# NCERT Solutions for Class 10th Maths Chapter 4 Quadratic Equations

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Page No: 73**Chapter 4 Quadratic Equations NCERT Class 10 Maths**Â Solutions that will help in achieving more marks. You don't have to wander and waste your precious time in finding bestÂ CBSE NCERT Solutions. It will help you a lot in understanding the concepts in practical way. These solutions are prepared by subject matter experts that will be useful in scoring more marks in the examinations. Through these**Class 10 Chapter 4 NCERT Solutions**you can complete your homework on time and also assess your skills in a better manner.**Exercise 4.1**

**1. Check whether the following are quadratic equations:**

(i) (

*x*Â + 1)

^{2}= 2(

*x*- 3)

(ii)

*x*

^{2}- 2

*x*= (-2)(3 -

*x*)

(iii) (

*x*- 2)(

*x*Â

*x*- 1)(

*x*+ 3)

(iv) (

*x*- 3)(2

*x*Â + 1) =

*x*(

*x*Â + 5)

(v) (2

*x*- 1)(

*x*- 3) = (

*x*Â + 5)(

*x*- 1)

(vi)

*x*

^{2}Â + 3

*x*Â + 1 = (

*x*- 2)

^{2}

(vii) (

*x*Â + 2)

^{3}= 2

*x*(

*x*

^{2}- 1)

(viii)

*x*

^{3}- 4

*x*

^{2}-

*x*Â + 1 = (

*x*- 2)

^{3}

**Answer**

(i) (

*x*Â + 2)

^{2}Â = 2(

*x*Â - 3)

â‡’

*x*

^{2}Â + 2

*x*Â + 1 = 2

*x*- 6

â‡’

*x*

^{2}Â + 7 = 0

It is of the form

*ax*

^{2}Â +

*bx*Â +

*c*= 0.

Hence, the given equation is quadratic equation.

(ii)Â

*x*

^{2}Â - 2

*x*Â = (-2)(3 -Â

*x*)

â‡’Â

*x*

^{2Â }-

^{Â }2

*x = -*6Â + 2

*xÂ*

â‡’Â

*x*

^{2Â }- 4

*x*Â + 6 = 0

It is of the formÂ

*ax*

^{2}Â +Â

*bx*Â +Â

*c*Â = 0.

Hence, the given equation is quadratic equation.

(iii) (

*x*- 2)(

*x*Â + 1) = (

*x*Â - 1)(

*x*Â + 3)

â‡’Â

*x*

^{2Â }-

*x*- 2 =Â

*x*

^{2Â }+ 2

*x*Â - 3

â‡’ 3

*x*- 1 =0

It is not of the formÂ

*ax*

^{2}Â +Â

*bx*Â +Â

*c*Â = 0.

Hence, the given equation is not a quadratic equation.

(iv) (

*x*Â - 3)(2

*x*Â + 1) =Â

*x*(

*x*Â + 5)

â‡’ 2

*x*

^{2Â }- 5

*x*- 3 =Â

*x*

^{2Â }+ 5

*x*

â‡’ Â

*x*

^{2Â }- 10

*x*- 3 = 0

It is of the formÂ

*ax*

^{2}Â +Â

*bx*Â +Â

*c*Â = 0.

Hence, the given equation is quadratic equation.

(v) (2

*x*Â - 1)(

*x*Â - 3) = (

*x*Â + 5)(

*x*Â - 1)

â‡’ 2

*x*

^{2Â }- 7

*xÂ +*Â 3 =Â

*x*

^{2Â }+ 4

*x*- 5

â‡’Â

*x*

^{2Â }- 11

*x*Â +Â 8 = 0

It is of the formÂ

*ax*

^{2}Â +Â

*bx*Â +Â

*c*Â = 0.

Hence, the given equation is quadratic equation.

(vi)Â

*x*

^{2}Â + 3

*x*Â + 1 = (

*x*Â - 2)

^{2}

â‡’Â

*x*

^{2}Â + 3

*x*Â + 1Â =Â

*x*

^{2}Â + 4Â - 4

*x*

â‡’ 7

*x*Â - 3 = 0

It is not of the formÂ

*ax*

^{2}Â +Â

*bx*Â +Â

*c*Â = 0.

Hence, the given equation is not a quadratic equation.

(vii) (

*x*Â + 2)

^{3}Â = 2

*x*(

*x*

^{2}Â - 1)

â‡’

*x*

^{3}Â + 8Â +

*x*

^{2}Â + 12

*x*= 2

*x*

^{3}Â - 2

*x*

â‡’

*x*

^{3}Â + 14

*x*- 6

*x*

^{2}- 8 = 0

It is not of the formÂ

*ax*

^{2}Â +Â

*bx*Â +Â

*c*Â = 0.

Hence, the given equation is not a quadratic equation.

(viii)Â

*x*

^{3}Â - 4

*x*

^{2}Â -Â

*x*Â + 1 = (

*x*Â - 2)

^{3}

â‡’ Â

*x*

^{3}Â - 4

*x*

^{2}Â -

*x*Â + 1Â =

*x*

^{3}Â - 8 - 6

*x*

^{2Â }+ 12

*x*

â‡’ 2

*x*

^{2}- 13

*x*+ 9 = 0

It is of the formÂ

*ax*

^{2}Â +Â

*bx*Â +Â

*c*Â = 0.

Hence, the given equation is quadratic equation.

**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.

**Answer**

Let the breadth of the rectangular plot =

*x*m

Hence, the length of the plot is (2

*x*+ 1) m.

Formula of area of rectangle = lengthÂ Ã—Â breadth = 528 m

^{2}

Putting the value of length and width, we get

(2

*xÂ*+ 1) Ã—

*x*= 528

â‡’ 2

*x*

^{2}+

*x*=528

â‡’ 2

*x*

^{2}+

*x*- 528 = 0

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

**Answer**

Let the first integer number =

*x*

Next consecutive positive integer will =

*x*+ 1

Product of both integers =

*x*Â Ã—Â (

*x*+1) = 306

â‡’Â

*x*

^{2Â }+

*x*= 306

â‡’Â

*x*

^{2Â }+

*x*- 306 = 0

(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.

**Answer**

Let take Rohan's age =

*x*years

Hence, his mother's age =

*x*+ 26

3 years from now

Rohan's age =

*x*+ 3

Age of Rohan's mother will =

*x*+ 26 + 3 =

*x*+ 29

The product of their ages 3 years from now will be 360 so that

(

*x*+ 3)(

*x*+ 29) = 360

â‡’Â

*x*

^{2}Â + 29

*x*+ 3

*x*+ 87 = 360

â‡’Â

*x*

^{2}Â + 32

*x*+ 87 - 360 = 0

â‡’Â

*x*

^{2}Â + 32

*x*- 273 = 0

(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**

Let the speed of train be

*x*km/h.

Time taken to travel 480 km = 480/

*x*km/h

*x*- 8) 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 = (480/

*x*Â + 3) km/h

Speed Ã— Time = Distance

(*x*- 8)(480/

*x*Â + 3) = 480

â‡’ 480Â + 3

*x*- 3840/

*x*- 24 = 480

â‡’ 3

*x*- 3840/

*x*= 24

â‡’ 3

*x*

^{2Â }- 8

*x*Â - 1280 = 0

Page No: 76

**Exercise 4.2**

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

(i)Â

*x*

^{2}Â â€“ 3

*x*â€“ 10 = 0

(ii) 2

*x*

^{2}Â +

*x*â€“ 6 = 0

(iii) âˆš2Â

*x*

^{2}Â + 7

*x*+ 5âˆš2 = 0

(iv) 2

*x*

^{2}Â â€“

*x*+ 1/8 = 0

(v) 100

*x*

^{2}Â â€“ 20

*x*+ 1 = 0

**Answer**

(i)Â

*x*

^{2}Â â€“ 3

*x*Â â€“ 10

=Â

*x*

^{2}- 5

*x*+ 2

*x*- 10

=

*x*(

*xÂ*- 5)Â + 2(

*x*- 5)

= (

*x*- 5)(

*x*+Â 2)

Roots of this equation are the values for whichÂ (

*x*Â - 5)(

*x*Â +Â 2) = 0

âˆ´

â‡’Â *x*- 5 = 0 or*x*Â + 2 = 0*x*= 5 or

*x*Â = -2

(ii) 2

*x*

^{2}Â +Â

*x*Â â€“ 6

= 2

*x*

^{2}+ 4

*x*- 3

*x*- 6

= 2

*x*(

*x*+Â 2) - 3(

*x*Â + 2)

= (

*x*+ 2)(2

*x*- 3)

Roots of this equation are the values for which (

*x*Â + 2)(2

*x*Â - 3) = 0

âˆ´Â

â‡’Â *x*Â + 2Â = 0 orÂ 2*x*Â - 3 = 0*x*Â = -2 or

*x*= 3/2

(iii) âˆš2Â

*x*

^{2}Â + 7

*x*Â + 5âˆš2

= âˆš2Â

*x*

^{2Â }+ 5

*x*+ 2

*x*+ 5âˆš2

=

*x*(âˆš2

*x*Â + 5)Â +Â âˆš2(âˆš2

*x*Â + 5)= (âˆš2

*x*Â + 5)(

*xÂ*+Â âˆš2)

Roots of this equation are the values for which (âˆš2

*x*Â + 5)(

*x*Â +Â âˆš2) = 0

âˆ´Â âˆš2

â‡’Â *x*Â + 5Â = 0 orÂ*x*Â +Â âˆš2Â = 0*x*Â = -5/âˆš2Â orÂ

*x*Â = -âˆš2

(iv) 2

*x*

^{2}Â â€“Â

*x*Â + 1/8

= 1/8 (16

*x*

^{2 Â }- 8

*x*+ 1)

= 1/8 (16

*x*

^{2 Â }- 4

*x*Â -4

*x*Â + 1)

= 1/8 (4

*x*(4

*x*

^{Â Â }- 1) -1(4

*x*Â -Â 1))

= 1/8(4

*x*- 1)

^{2}

Roots of this equation are the values for which (4

*xÂ*- 1)

^{2}= 0

âˆ´ (4

â‡’ *x*- 1) = 0 or (4*x*- 1) = 0*x*= 1/4 or

*x*= 1/4

(v) 100

*x*

^{2}Â â€“ 20

*x*Â + 1

= 100

*x*

^{2}Â â€“ 10

*x*Â - 10

*x*Â + 1

= 10

*x*(10

*x*- 1) -1(10

*x*- 1)

= (10

*x*- 1)

^{2}

Roots of this equation are the values for whichÂ (10

*x*Â - 1)

^{2}= 0

âˆ´ (10

*x*- 1) = 0 or (10

*x*- 1) = 0

â‡’

*x*= 1/10 or

*x*= 1/10

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.

**Answer**

Let the number of John's marbles be

Therefore, number of Jivanti's marble = 45 -

After losing 5 marbles,

Number of John's marbles =

Number of Jivanti's marbles = 45 -

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

*x*.Therefore, number of Jivanti's marble = 45 -

*x*After losing 5 marbles,

Number of John's marbles =

*x*- 5Number of Jivanti's marbles = 45 -

*x*- 5 = 40 -*x*It is given that the product of their marbles is 124.

âˆ´ (

â‡’Â *x*- 5)(40 -*x*) = 124*x*

^{2}Â â€“ 45

*x*Â + 324 = 0

â‡’Â

*x*

^{2}Â â€“ 36

*x*- 9

*x*Â + 324 = 0

â‡’

*x*(

*x*- 36) -9(

*x*- 36) = 0

â‡’ (

*x*- 36)(

*x*- 9) = 0

Either

*x*- 36 = 0 or

*x*- 9 = 0

â‡’Â

*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) 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**

Let the number of toys produced be

*x*.

âˆ´ Cost of production of each toy = Rs (55 -

*x*)

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

âˆ´

*x*(55 -*x*) = 750
â‡’Â

*x*^{2}Â â€“Â 55*x*Â + 750 = 0
â‡’Â

*x*^{2}Â â€“ 25*x*-Â 30*x*Â + 750 = 0
â‡’

*x*(*x*- 25) -30(*x*- 25) = 0
â‡’ (

Either, *x*- 25)(*x*- 30) = 0*x*-25 = 0 or

*x*- 30 = 0

â‡’Â

*x*= 25 or

*x*= 30

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

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.

Therefore,

*x*(27 -

*x*) = 182

â‡’Â

*x*

^{2}Â â€“ 27

*x*Â - 182 = 0

â‡’Â

*x*

^{2}Â â€“ 13

*x*- 14

*x*Â + 182 = 0

â‡’

*x*(

*x*- 13) -14(

*x*- 13) = 0

â‡’ (

*x*- 13)(

*x*-14) = 0

Either

*x*= -13Â = 0 or

*x*- 14 = 0

â‡’Â

*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.

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

**Answer**

Let the consecutive positive integers be

*x*and

*x*+ 1.

Therefore,Â

*x*

^{2}Â + (

*x*Â + 1)

^{2}= 365

â‡’Â

*x*

^{2Â }+Â

*x*

^{2Â }+ 1 + 2

*x*= 365

â‡’ 2

*x*

^{2}Â + 2x - 364 = 0

â‡’Â

*x*

^{2Â }+Â

*x*- 182

*= 0*

â‡’Â

*x*

^{2Â }+ 14

*x*- 13

*x*Â - 182 = 0

â‡’

*x*(

*x*+Â 14) -13(

*x*+Â 14) = 0

â‡’ (

*x*Â + 14)(

*x*- 13) = 0

Either

*x*+ 14 = 0 or

*x*- 13 = 0,

â‡’Â

Since the integers are positive,

âˆ´

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

*x*= - 14 or*x*= 13Since the integers are positive,

*x*can only be 13.âˆ´

*x*+ 1 = 13 + 1 = 14Therefore, 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.

**Answer**

Let the base of the right triangle be

*x*cm.

Its altitude = (

*x*- 7) cm

From Pythagoras theorem, we have

Base

^{2}Â + Altitude

^{2}= Hypotenuse

^{2}

âˆ´Â

*x*

^{2Â }+ (

*x*- 7)

^{2}= 132

â‡’Â

*x*

^{2Â }+Â

*x*

^{2Â }+ 49 - 14

*x*= 169

â‡’ 2

*x*

^{2Â }- 14

*x*- 120 = 0

â‡’

*x*

^{2Â }- 7

*x*Â - 60 = 0

â‡’Â

*x*

^{2Â }- 12

*xÂ*+ 5

*x*- 60 = 0

â‡’Â

*x*(

*x*- 12) + 5(

*xÂ*- 12) = 0

â‡’ (

*x*- 12)(

*x*Â + 5) = 0

Either

*x*- 12 = 0 or

*x*+ 5 = 0,

â‡’

*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.

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.âˆ´

*x*(2

*x*Â + 3) = 0

â‡’ 2

*x*

^{2Â }+ 3

*x*- 90 = 0

â‡’ 2

*x*

^{2Â }+ 15

*x*-12

*x*- 90 = 0

â‡’

*x*(2

*x*Â + 15) -6(2

*x*Â + 15) = 0

â‡’ (2

*x*Â + 15)(

*x*- 6) = 0

Either 2

*x*+ 15 = 0 or

*x*- 6 = 0

â‡’

*x*= -15/2 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 Ã— 6 + 3 = Rs 15.

Hence, number of articles produced = 6

Cost of each article = 2 Ã— 6 + 3 = Rs 15.

Page No: 87

**Exercise 4.3**

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

(i) 2

(ii)Â 2*x*^{2}Â â€“ 7*x*+3 = 0*x*

^{2}Â +

*x*â€“ 4 = 0

(iii)Â 4

*x*

^{2}Â + 4âˆš3

*x*+ 3 = 0

(iv)Â 2

(i) 2

â‡’ 2

On dividing both sides of the equation by 2, we get

â‡’Â

â‡’Â

On adding (7/4)

â‡’ (

*x*^{2}Â +*x*+ 4 = 0**Answer**(i) 2

*x*^{2}Â â€“Â 7*x*Â + 3 = 0â‡’ 2

*x*^{2}Â â€“Â 7*x*= - 3On dividing both sides of the equation by 2, we get

â‡’Â

*x*^{2}Â â€“ 7*x*/2 Â = -3/2â‡’Â

*x*^{2}Â â€“ 2 Ã—*x*Ã— Â 7/4 = -3/2On adding (7/4)

^{2}to both sides of equation, we getâ‡’ (

*x*)^{2Â }- 2 Ã—*x*Ã— 7/4 + (7/4)^{2}= (7/4)^{2}- 3/2*x*- 7/4)

^{2}= 49/16 - 3/2

â‡’ (

*x*- 7/4)

^{2}= 25/16

â‡’ (

*x*Â - 7/4) =Â Â± 5/4

â‡’

*x*= 7/4Â Â± 5/4

â‡’

*x*= 7/4Â + 5/4 orÂ

*x*Â = 7/4 - 5/4

â‡’Â

*x*Â = 12/4 orÂ

*x*Â = 2/4

â‡’

*x*= 3 or 1/2

(ii) 2

*x*

^{2}Â +Â

*x*Â â€“ 4 = 0

â‡’ 2

*x*

^{2}Â +Â

*xÂ*= 4

On dividing both sides of the equation, we get

â‡’Â

*x*

^{2}Â +

*x*/2 = 2

On adding (1/4)

^{2Â }to both sides of the equation, we get

â‡’ (

*x*)

^{2Â }+Â 2 Ã—Â

*x*Â Ã— 1/4 + (1/4)

^{2}Â = 2Â + (1/4)

^{2}

â‡’ (

*x*Â + 1/4)

^{2}Â = 33/16

â‡’

*x*Â + 1/4 = Â± âˆš33/4

â‡’Â

*x*= Â± âˆš33/4 - 1/4

â‡’

*x*= Â± âˆš33-1/4

â‡’Â

*x*Â = âˆš33-1/4 orÂ

*x*Â = -âˆš33-1/4

(iii) 4

*x*

^{2}Â + 4âˆš3

*x*Â + 3 = 0

â‡’ (2

*x*)

^{2}Â + 2 Ã— 2

*x*Ã— âˆš3 + (âˆš3)

^{2}Â = 0

â‡’ (2

*x*Â + âˆš3)

^{2}Â = 0

â‡’ (2

*x*Â + âˆš3) = 0 and (2

*x*Â + âˆš3) = 0

â‡’

*x*= -âˆš3/2 orÂ

*x*Â = -âˆš3/2

(iv) 2

*x*

^{2}Â +Â

*x*Â + 4 = 0

â‡’ 2

*x*

^{2}Â +Â

*x*= -4

On dividing both sides of the equation, we get

â‡’Â

*x*

^{2}Â + 1/2

*x*= 2

â‡’Â

*x*

^{2}Â + 2Â Ã—

*xÂ*Ã— 1/4 = -2

On adding (1/4)

^{2Â }to both sides of the equation, we get

â‡’ (

*x*)

^{2Â }+Â 2 Ã—Â

*x*Â Ã— 1/4 + (1/4)

^{2}Â = (1/4)

^{2Â }- 2

â‡’ (

*x*Â + 1/4)

^{2}Â = 1/16 - 2

â‡’ (

*x*Â + 1/4)

^{2}Â = -31/16

However, the square of number cannot be negative.

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

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

**Answer**

(i) 2

*x*

^{2}Â â€“Â 7

*x*Â + 3 = 0

On comparing this equation with

*ax*

^{2}Â +

*bxÂ*+

*c*= 0, we get

*a*= 2,

*b*= -7 and

*c*= 3

By using quadratic formula, we get

*x*= -

*b*Â±âˆš

*b*

^{2}- 4

*ac*/2a

â‡’

*x*= 7Â±âˆš49 - 24/4

â‡’Â

*x*Â = 7Â±âˆš25/4

â‡’Â

*x*Â = 7Â±5/4

â‡’Â

*x*Â = 7+5/4 orÂ

*x*Â = 7-5/4

â‡’Â

*x*Â = 12/4 or 2/4

âˆ´ Â

*x*Â =Â 3 or 1/2

(ii) 2

*x*

^{2}Â +Â

*x*Â - 4 = 0

On comparing this equation withÂ

*ax*

^{2}Â +Â

*bxÂ*+Â

*c*Â = 0, we get

a = 2, b = 1 andÂ

*c*= -4

By using quadratic formula, we get

*x*Â = -

*b*Â±âˆš

*b*

^{2}Â - 4

*ac*/2a

â‡’

*x*Â = -1Â±âˆš1+32/4

â‡’

*x*Â = -1Â±âˆš33/4

âˆ´Â

*x*Â = -1+âˆš33/4 orÂ

*x*Â = -1-âˆš33/4

(iii) 4

*x*

^{2}Â +Â 4âˆš3

*x*Â + 3 = 0

On comparing this equation withÂ

*ax*

^{2}Â +Â

*bxÂ*+Â

*c*Â = 0, we get

*a*=Â 4,

*b*=Â 4âˆš3 andÂ c = 3

By using quadratic formula, we get

*x*Â = -

*b*Â±âˆš

*b*

^{2}Â - 4

*ac*/2a

â‡’Â

*x*Â = -4âˆš3Â±âˆš48-48/8

â‡’Â

*x*Â = -4âˆš3Â±0/8

âˆ´Â

*x*Â = -âˆš3/2 orÂ

*x*Â = -âˆš3/2

(iv)Â 2

*x*

^{2}Â +Â

*x*Â + 4 = 0

On comparing this equation withÂ

*ax*

^{2}Â +Â

*bxÂ*+Â

*c*Â = 0, we get

*a*= 2,

*b*= 1 andÂ

*c*= 4

By using quadratic formula, we get

*x*Â = -

*b*Â±âˆš

*b*

^{2}Â - 4

*ac*/2a

â‡’Â

*x*Â = -1Â±âˆš1-32/4

â‡’Â

*x*Â = -1Â±âˆš-31/4

The square of a number can never be negative.

âˆ´There is no real solution of this equation.

Page No: 88

**3. Find the roots of the following equations:**

(i)Â

*x*-1/

*x*= 3,

*x*Â â‰ 0

(ii) 1/

*x*+4 - 1/

*x*-7 = 11/30,

*x*= -4, 7

**Answer**

(i)Â

*x*-1/

*x*Â = 3

â‡’Â

*x*

^{2}Â - 3

*x*-1 = 0

On comparing this equation withÂ

*ax*

^{2}Â +Â

*bxÂ*+Â

*c*Â = 0, we get

*a*Â = 1,Â

*b*Â = -3 andÂ

*c*Â = -1

By using quadratic formula, we get

*x*Â = -

*b*Â±âˆš

*b*

^{2}Â - 4

*ac*/2a

â‡’Â

*x*Â =Â 3Â±âˆš9+4/2

â‡’Â

*x*Â = 3Â±âˆš13/2

âˆ´Â

*x*Â = 3+âˆš13/2 orÂ

*x*Â = 3-âˆš13/2

(ii) 1/

*x*+4 - 1/

*x*-7 = 11/30

â‡’

*x*-7-

*x*-4/(

*x*+4)(

*x*-7) = 11/30

â‡’ -11/(

*x*+4)(

*x*-7) = 11/30

â‡’ (

*x*+4)(

*x*-7) = -30

â‡’Â

*x*

^{2}Â - 3

*x*- 28 = 30

â‡’Â

*x*

^{2}Â - 3

*x*Â + 2 = 0

â‡’Â

*x*

^{2}Â - 2

*x*-

*x*Â + 2 = 0

â‡’

*x*(

*x*- 2) - 1(

*x*- 2) = 0

â‡’ (

*x*Â - 2)(

*x*- 1) = 0

â‡’

*x*= 1 or 2

4.Â The sum of the reciprocals of Rehman's ages, (in years) 3 years ago and 5 years from now is 1/3. 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 1/3.