}) 62

on both sides

\Rightarrow x^{2} + 3 x + (\frac{3}{2})^{2} = \frac{- 3}{4} + (\frac{3}{2})^{2}

\Rightarrow (x + \frac{3}{2})^{2} = \frac{- 3}{4} + \frac{3}{4}

\Rightarrow (x + \frac{3}{2})^{2} = 0

x = \frac{- 3}{2}, \frac{- 3}{2}

same roots.

d) .

2 x^{2} + x + 4 = 0

\Rightarrow x^{2} + \frac{x}{2} + 2 = 0

x^{2} + \frac{x}{2} = - 2

Adding

(\frac{1}{4})^{2}

on both sides

\Rightarrow x^{2} + \frac{x}{2} + (\frac{1}{4})^{2} = - 2 + (\frac{1}{4}) 62

\Rightarrow (x + \frac{1}{4})^{2} = - 2 + \frac{1}{1 6}

\Rightarrow (x + \frac{1}{4})^{2} = - \frac{- 3 1}{1 6}

Hence, solved.

Question 10

Find the roots of the quadratic equation by applying the quadratic formula

2 x^{2} - 7 x + 3 = 0

A

3, - \frac{1}{2}

B

- 3, \frac{1}{2}

C

3, \frac{1}{2}

D

None of these

Easy

Solution

Verified by Toppr

Answer

Given equation is

2 x^{2} - 7 x + 3 = 0

Hence,

a = 2, b = - 7, c = 3

Therefore,

x = \frac{- b \pm b ^{2} - 4 a c}{2 a}

= \frac{7 \pm 2 5}{4}

= \frac{7 \pm 5}{4}

Thus,

x = 3

and

x = \frac{1}{2}

Answer

Given equation is

2 x^{2} - 7 x + 3 = 0

Hence,

a = 2, b = - 7, c = 3

Therefore,

x = \frac{- b \pm b ^{2} - 4 a c}{2 a}

= \frac{7 \pm 2 5}{4}

= \frac{7 \pm 5}{4}

Thus,

x = 3

and

x = \frac{1}{2}

NCERT Solutions for Class 10 Maths Chapter 4 : Quadratic Equations

Download PDF of NCERT Solutions for Class 10 Maths Chapter 4 : Quadratic Equations

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NCERT Solutions for Class 10 Maths Chapter 4 : Quadratic Equations

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations – Brief Overview

1. Quadratic Equations

2. Zeroes/Roots of a Quadratic Equation

3. Solution of a Quadratic Equation by Factorisation Method

4. Solution of a Quadratic Equation by Completing the Square Method

5. Quadratic Formula

6. Nature of Roots

Frequently Asked Questions on NCERT Class 10 Maths Chapter 4 : Quadratic Equations