**Quadratic Equations**. In the beginning, an example of a rectangular prayer hall is given and from the given situation length and breadth of a rectangular hall is obtained. This example is given to show how the quadratic equations can be used to solve real-life problems.

**is the**

*ax*^{2 }+*bx*+*c*= 0**standard form of a quadratic equation**where

**In the first exercise, students will learn how to check whether the given equation is**

*a*$\ne $ 0.**quadratic or not**and representing the same. Various word problems are given and students need to form quadratic equation from the given word problem.

The next section is about the topic-

**Solution of Quadratic equation by factorisation**.

**Roots**of the equation can be found by

**factorising the equation**into two

**linear factors**and then equating each

**factor**to

**zero**. In exercise 4.2 students have to find

**roots**of the following equation by

**factorisation**method.

The next part is about the solution of a

**Quadratic equation by completing the squares**. This concept is made more understandable by the illustration of figures. Problems given in the next exercise is based on the same concept.

The next method given is the

**use of**

**quadratic formula**. This is the most important method used to solve the

**quadratic equation.**

In exercise 4.3 students will find several word problems that will help them to make their concept more clear.

Thus to conclude there are three methods to solve a

**quadratic equation:**

**FACTORISATION METHOD****COMPLETING SQUARES****QUADRATIC FORMULA**

**nature of roots**is discussed. Based on the value of

**discriminant,**a

**quadratic equation**can have:

**TWO DISTINCT REAL ROOTS**:**Discriminant**is**greater than 0**.**TWO REAL EQUAL ROOTS**:**Discriminant**is**equal to 0**.**NO REAL ROOTS**:**Discriminant**is**less than 0**.

**quadratic equation**more clearly.

In the end, for recapitulation, a summary of the chapter is given.

#### Page No 73:

#### Question 1:

Check whether the following are quadratic equations:

#### Answer:

It is of the form .

Hence, the given equation is a quadratic equation.

It is of the form .

Hence, the given equation is a quadratic equation.

It is not of the form .

Hence, the given equation is not a quadratic equation.

It is of the form .

Hence, the given equation is a quadratic equation.

It is of the form .

Hence, the given equation is a quadratic equation.

It is not of the form .

Hence, the given equation is not a quadratic equation.

It is not of the form .

Hence, the given equation is not a quadratic equation.

It is of the form .

Hence, the given equation is a quadratic equation.

##### Video Solution for quadratic equations (Page: 73 , Q.No.: 1)

NCERT Solution for Class 10 math - quadratic equations 73 , Question 1

#### Page No 73:

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

#### Answer:

(i) Let the breadth of the plot be *x* m.

Hence, the length of the plot is (2*x* + 1) m.

Area of a rectangle = Length × Breadth

∴ 528 = *x *(2*x* + 1)

(ii) Let the consecutive integers be *x* and *x* + 1.

It is given that their product is 306.

∴

(iii) Let Rohan’s age be *x*.

Hence, his mother’s age = *x* + 26

3 years hence,

Rohan’s age = *x* + 3

Mother’s age = *x* + 26 + 3 = *x* + 29

It is given that the product of their ages after 3 years is 360.

(iv) Let the speed of train be *x* km/h.

Time taken to travel 480 km =

In second condition, let the speed of train = km/h

It is also given that the train will take 3 hours to cover the same distance.

Therefore, time taken to travel 480 km = hrs

Speed × Time = Distance

$\Rightarrow 480+3x-\frac{3840}{x}-24=480$

$\Rightarrow 3x-\frac{3840}{x}=24$

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

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

##### Video Solution for quadratic equations (Page: 73 , Q.No.: 2)

NCERT Solution for Class 10 math - quadratic equations 73 , Question 2

#### Page No 76:

#### Question 1:

Find the roots of the following quadratic equations by factorisation:

#### Answer:

Roots of this equation are the values for which= 0

∴ = 0 or = 0

i.e., *x* = 5 or *x *= −2

Roots of this equation are the values for which= 0

∴ = 0 or = 0

i.e., *x* = −2 or *x *=

Roots of this equation are the values for which = 0

∴ = 0 or = 0

i.e., *x *= or* x* =

Roots of this equation are the values for which = 0

Therefore,

i.e.,

Roots of this equation are the values for which = 0

Therefore,

i.e.,

##### Video Solution for quadratic equations (Page: 76 , Q.No.: 1)

NCERT Solution for Class 10 math - quadratic equations 76 , Question 1

#### Page No 76:

#### Question 2:

(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. 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. Find out the number of toys produced on that day.

#### Answer:

(i) Let the number of John’s marbles be *x*.

Therefore, number of Jivanti’s marble = 45 − *x*

After losing 5 marbles,

Number of John’s marbles = *x* − 5

Number of Jivanti’s marbles = 45 − *x* − 5 = 40 − *x*

It is given that the product of their marbles is 124.

Either = 0 or *x* − 9 = 0

i.e., *x* = 36 or *x* = 9

If the number of John’s marbles = 36,

Then, number of Jivanti’s marbles = 45 − 36 = 9

If number of John’s marbles = 9,

Then, number of Jivanti’s marbles = 45 − 9 = 36

(ii) Let the number of toys produced be *x*.

∴ Cost of production of each toy = Rs (55 − *x*)

It is given that, total production of the toys = Rs 750

Either = 0 or *x* − 30 = 0

i.e., *x* = 25 or *x* = 30

Hence, the number of toys will be either 25 or 30.

##### Video Solution for quadratic equations (Page: 76 , Q.No.: 2)

NCERT Solution for Class 10 math - quadratic equations 76 , Question 2

#### Page No 76:

#### Question 3:

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

#### Answer:

Let the first number be *x* and the second number is 27 − *x*.

Therefore, their product = *x* (27 − *x*)

It is given that the product of these numbers is 182.

Either = 0 or *x* − 14 = 0

i.e., *x* = 13 or *x* = 14

If first number = 13, then

Other number = 27 − 13 = 14

If first number = 14, then

Other number = 27 − 14 = 13

Therefore, the numbers are 13 and 14.

##### Video Solution for quadratic equations (Page: 76 , Q.No.: 3)

NCERT Solution for Class 10 math - quadratic equations 76 , Question 3

#### Page No 76:

#### Question 4:

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

#### Answer:

Let the consecutive positive integers be *x* and *x* + 1.

Either *x* + 14 = 0 or *x* − 13 = 0, i.e., *x*_{ }= *−*14 or *x*_{ }= 13

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

∴ *x* + 1 = 13 + 1 = 14

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

##### Video Solution for quadratic equations (Page: 76 , Q.No.: 4)

NCERT Solution for Class 10 math - quadratic equations 76 , Question 4

#### Page No 76:

#### Question 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.

#### Answer:

Let the base of the right triangle be *x* cm.

Its altitude = (*x* − 7) cm

Either *x* − 12 = 0 or *x* + 5 = 0, i.e., *x*_{ }= 12 or *x*_{ }= −5

Since sides are positive, *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.

##### Video Solution for quadratic equations (Page: 76 , Q.No.: 5)

NCERT Solution for Class 10 math - quadratic equations 76 , Question 5

#### Page No 76:

#### Question 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.

#### Answer:

Let the number of articles produced be *x*.

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

It is given that the total production is Rs 90.

Either 2*x* + 15 = 0 or *x* − 6 = 0, i.e., *x*_{ }= or *x*_{ }= 6

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

Hence, number of articles produced = 6

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

##### Video Solution for quadratic equations (Page: 76 , Q.No.: 6)

NCERT Solution for Class 10 math - quadratic equations 76 , Question 6

#### Page No 87:

#### Question 1:

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

#### Answer:

##### Video Solution for quadratic equations (Page: 87 , Q.No.: 1)

NCERT Solution for Class 10 math - quadratic equations 87 , Question 1

#### Page No 87:

#### Question 2:

Find the roots of the quadratic equations given in Q.1 above by applying the quadratic formula.

#### Answer:

##### Video Solution for quadratic equations (Page: 87 , Q.No.: 2)

NCERT Solution for Class 10 math - quadratic equations 87 , Question 2

#### Page No 88:

#### Question 3:

Find the roots of the following equations:

#### Answer:

##### Video Solution for quadratic equations (Page: 88 , Q.No.: 3)

NCERT Solution for Class 10 math - quadratic equations 88 , Question 3

#### Page No 88:

#### Question 4:

The sum of the reciprocals of Rehman’s ages, (in years) 3 years ago and 5 years from now is. Find his present age.

#### Answer:

Let the present age of Rehman be *x* years.

Three years ago, his age was (*x* − 3) years.

Five years hence, his age will be (*x* + 5) years.

It is given that the sum of the reciprocals of Rehman’s ages 3 years ago and 5 years from now is.

However, age cannot be negative.

Therefore, Rehman’s present age is 7 years.

##### Video Solution for quadratic equations (Page: 88 , Q.No.: 4)

NCERT Solution for Class 10 math - quadratic equations 88 , Question 4

#### Page No 88:

#### Question 5:

In a class test, the sum of Shefali’s marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product of their marks would have been 210. Find her marks in the two subjects.

#### Answer:

Let the marks in Maths be *x*.

Then, the marks in English will be 30 − *x*.

According to the given question,

If the marks in Maths are 12, then marks in English will be 30 − 12 = 18

If the marks in Maths are 13, then marks in English will be 30 − 13 = 17

##### Video Solution for quadratic equations (Page: 88 , Q.No.: 5)

NCERT Solution for Class 10 math - quadratic equations 88 , Question 5

#### Page No 88:

#### Question 6:

The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 metres more than the shorter side, find the sides of the field.

#### Answer:

Let the shorter side of the rectangle be *x* m.

Then, larger side of the rectangle = (*x* + 30) m

However, side cannot be negative. Therefore, the length of the shorter side will be

90 m.

Hence, length of the larger side will be (90 + 30) m = 120 m

##### Video Solution for quadratic equations (Page: 88 , Q.No.: 6)

NCERT Solution for Class 10 math - quadratic equations 88 , Question 6

#### Page No 88:

#### Question 7:

The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.

#### Answer:

Let the larger and smaller number be *x* and *y* respectively.

According to the given question,

However, the larger number cannot be negative as 8 times of the larger number will be negative and hence, the square of the smaller number will be negative which is not possible.

Therefore, the larger number will be 18 only.

Therefore, the numbers are 18 and 12 or 18 and −12.

##### Video Solution for quadratic equations (Page: 88 , Q.No.: 7)

NCERT Solution for Class 10 math - quadratic equations 88 , Question 7

#### Page No 88:

#### Question 8:

A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.

#### Answer:

Let the speed of the train be *x* km/hr.

Time taken to cover 360 km hr

According to the given question,

However, speed cannot be negative.

Therefore, the speed of train is 40 km/h

##### Video Solution for quadratic equations (Page: 88 , Q.No.: 8)

NCERT Solution for Class 10 math - quadratic equations 88 , Question 8

#### Page No 88:

#### Question 9:

Two water taps together can fill a tank in hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.

#### Answer:

Let the time taken by the smaller pipe to fill the tank be *x* hr.

Time taken by the larger pipe = (*x* − 10) hr

Part of tank filled by smaller pipe in 1 hour =

Part of tank filled by larger pipe in 1 hour =

It is given that the tank can be filled in hours by both the pipes together. Therefore,

Time taken by the smaller pipe cannot be = 3.75 hours. As in this case, the time taken by the larger pipe will be negative, which is logically not possible.

Therefore, time taken individually by the smaller pipe and the larger pipe will be 25 and 25 − 10 =15 hours respectively.

##### Video Solution for quadratic equations (Page: 88 , Q.No.: 9)

NCERT Solution for Class 10 math - quadratic equations 88 , Question 9

#### Page No 88:

#### Question 10:

An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speeds of the express train is 11 km/h more than that of the passenger train, find the average speed of the two trains.

#### Answer:

Let the average speed of passenger train be *x* km/h.

Average speed of express train = (*x* + 11) km/h

It is given that the time taken by the express train to cover 132 km is 1 hour less than the passenger train to cover the same distance.

Speed cannot be negative.

Therefore, the speed of the passenger train will be 33 km/h and thus, the speed of the express train will be 33 + 11 = 44 km/h.

##### Video Solution for quadratic equations (Page: 88 , Q.No.: 10)

NCERT Solution for Class 10 math - quadratic equations 88 , Question 10

#### Page No 88:

#### Question 11:

Sum of the areas of two squares is 468 m^{2}. If the difference of their perimeters is 24 m, find the sides of the two squares.

#### Answer:

Let the sides of the two squares be *x* m and *y* m. Therefore, their perimeter will be 4*x* and 4*y* respectively and their areas will be *x*^{2} and *y*^{2 }respectively.

It is given that

4*x* − 4*y* = 24

*x* − *y* = 6

*x *= *y* + 6

However, side of a square cannot be negative.

Hence, the sides of the squares are 12 m and (12 + 6) m = 18 m

##### Video Solution for quadratic equations (Page: 88 , Q.No.: 11)

NCERT Solution for Class 10 math - quadratic equations 88 , Question 11

#### Page No 88:

#### Question 1:

Find the nature of the roots of the following quadratic equations.

If the real roots exist, find them;

(I) 2*x*^{2
}−3*x *+ 5 = 0

(II)

(III) 2*x*^{2
}− 6*x *+ 3 = 0

#### Answer:

We know
that for a quadratic equation *ax*^{2 }+ *bx *+ *c*
= 0, discriminant is *b*^{2 }− 4*ac.*

(A) If *b*^{2
}− 4*ac* > 0 →
two distinct real roots

(B) If *b*^{2
}− 4*ac* = 0 →
two equal real roots

(C) If *b*^{2
}− 4*ac* < 0 →
no real roots

(I) 2*x*^{2
}−3*x *+ 5 = 0

Comparing this equation with *ax*^{2 }+ *bx *+ *c*
= 0, we obtain

*a* = 2, *b* = −3, *c* = 5

Discriminant = *b*^{2 }− 4*ac *= (− 3)^{2
}− 4 (2) (5) = 9 − 40

= −31

As *b*^{2 }− 4*ac* < 0,

Therefore, no real root is possible for the given equation.

(II)

Comparing this equation with *ax*^{2 }+ *bx *+ *c*
= 0, we obtain

Discriminant

= 48 − 48 = 0

As *b*^{2 }− 4*ac* = 0,

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

And the roots will be and .

Therefore, the roots are and.

(III) 2*x*^{2
}− 6*x *+ 3 = 0

Comparing this equation with *ax*^{2 }+ *bx *+ *c*
= 0, we obtain

*a* = 2, *b* = −6, *c* = 3

Discriminant = *b*^{2 }− 4*ac *= (− 6)^{2
}− 4 (2) (3)

= 36 − 24 = 12

As *b*^{2 }− 4*ac* > 0,

Therefore, distinct real roots exist for this equation as follows.

Therefore, the roots are or .

#### Page No 91:

#### Question 2:

Find the values of *k* for each of the following quadratic equations, so that they have two equal roots.

(I) 2*x*^{2 }+ *kx *+ 3 = 0

(II) *kx* (*x* − 2) + 6 = 0

#### Answer:

We know that if an equation *ax*^{2 }+ *bx *+ *c* = 0 has two equal roots, its discriminant

(*b*^{2 }− 4*ac*) will be 0.

(I) 2*x*^{2 }+ *kx *+ 3 = 0

Comparing equation with *ax*^{2 }+ *bx *+ c = 0, we obtain

*a* = 2, *b* = *k*, *c* = 3

Discriminant = *b*^{2 }− 4*ac *= (*k*)^{2}− 4(2) (3)

= *k*^{2 }− 24

For equal roots,

Discriminant = 0

*k*^{2 }− 24 = 0

*k*^{2}_{ }= 24

(II) *kx *(*x* − 2) + 6 = 0

or *kx*^{2} − 2*kx* + 6 = 0

Comparing this equation with *ax*^{2} + *bx* + *c* = 0, we obtain

*a* = *k*, *b* = −2*k*, *c* = 6

Discriminant = *b*^{2} − 4*ac* = (− 2*k*)^{2} − 4 (*k*) (6)

= 4*k*^{2} − 24*k*

For equal roots,

*b*^{2} − 4*ac* = 0

4*k*^{2} − 24*k* = 0

4*k* (*k* − 6) = 0

Either 4*k* = 0 or *k* = 6 = 0

*k* = 0 or *k* = 6

However, if *k* = 0, then the equation will not have the terms ‘*x*^{2}’ and ‘*x*’.

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

##### Video Solution for quadratic equations (Page: 91 , Q.No.: 2)

NCERT Solution for Class 10 math - quadratic equations 91 , Question 2

#### Page No 91:

#### Question 3:

Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m^{2}?

If so, find its length and breadth.

#### Answer:

Let the breadth of mango grove be *l*.

Length of mango grove will be 2*l*.

Area of mango grove = (2*l*) (*l*)

= 2*l*^{2}

Comparing this equation with *al*^{2} + *bl *+ *c* = 0, we obtain

*a* = 1 *b* = 0, *c* = 400

Discriminant = *b*^{2} − 4*ac* = (0)^{2} − 4 × (1) × (− 400) = 1600

Here, *b*^{2} − 4*ac* > 0

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

However, length cannot be negative.

Therefore, breadth of mango grove = 20 m

Length of mango grove = 2 × 20 = 40 m

##### Video Solution for quadratic equations (Page: 91 , Q.No.: 3)

NCERT Solution for Class 10 math - quadratic equations 91 , Question 3

#### Page No 91:

#### Question 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 ages in years was 48.

#### Answer:

Let the age of one friend be *x* years.

Age of the other friend will be (20 − *x*) years.

4 years ago, age of 1^{st} friend = (*x* − 4) years

And, age of 2^{nd} friend = (20 − *x *− 4)

= (16 − *x*) years

Given that,

(*x* − 4) (16 − *x*) = 48

16*x* − 64 − *x*^{2} + 4*x* = 48

− *x*^{2} + 20*x* − 112 = 0

*x*^{2} − 20*x* + 112 = 0

Comparing this equation with *ax*^{2} + *bx* + *c* = 0, we obtain

*a* = 1, *b* = −20, *c *= 112

Discriminant = *b*^{2} − 4*ac* = (− 20)^{2} − 4 (1) (112)

= 400 − 448 = −48

As *b*^{2} − 4*ac* < 0,

Therefore, no real root is possible for this equation and hence, this situation is not possible.

##### Video Solution for quadratic equations (Page: 91 , Q.No.: 4)

NCERT Solution for Class 10 math - quadratic equations 91 , Question 4

#### Page No 91:

#### Question 5:

Is it possible to design a rectangular park of perimeter 80 and area 400 m^{2}? If so find its length and breadth.

#### Answer:

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

Perimeter = 2 (*l* + *b*) = 80

*l* + *b* = 40

Or, *b* = 40 −* l*

Area = *l* × *b* = *l* (40 − *l*) = 40*l** *− *l*^{2}

40*l** *− *l*^{2} = 400

*l*^{2} − 40*l* + 400 = 0

Comparing this equation with

*al*^{2} + *bl *+ c = 0, we obtain

*a* = 1, *b* = −40, *c* = 400

Discriminant = *b*^{2} − 4*ac* = (− 40)^{2} −4 (1) (400)

= 1600 − 1600 = 0

As *b*^{2} − 4*ac* = 0,

Therefore, this equation has equal real roots. And hence, this situation is possible.

Root of this equation,

Therefore, length of park, *l* = 20 m

And breadth of park, *b* = 40 − *l* = 40 − 20 = 20 m

##### Video Solution for quadratic equations (Page: 91 , Q.No.: 5)

NCERT Solution for Class 10 math - quadratic equations 91 , Question 5

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