NCERT Solutions for Class 10 maths Chapter 4 Quadratic Equation - Future Study Point

NCERT Solutions for Class 10 maths Chapter 4 Quadratic Equation

quadratic equation ncert solutions

NCERT Solutions for Class 10 maths Chapter 4 Quadratic Equation

quadratic equation ncert solutions

NCERT Solutions for Class 10 maths Chapter 4 Quadratic Equation is one of the most important study inputs for clearing your maths concept.NCERT Solutions for Class 10 maths Chapter 4 Quadratic Equation are created here by an expert of maths for the purpose of helping the class 10 students for term-2 CBSE Board exam because Chapter 4 Quadratic Equation is one of the important chapter that is the part of CBSE Board 2021-22 curriculum. It is mandatory to study NCERT Solutions for Class 10 maths for clearing the basic of maths which is required to solve the questions of question paper. Here you can study important MCQs and important subjective questions related to the chapter 4-quadratic equations along the  NCERT Solutions for Class 10 maths Chapter 4 Quadratic Equation.Studying all inputs MCQs,important questions and NCERT solutions guarantees you to solve all questions of Quadratic Equations in the term 2 CBSE Board exam 2021-22.

NCERT Solutions for Class 10 maths Chapter 4 Quadratic Equation

Exercise 4.1

 


Q1.Check whether the following are quadratic equations

(i) (x +1)² = 2( x -3)

(ii) x – 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

(i) (x +1)² = 2( x -3)

Expanding the LHS, we get the term x²,therefore it is a quadratic equation

(ii) x – 2x =(-2)(3 – x)

There is no any quadratic term containing x², therefore it is not a quadratic equation

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

Simplifying LHS and RHS,we get quadratic term x² in both sides of the same coefficients, cancelling the term x²,we find the resultant equation is not a quadratic equation.

(iv) ( x -3)(2x + 1) =  x (x + 5)

Simplifying LHS and RHS,we get quadratic term 2x² in LHS and x² in RHS, we find the resultant equation contains the term 2x²-x²=x²,so it is a quadratic equation.

(v) ( 2x -1)(x -3) =   (x + 5)(x – 1)

Simplifying LHS and RHS,we get quadratic term 2x² in LHS and x² in RHS, we find the resultant equation contains the term 2x²-x²=x²,so it is a quadratic equation.

(vi) x²+3x +1 =  (x – 2)²

Expanding RHS,we get the quadratic term x² which cancelled by the LHS term x²,hence the resultant equation is not a quadratic equation.

(vii) (x +2)³=  2x(x² – 1)

Expanding LHS,we get the cubic term x³ ,hence the resultant equation is not a quadratic equation.

(viii) x³ -4x²-x +1 = (x – 2)³

Expanding RHS,we get the cubic term x³ which is cancelled by the LHS term x³ ,hence the resultant equation is  a quadratic equation.

NCERT Solutions for Class 10 maths Chapter 4 Quadratic Equation

Represent the following situations in the form of quadratic equation

(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.

Ans. Let the breadth of the rectangular plot is x m then according to the question length becomes 2x + 1

The given area of a rectangular plot is 528 m²

x(2x + 1) = 528

2x² + x – 528  =0

Therefore required quadratic equation is 2x² + x – 528  =0

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

Ans.Let one of the positive integers is x then another successive positive integer is x +1

x(x + 1) = 306

x² + x – 306 =0

Therefore required quadratic equation is x² + 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.

Ans. Let Rohan’s age is x years then according to the question Rohan’s mother age is x +26

The age of Rohan 3 years from now is x + 3

The age of Rohan’s mother  3 years from now is x + 26 +3 =x +29

Since product of their ages (in years) 3 years from now is 360

∴ (x + 3)(x +29) = 360

x² +(3 +29)x +29×3 =360

x² + 32x +87 – 360 =0

x² + 32x – 273=0

Therefore required quadratic equation is x² + 32x – 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.

Ans.Let the speed of the train is x km/h.

The distance covered by the train is 480 km

Time taken with the speed of x km/h = distance/time =480/x

Time taken with the speed of (x-8) km/h = 480/(x-8)

Since it is given to us

Time taken with the speed of (x-8) km/h = Time taken with the speed of x km/h +3 h

480/(x-8) = 480/x + 3

480/(x-8) – 480/x = 3

[480x -480x+ 3840] = 3x(x-8)

3x² – 24x – 3840 = 0

x² – 8x -1280 =0

Therefore required quadratic equation is x² – 8x -1280 =0

NCERT Solutions for Class 10 maths Chapter 4 Quadratic Equation

Exercise 4.2 

NCERT Solutions for Class 10 maths Chapter 4 Quadratic Equation updated for 2023-24

See the Video

 

Solutions of Exercise 4.2 Question-1 Quadratic Equation

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

Ans.(i) x² – 3x -10 = 0

Factorizing the equation

x²- 5x +2x -10 = 0

x(x – 5) + 2(x – 5) =0

(x – 5)(x + 2) =0

x =5,-2

Ans.(ii) 2x² + x- 6 = 0

Factorizing the equation

2x² + 4x- 3x-6 = 0

2x(x + 2) – 3(x + 2) =0

(x + 2)(2x – 3) = 0

x =-2,3/2

Ans.(iii) √2 x² + 7x + 5√2 = 0

Factorizing the equation

√2 x² + 5x +2x+ 5√2 = 0

x(√2 x +5) + √2(√2 x +5) =0

(√2 x +5)(x +√2) =0

x =-5/√2,-√2

Ans.(iv) 2x² – x +1/8 = 0

Multiplying the quadratic equation by 8

16x² – 8x + 1 =0

16x² – 4x -4x+ 1 =0

4x(4x -1) -1(4x – 1) =0

(4x -1)(4x -1)=0

x =1/4,1/4

Ans.(v) 100x² -20x + 1 = 0

Factorizing the equation

100x² -10x -10x+ 1 = 0

10x(10x – 1) – 1(10x – 1) =0

(10x – 1)(10x – 1) =0

x = 1/10, 1/10

Solutions of Exercise 4.2 Question-1 Quadratic Equation

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.

Ans.Let the number of marbles  John had to start with be x

Total marbles they have 45 then number of marbles Jivanti have (45 -x)

Both of them lost 5 marbles, now John has the remaining marbles x -5 and Jivanti has 45 -x -5 = 40 -x

Product of the number of marbles they now have = 124

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

40x – x² -200+5x =124

-x² + 40x +5x- 200=124

-x² + 40x +5x- 200-124=0

-x² + 45x -324 =0

x² – 45x +324 =0

x² – 36x -9x+324 =0

x( x – 36) -9( x – 36) =0

( x – 36)( x – 9) =0

x =36,9

Hence, if John’s marbles are 36 then  Jivanti has (45 -36=9)

If John’s marbles are 9 then  Jivanti has (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 ` 750. We would like to find out the number of toys produced on that day.

Ans. Let the number of toys produced on that day is x

According to question,the cost of toy is x -55

On a particular day, the total cost of production was  750.

Since no. of toys produced on a particular day =x(x -55)

According to question

x(x -55) =750

x² -55x -750 =0

x² -30x -25x-750 =0

x( x – 30) – 25(x – 30) =0

( x – 30)( x – 25) =0

x =30,25

Hence the number of toys produced on that day is either 30 or 25

Solutions of Exercise 4.2 Question-3 Quadratic Equation

Q3. Find two numbers whose sum is 27 and the product is 182.

Ans. Let one of the number is x

Since sum of both numbers given is 27 then another number is 27 – x

The product of both numbers given is 182

x(27 – x) = 182

27x – x² – 182 = 0

– x² + 27x – 182 = 0

x²-27x +182 = 0

x²-13x -14x+182 = 0

x(x – 13) – 14(x – 13) =0

(x – 13) (x – 14) =0

x =13,14

Therefore if one of the number is 13 then another number is 27 – 13 =14 and  if one of the number is 14 then another number is 27 – 14 =13

Solutions of Exercise 4.2 Question-4 Quadratic Equation

Q4.Find two consecutive positive integers, sum of whose squares is 365.

Ans. Let one of the positive integer is x then another is x +1

According to question

x² + (x +1)² = 365

x² + x² + 1 + 2x – 365 =0

2x² + 2x -364 = 0

x² + x – 182 = 0

x² + 14x – 13x -182 = 0

x( x + 14) – 13( x + 14) =0

( x + 14)( x – 13) =0

x =-14, 13

Neglecting the value x =-14 since x is a positive integer

∴ One of the postive integer is 13 and another is x +1 =13 +1 =14

Solutions of Exercise 4.2 Question-5 Quadratic Equation

Q5. The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides

Ans. Let the base of the triangle is x then altitude is x – 7

The hypotenuse of the triangle is given 13 cm

Applying the pythogorus theorem

(x – 7)² + x² = 13²

x² +49 -14x + x² = 169

2x² -14x  +49 – 169 = 0

2x² -14x   – 120 = 0

x² -7x   – 60= 0

x² -12x  +5x – 60= 0

x( x – 12) + 5( x – 12) =0

( x – 12)( x + 5) =0

x = 12, -5

Neglecting the value x =-5 since x is side of triangle which can’t be negative

Therefore base of the triangle is 12 cm and altitude is 12 -7 = 5 cm

Solutions of Exercise 4.2 Question-6 Quadratic Equation

Q6. 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.

Ans. Let the number of articles produced on that day = x

Since the cost of each article was observed on that day 3 more than twice the number of articles produced

Therefore the cost of each article is = 2x + 3

The total cost of production as per the question=x(2x + 3)

It is given in question the total cost of production on that day= Rs 90

According to question

x(2x + 3) = 90

2x² + 3x –90 = 0

2 × 90 = 2×3×3×5 ×2 =15×12

2x² + 15x –12x –90 = 0

x(2x + 15) –6(2x +15) =0

(2x + 15)(x –6) = 0

x = -15/2, 6

The number of articles produced can’t be negative so the number of articles produced on a day is  6 and the cost of each article is   (2x +3 )=Rs(2×6 + 3) = Rs. 15

NCERT Solutions for Class 10 maths Chapter 4 Quadratic Equation

Exercise 4.3 

Solutions of Exercise 4.3 Question-1 Quadratic Equation

See the Video and subsribe us in Utube NCERT Solutions for Class 10 maths Chapter 4 Quadratic Equation updated for 2023-24

Q1. Find the roots of the following quadratic equations, if they exist, by the method of completing the square:

(i) 2x ²- 7x +3 = 0

(ii) 2x²+ x -4 = 0
(iii) 4x²+ 4√3x + 3 = 0

(iv) 2x²+ x + 4 = 0

Ans.(i) 2x ²- 7x +3 = 0

Multiplying the quadratic equation by 2

4x ²- 14x +6 = 0

(2x)² – 14x +6 =0

Comparing this equation with comlete square polynomial (a +b)² = a² + 2ab + b²

2ab = 14x

2(2x)b = 14x

b = 14/4 = 7/2

Adding and subtracting (7/2)²  in the RHS of the equation

(2x)² – 14x +(7/2)²-(7/2)²+6 =0

(2x)² – 14x +(7/2)²-49/4+6 =0

(2x – 7/2)² – 25/4=0

(2x – 7/2)² – (5/2)²=0

(2x 7/2+ 5/2)(2x – 7/2- 5/2)=0

(2x -1)(2x -6) =0

x = 1/2, 3

(ii) 2x²+ x -4 = 0

Multiplying the quadratic equation by 2

4x²+ 2x -8 = 0

(2x)² + 2x -8 = 0

Comparing this equation with comlete square polynomial (a +b)² = a² + 2ab + b²

2ab = 2x

2(2x)b =2x

b = 1/2

Adding and subtracting (1/2)²  in the RHS of the equation

(2x)² + 2x +(1/2)² -(1/2)² -8 = 0

(2x)² + 2x +(1/2)²-1/4-8 =0

(2x +1/2)² – 33/4=0

(2x +1/2)² – (√33/2)²=0

(2x + 1/2+ √33/2)(2x + 1/2-√33/2)=0

[2x +(√33 +1)/2][2x +(-√33 +1)/2] =0

x = -(√33 +1)/4 or x = (√33 -1)/4

x = (-√33 -1)/4 or x = (√33 -1)/4

(iii) 4x²+ 4√3x + 3 = 0

(2x)² + 4√3x + 3 = 0

Comparing this equation with comlete square polynomial (a +b)² = a² + 2ab + b²

2ab = 4√3x

2(2x)b =4√3x

b = √3

Adding and subtracting (√3)²  in the RHS of the equation

(2x)² + 4√3x + (√3)²  – (√3)²+3 = 0

(2x + √3)² – (√3)²+3 =0

(2x + √3)² =0

(2x + √3)(2x + √3) =0

x = -√3/2 or x =-√3/2

(iv) 2x²+ x + 4 = 0

Multiplying the quadratic equation by 2

4x²+ 2x +8 = 0

(2x)² + 2x +8 = 0

Comparing this equation with comlete square polynomial (a +b)² = a² + 2ab + b²

2ab = 2x

2(2x)b =2x

b = 1/2

Adding and subtracting (1/2)²  in the RHS of the equation

(2x)² + 2x +(1/2)² -(1/2)² +8 = 0

(2x)² + 2x +(1/2)²-1/4+8 =0

(2x +1/2)² + 31/4=0

(2x +1/2)² = -31/4

Square of a number can’t be negative, therefore the roots of the given quadratic equation can’t be real

See the video for clearing your doubt 

 

Q2.Find the roots of the quadratic equations given in Q.1 above by applying the quadratic formula.

(i) 2x2 – 7x + 3 = 0

Ans.On comparing the given equation with standard quadratic equation ax² +bx +c =0

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

Applying the formula of quadratic equation

 

x =(7 +5)/4 =3

x = (7 -5)/4 =2/4 = 1/2

(ii) 2x²+ x -4 = 0

Ans.On comparing the given equation with standard quadratic equation ax² +bx +c =0

a =2, b = 1 and c =-4

Applying the formula of quadratic equation

Hence the roots are x = (-√33 -1)/4 or x = (√33 -1)/4

(iii) 4x²+ 4√3x + 3 = 0

Ans.On comparing the given equation with standard quadratic equation ax² +bx +c =0

a =4, b = 4√3 and c =3

Applying the formula of quadratic equation

x = (-4√3+0/8) =-√3/2 or x = (-4√3-0/8) =-√3/2

(iv) 2x²+ x + 4 = 0

Ans.On comparing the given equation with standard quadratic equation ax² +bx +c =0

a =2, b = 1 and c =4

Applying the formula of quadratic equation

∴ x = (-1 +√-31)/4 or  x = (-1 -√-31)/4

Q3.Find the roots of the following equations:

(i) x-1/x = 3, x ≠ 0
(ii) 1/(x+4) – 1/(x-7) = 11/30, x≠ -4, 7

Ans.(i) x-1/x = 3, x ≠ 0

Multiplying the equation by x

x² – 1 =3x

x² – 3x -1 =0

x = (3+√13)/2 or x = (3-√13)/2 

(ii) 1/(x+4) – 1/(x-7) = 11/30

Simplifying it

(x -7 -x -4)/(x+4)(x-7) = 11/30

-11/(x² -3x -28) = 11/30

x² -3x – 28=-30

x² -3x – 28 +30 =0

x² -3x + 2 =0

x² -2x -x+ 2 =0

x( x – 2) -1(x – 2) =0

( x – 2)(x – 1) =0

x =2,1

The required roots of the quadratic equation are x =2 or x =1

Q4.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.

Ans. Let the present age of Rahman is x

3 years ago Rehman’s age is x – 3

3 years from now Rehman’s age is x +5

According to question the sum of reciprocals of Rehman’s ages, 3 years ago and 5 years from now is 1/3

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

Simplifying it

(x +5 + x -3)/(x -3)(x +5)  =1/3

(2x +2)/(x²+2x -15) = 1/3

x²+2x -15 = 6x +6

x²+2x -15 – 6x -6=0

x²-4x -21 =0

x²-7x+3x -21 =0

x(x -7) + 3(x – 7) =0

(x -7)(x +3) =0

x = 7, -3

Neglecting x =-3 ,since age can’t be negative 

Therefore Rehman’s age is 7 years

Solutions of Exercise 4.3 Question-5 Quadratic Equation

Q5.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.

Ans. Let Shefali’s marks in Mathematics be x

Then her marks in English will be 30 – x

According to question

(x +2)(30 – x -3) =210

(x +2)(27 -x) =210

27x -x² +54 -2x =210

-x² + 25x +54 – 210 =0

-x² + 25x – 156 =0

x² – 25x + 156 =0

x² – 12x-13x + 156 =0

x(x -12) – 13(x – 12) =0

(x -12)(x – 13) =0

x =12,13

If Shefali’s marks in Mathematics are 12 then marks in english are 30-12=18

If Shefali’s marks in Mathematics are 13 then marks in english are 30-13=17

Solutions of Exercise 4.3 Question-6 Quadratic Equation

Q6.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.

Ans.Let the length of the shorter side of the rectangle is x m

Then the diagonal of a rectangular field is (x +60) m

Also longer side of the rectangle is (x +30) m

Applying the Pythogorus theorem

(x +60)² = x² + (x +30) ²

x² + 120x +3600 = x² + x² + 60x +900

x² + 60x -120x+900 -3600 =0

x² -60x -2700 =0

x² -90x+30x -2700 =0

x(x -90) + 30(x – 90) =0

(x -90)(x +30) =0

x =90,-30

Neglecting x =-30 because length of side can’t be negative

Hence length of shorter side of the field is 90 m ,length of diagonal is 90 +60 =150 m, length of longer side is 90 +30 =120 m

Solutions of Exercise 4.3 Question-7 Quadratic Equation

Q7.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.

Ans. Let one of the number is x

Then its square is x²

Difference of squares of two numbers is given 180

Square of another number is = x²-180

Let the larger number is x

According to the question

x²-180 = 8x

x² -8x – 180 = 0

x² – 8x – 180 =0

x² – 18x +10x -180 =0

x(x- 18)+10(x – 18) =0

(x- 18)(x +10) =0

x = 18,-10

Neglecting x = -10, because it is given that the square of the smaller number is 8 times the larger number,so square of a number can’t be negative

Therefore larger number is 18 

Then smaller number is =±√(x²-180) =±√(18²-180)=±√(324-180)=±√144=±12

Solutions of Exercise 4.3 Question-8 Quadratic Equation

Q8.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.

Ans. Let the speed of the train is x km/h

The distance given is 360 km

Time taken with the speed of x =distance/time =360/x

The modified speed is (x + 5) km/h

Time taken with the modified speed of x+5 =distance/time =360/(x+5)

According to question

360/(x+5) =360/x -1

360/(x+5)-360/x =-1

(360x -360x -1800)/x(x+5) =-1

-x² -5x = -1800

x² + 5x =1800

x² + 5x -1800=0

x² + 45x -40x-1800=0

x(x +45) -40(x -45) =0

(x +45)(x -40) =0

x =-45,40

Speed can’t be negative, therefore the speed of the train is 40 km/h

Solutions of Exercise 4.3 Question-9 Quadratic Equation

Q9.Two water taps together can fill a tank in 9(3/8) 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.

Ans. Let the smaller tap is capable to fill the tank in x h

In 1 h smaller tap can fill the tank = (1/x) part

Then the tank filled up by the larger tap is (x -10) h

In 1 h larger tap can fill the tank = (1/(x-10) part

Both taps are capable to fill the tank in 9(3/8)=75/8 h

In 1 h both taps can fill the tank = 8/75 part

Therefore as per the condition in the question

1/x + 1/(x-10) = 8/75

(x-10 +x)/x(x-10) = 8/75

(2x-10)75 = 8x² -80x

150x+ 750 = 8x² -80x

8x² -80x -150x +750=0

8x² -230x +750=0

8x² -200x-30x +750=0

8x(x – 25) – 30( x – 25) =0

(x – 25)(8x – 30) =0

x =25, 30/8 =3.75

If the smaller tank fill the tank in 25 h then the larger tank will fill the tank in 25-10=15h

If the smaller tank fill the tank in 3.75 h then the larger tank will fill the tank in 3.75-10=negative value which is impossible

Solutions of Exercise 4.3 Question-10 Quadratic Equation

Q10.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.

Ans.Let the speed of the passenger train is x km/h

The distance travelled by the trains is 132 km

Time taken by the  the passenger train is =distance/speed=132/x h

The speed of the express train is (x+11) km/h

Time taken by  the express train is =distance/speed=132/(x+11) h

According to the question

132/(x+11) = 132/x – 1

132/(x+11) – 132/x = – 1

(132x -132x – 1452)/( x+11)x = -1

-1452 = -x2– 11x

X2 + 11x – 1452 = 0

X2 + 44x –33x- 1452 = 0

X(x + 44) -33(x +44) =0

(x + 44)(x – 33) =0

X =-44,33

Neglecting x =-44 because speed of the train can’t be negative

Therefore average speed of the passenger train is 33 km/h and average speed of express train is 33 +11 = 44 km/h

Therefore the smaller tank fill the tank in 25 h then the larger tank will fill the tank in 15h

Solutions of Exercise 4.3 Question-11 Quadratic Equation

Q11. Sum of the areas of two squares is 468 m2. If the difference of their perimeters is 24 m, find the sides of the two squares.

Ans.Let the side of the a square is x m

Then perimeter of the square is 4x

Perimeter of another square is 4x -24

Then side of another square is (4x -24)/4 = x -6

Since sum of the areas of two squares is 468 m2

x2 +(x -6)2 = 468

x2+ x2+36 -12x = 468

2x2 -12x – 468 +36 =0

2x2 -12x – 432 =0

x2 -6x – 216 =0

x2 -18x+12x – 216 =0

x(x -18) +12(x – 18) =0

(x -18)(x +12) =0

X = 18, -12

Neglecting the value of  x= -12

Therefore the side of one of the square is 18 m

Then side of another square is = x -6=18 -6 =12 m

NCERT Solutions for Class 10 maths Chapter 4 Quadratic Equation

Exercise 4.4 

Solutions of Exercise 4.4 Question-1 Quadratic Equation

 Q1.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

Ans. (i) 2x2 -3x + 5 = 0

D=b2– 4ac =(-3)2– 4(2)5 = 9 -40 =-31

Since D < 0,therefore roots of the equation are not real

(ii)   3x2 – 4√3x + 4 = 0

Ans.a =3,b = – 4√3, c =4

D=b2– 4ac =(- 4√3)2– 4(3)4 = 48 -48 = 0

Therefore roots of the equation are equal

x = -b/2a = (- 4√3)/2(3) = – 2/√3 or x = – 2/√3

(iii) 2x2 – 6x + 3 = 0

D=b2– 4ac =(-6)2– 4(2)3 = 36 – 24 =-12

x = (-b± √ D)/2a = [-(-6)±√12)/2×2=(6 ±√12)/4

Q2.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

Ans. (i) 2x2 + kx + 3 = 0

D=b2– 4ac = k2-4(2)3 = k2-24=0

K2 = 24

K = ±√24 = ±2√6

(ii) kx (x – 2) + 6 = 0

Kx2– 2kx +6 =0

D=b2– 4ac = (-2k)2-4k(6) = 4k2-24k=0

4k2-24k =0

4k(k -6) =0

K=0, k=6

Q3.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.

Ans. Let the length of the rectangular mango grove is x

Then its length is 2x

Area of  rectangular mango grove = Length x Breadh =x (2x) = 2x2

According to question

2x2= 800

x2 = 400

x = +-20

Neglecting the sign x= – 20, therefore breadh of the rectangle is 20 m

And length of the rectangle is = 2x =2(20) =40

Q4.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.

Ans. Let the age of one of the friend is x years

Since sum of the ages of two friends  given is 20 years,then age of another friend is 20 –x

Four years ago the age of one friend is (x – 4) years

Four years ago the age of another friend is 20-x-4= (16 -x ) years

Four years ago, the product of their ages given 48.

Therefore

(x – 4) (16 -x ) = 48

16 x – x2 – 64 + 4x = 48

16 x – x2 – 64 + 4x – 48=0

-x2 + 20x – 112=0

X2 – 20x + 112 =0

A = 1,b = -20, c = 112

D=b2– 4ac =(-20)2 -4(1)112 =400 -448= – 48

D < 0

Therefore the roots of quadratic equation are not real

Hence given situation in the question is not possible

Q5. Is it possible to design a rectangular park of perimeter 80 m and area 400 m2? If so find its length and breadth.

Ans. Let the breadh of a rectangular park is x m

Perimeter of the given park =2(length +breadh) = 80

Therefore length of the given park is = (80 – 2x)/2 = (40 – x) m

Area of the given park = 400 m2

X( 40-x) =400

40x – x2-400 =0

40 x –x2 – 400 = 0

x2 – 40x + 400 = 0

D=b2– 4ac =(-40)2 -4(1)400 =1600 – 400=  0

D = 0

Therefore the roots of the quadratic equation are  real and equal

Hence  it is possible to design  such a rectangular park

Both of the roots of the equation are  x= –b/2a = -(-40)/2= 20

Breadh of the rectangular park is 20 m and its length 40 -20 =20 m

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