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NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equation

Last Updated : 20 Mar, 2024
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NCERT Solutions Class 10 Maths Chapter 4 Quadratic Equations is a resource created by the team at GFG to help students clear their doubts while solving problems from the NCERT textbook. It helps clear the frustration caused by being stuck on a problem for a while. The NCERT Solutions for Class 10 Maths covers all the questions in the exercise of the NCERT textbook for this chapter.

In this NCERT Class 10 Maths Chapter 4 Solutions, students are introduced to the world of quadratic equations, although quadratic equations are introduced to students in Class 9 under the chapter named “Polynomials” there was no separation of this from polynomials but in reality, there are many applications of quadratic equation, therefore it is studied here in class 10 maths as a stand-alone chapter. In this chapter, students learn about quadratic equations, various methods to solve quadratic equations, the nature of roots based on discriminants for quadratic equations, and many real-life based problems which can be modelled in the form of quadratic equations and then solved to get the required solution.

List of Exercises in NCERT Class 10 Maths Chapter 4 – Quadratic Equation

These Solutions cover all four exercises of the NCERT Class 10 Maths Chapter 4, which are as follows:

Here, you can also find all of the solutions for the NCERT exercises for CBSE Class 10 Maths.

NCERT Solutions Class 10 Maths Chapter 4 – Quadratic Equation: Exercise 4.1

In this Exercise, problems are based on the basic introductory concept of quadratic equations such as whether a given equation is a quadratic equation or not. Other than this some problems here involve mathematical representation of real-world scenarios which can be modelled in terms of quadratic equations.

Question 1. Check whether the following are quadratic equations:

(i) (x + 1)2 = 2(x – 3) 

Solution:

Here,

LHS = (x + 1)2

= x2 + 2x + 1        (Using identity (a+b)2 = a2 + 2ab + b2)

and, RHS = 2(x–3)

= 2x – 6

As, LHS = RHS

x2 + 2x + 1 = 2x – 6

x2 + 7 = 0 ………….(I)

As, the eqn. (I) is in the form of ax2 + bx + c = 0 where (a≠0).

Hence, the equation is QUADRATIC equation because highest power of x is 2.

(ii) x2 – 2x = (–2) (3 – x)

Solution:

Here,

LHS = x2 – 2x

and, RHS = (–2) (3 – x)

= 2x–6

As, LHS = RHS

x2 – 2x = 2x – 6

x2 – 4x + 6 = 0 ………….(I)

As, the eqn. (I) is in the form of ax2 + bx + c = 0 where (a≠0).

Hence, the equation is QUADRATIC equation because highest power of x is 2.

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

Solution:

Here,

LHS = (x – 2)(x + 1)

= x2 + (–2+1)x + (–2)(1)    (Using identity (x+a) (x+b) = x2 + (a+b)x + ab)

= x2 – x – 2

and, RHS = (x – 1)(x + 3) 

= x2 + (–1+3)x + (–1)(3)   (Using identity (x+a) (x+b) = x2 + (a+b)x + ab)

= x2 + 2x – 3

As, LHS = RHS

x2 – x – 2 = x2 + 2x – 3

3x – 1 = 0 ………….(I)

As, the eqn. (I) is not in the form of ax2 + bx + c = 0 because (a=0).

Hence, the equation is NOT QUADRATIC equation because highest power of x is 1.

(iv) (x – 3)(2x +1) = x(x + 5)

Solution:

Here,

LHS = (x – 3)(2x +1)

= 2x2 + x +(–3)(2x) + (–3)(1)

= 2x2 – 5x – 3

and, RHS = x(x + 5)

= x2 + 5x

As, LHS = RHS

2x2 – 5x – 3 = x2 + 5x

x2 – 10x – 3 = 0 ………….(I)

As, the eqn. (I) is in the form of ax2 + bx + c = 0 where (a≠0).

Hence, the equation is QUADRATIC equation because highest power of x is 2.

(v) (2x – 1)(x – 3) = (x + 5)(x – 1) 

Solution:

Here,

LHS = (2x – 1)(x – 3)

= 2x2 + (2x)(–3) +(–1)(x) + (–1)(–3)

= 2x2 – 7x + 3

and, RHS = (x + 5)(x – 1) 

= x2 + 5(x) + (–1)(x) + (5)(–1)     (Using identity (x+a) (x+b) = x2 + (a+b)x + ab)

= x2 + 4(x) – 5

As, LHS = RHS

2x2 – 7x + 3 = x2 + 4(x) – 5

x2 – 11x + 8 = 0 ………….(I)

As, the eqn. (I) is in the form of ax2 + bx + c = 0 where (a≠0).

Hence, the equation is QUADRATIC equation because highest power of x is 2.

(vi) x2 + 3x + 1 = (x – 2)2

Solution:

Here,

LHS = x2 + 3x + 1

and, RHS = (x – 2)2

= x2 – 4x + 4      (Using identity (ab)2 = a2 2ab + b2)

As, LHS = RHS

x2 + 3x + 1 = x2 – 4x + 4

7x – 3 = 0 ………….(I)

As, the eqn. (I) is not in the form of ax2 + bx + c = 0 because (a=0).

Hence, the equation is NOT QUADRATIC equation because highest power of x is 1.

(vii) (x + 2)3 = 2x (x2 – 1) 

Solution:

Here,

LHS = (x + 2)3

= x3 + 23 + 3x(2)(x+2)   (Using identity (x+y)3 = x3 + y3 + 3xy(x+y))

= x3 + 8 + 6x2 +12x

and, RHS = 2x (x2 – 1) 

= 2x3  – 2x      

As, LHS = RHS

x3 + 8 + 6x2 +12x = 2x3  – 2x 

x3 – 6x2 – 14x – 8 = 0 ………….(I)

As, the eqn. (I) is not in the form of ax2 + bx + c = 0 where (a≠0).

Hence, the equation is NOT QUADRATIC equation because highest power of x is 3.

(viii) x3 – 4x2 – x + 1 = (x – 2)3

Solution:

Here,

LHS = x3 – 4x2 – x + 1

and, RHS = (x – 2)3

= x3 – 23 – 3x(2)(x–2)     (Using identity (xy)3 = x3 y3 3xy(x-y))

= x3 – 8 – 6x2 +12x

As, LHS = RHS

x3 – 4x2 – x + 1 = x3 – 8 – 6x2 +12x

2x2 – 13x + 9 = 0 ………….(I)

As, the eqn. (I) is in the form of ax2 + bx + c = 0 where (a≠0).

Hence, the equation is QUADRATIC equation because highest power of x is 2.

Question 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 meters) is one more than twice its breadth. We need to find the length and breadth of the plot.

Solution:

Let’s consider,

Breadth of the rectangular plot = b m

Then, length of the plot = (2b + 1) m.

As, Area of rectangle = length × breadth 

528 m2 = (2x + 1) × x

2x2 + x =528

2x2 + x – 528 = 0 ……………….(I)

As, the eqn. (I) is in the form of ax2 + bx + c = 0 where (a≠0).

Hence, the equation is QUADRATIC equation because highest power of x is 2.

Therefore, the length and breadth of plot, satisfies the quadratic equation, 2x2 + x – 528 = 0, which is the required representation of the problem mathematically.

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

Solution:

Let’s consider,

The first integer number = x

Next consecutive positive integer will be = x + 1

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

x2 + x = 306

x2 + x – 306 = 0 ……………….(I)

As, the eqn. (I) is in the form of ax2 + bx + c = 0 where (a≠0).

Hence, the equation is QUADRATIC equation because highest power of x is 2.

Therefore, the two integers x and x+1, satisfies the quadratic equation, x2 + x – 306 = 0, which is the required representation of the problem mathematically.

(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.

Solution:

Let’s consider,

Age of Rohan’s = x  years

So, 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 will be equal to 360, such that

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

x2 + 29x + 3x + 87 = 360

x2 + 32x + 87 – 360 = 0

x2 + 32x – 273 = 0 ……………….(I)

As, the eqn. (I) is in the form of ax2 + bx + c = 0 where (a≠0).

Hence, the equation is QUADRATIC equation because highest power of x is 2.

Therefore, the age of Rohan and his mother, satisfies the quadratic equation, x2 + 32x – 273 = 0, which is the required representation of the problem mathematically.

(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.

Solution:

Let’s consider,

The speed of train = x  km/hr

Time taken to travel 480 km = (480/x) hr

Time = Distance / Speed

Here, According to the given condition,

The speed of train = (x – 8) km/hr

As, the train will take 3 hours more to cover the same distance.

Hence, Time taken to travel 480 km = 480/(x+3) km/hr

As,

Speed × Time = Distance

(x – 8)(480/(x + 3) = 480

480 + 3x – 3840/x – 24 = 480

3x – 3840/x = 24

3x2 – 8x – 1280 = 0 ……………….(I)

As, the eqn. (I) is in the form of ax2 + bx + c = 0 where (a≠0).

Hence, the equation is QUADRATIC equation because highest power of x is 2.

Therefore, the speed of the train, satisfies the quadratic equation, 3x2 – 8x – 1280 = 0, which is the required representation of the problem mathematically.

NCERT Class 10 Maths Chapter 4 – Quadratic Equation: Exercise 4.2

In this exercise, questions related to the first method which is called splitting the middle term. Other than the basic solution of equations in this exercise some word problems based on real-world scenarios are also included, in which we need to formulate the quadratic equation first and then solve it to find the required solution.

Question 1. Find the roots of the following quadratic equations by factorization:

(i) x2– 3x – 10 = 0 

Solution:

Here, LHS = x2– 3x – 10 

= x2 – 5x + 2x – 10

= x(x – 5) + 2(x – 5)

= (x – 5)(x + 2)

The roots of this equation, x2 – 3x – 10 = 0 are the values of x for which 

(x – 5)(x + 2) = 0

Hence, x – 5 = 0 or x + 2 = 0

x = 5 or x = -2

(ii) 2x2 + x – 6 = 0

Solution:

Here, LHS = 2x2 + x – 6

= 2x2 + 4x – 3x – 6

= 2x(x + 2) – 3(x + 2)

= (2x– 3)(x + 2)

The roots of this equation, 2x2 + x – 6 = 0 are the values of x for which

(2x– 3)(x + 2) = 0

Hence, 2x– 3 = 0 or x + 2 = 0

⇒ x = 3/2 or x = –2

(iii) √2x2 + 7x + 5√2 = 0

Solution:

Here, LHS = √2x2 + 7x + 5√2

= √2x2 + 5x + 2x + 5√2

= x(√2x + 5) + √2(√2x + 5)

= (√2x + 5) (x +√2)

The roots of this equation, √2x2 + 7x + 5√2 = 0 are the values of x for which

(√2x + 5) (x +√2) = 0

Hence, √2x + 5 = 0 or x +√2 = 0

x = –5/√2 or x = –√2

(iv) 2x2 – x + 1/8 = 0

Solution:

Here, LHS = 2x2 – x + 1/8

= 1/8(16x2 – 8x + 1)

= 1/8(16x2 – 4x -4x + 1)

= 1/8(4x(4x-1) -1 (4x-1))

= 1/8 (4x-1) (4x-1) 

The roots of this equation, 2x2 – x + 1/8 = 0 are the values of x for which

1/8 (4x-1) (4x-1)  = 0

(4x-1)2 = 0

Hence, 4x-1 = 0 or 4x-1 = 0

⇒ x = 1/4 or x = 1/4

(v) 100x2 – 20x + 1 = 0

Solution:

Here, LHS = 100x2 – 20x + 1

= 100x2 – 10x – 10x + 1

= 10x(10x – 1) – 1(10x – 1)

= (10x – 1) (10x – 1)

The roots of this equation, 100x2 – 20x + 1 = 0 are the values of x for which

(10x – 1) (10x – 1) = 0

(10x – 1)2 = 0

Hence, 10x – 1 = 0 or 10x – 1 = 0

x = 1/10 or x = 1/10

Question 2. Solve 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.

Solution:

Let’s say, 

The number of marbles John have = x.

So, number of marble Jivanti have = 45 – x

After losing 5 marbles each,

Number of marbles John have = x – 5

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

Here, According to the given condition

(x – 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

Hence, 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) 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.

Solution:

 Let us say, 

Number of toys produced in a day be x.

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

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

So, 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

Hence, 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.

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

Solution:

Let’s say, 

First number be x and the second number is 27 – x.

Therefore, the product of two numbers will be:

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

Hence, x – 13 = 0 or x – 14= 0

x = 13 or x = 14

Hence, 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.

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

Solution:

Let’s say, 

Two consecutive positive integers be x and x + 1.

Here, According to the given condition,

x2 + (x + 1)2 = 365

x2 + x2 + 1 + 2x = 365

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

Hence, x – 13 = 0 or x + 14= 0

x = 13 or x = – 14

As, here it is said positive integers, so x can be 13, only.

So,

x = 13

and, x + 1 = 13 + 1 = 14

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

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.

Solution:

Let’s say, 

Base of the right triangle be x cm.

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

Base2 + Altitude2 = Hypotenuse2                       (Pythagoras theorem)

x2 + (x – 7)2 = 132

x2 + x2 + 49 – 14x = 169    (using identity (a-b)2 = a2 – 2ab + b2)

2x2 – 14x – 120 = 0

x2 – 7x – 60 = 0               (Dividing by 2)

x2 – 12x + 5x – 60 = 0

x(x – 12) + 5(x – 12) = 0

(x – 12)(x + 5) = 0

Hence, x – 12 = 0 or x + 5= 0

x = 12 or x = – 5

As, here side will be a  positive integers, so x can be 12, only.

Therefore, the base of the given triangle is 12 cm and,

the altitude of this triangle will be (12 – 7) cm = 5 cm.

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 90, find the number of articles produced and the cost of each article.

Solution:

Let’s say,

Number of articles produced be x.

So, cost of production of each article = ₹ (2x + 3)

Here, According to the given condition

x(2x + 3) = 90

2x2 + 3x – 90 = 0

2x2 + 15x -12x – 90 = 0

x(2x + 15) -6(2x + 15) = 0

(2x + 15)(x – 6) = 0

Hence, 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, 

So, x = 6.

Hence, number of articles produced = 6

Cost of each article = 2 × 6 + 3 = ₹ 15.

Quadratic Equation: Exercise 4.3

In this exercise, questions are based on the second method for solving quadratic equations which is called completing the square method and also the quadratic formula. Other than this, there are word problems based on real-world scenarios same as the previous exercise.

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

(i) 2x2– 7x + 3 = 0

Solution:

2x2 – 7x = – 3

Dividing by 2 on both sides, we get

x2[Tex]\frac{7x}{2}          [/Tex]= –[Tex]\frac{3}{2} [/Tex]

x2 -2 × x ×[Tex]\frac{7}{4}          [/Tex]= –[Tex]\frac{3}{2} [/Tex]

On adding ([Tex]\frac{7}{4}          [/Tex])2 to both sides of equation, we get

(x)2 – 2×x×[Tex]\frac{7}{4}          [/Tex]+([Tex]\frac{7}{4}          [/Tex])2 = ([Tex]\frac{7}{4}          [/Tex])2[Tex]\frac{3}{2} [/Tex]

(x-[Tex]\frac{7}{4}          [/Tex])2 = ([Tex]\frac{49}{16}          [/Tex]) – ([Tex]\frac{3}{2}          [/Tex]) (Using identity: a2 – 2ab + b2 = (a-b)2)

(x-[Tex]\frac{7}{4}          [/Tex])2 =[Tex]\frac{25}{16} [/Tex]

(x-[Tex]\frac{7}{4}          [/Tex])2 = ±[Tex]\frac{5}{4} [/Tex]

x =[Tex]\frac{7}{4} ± \frac{5}{4} [/Tex]

x =[Tex]\frac{7}{4} + \frac{5}{4}          [/Tex]or x =[Tex]\frac{7}{4} – \frac{5}{4} [/Tex]

x =[Tex]\frac{12}{4}          [/Tex]or x =[Tex]\frac{2}{4} [/Tex]

x = 3 or x =[Tex]\frac{1}{2} [/Tex]

(ii) 2x2+ x – 4 = 0

Solution:

2x2 + x = 4

Dividing both sides of the equation by 2, we get

x2 +[Tex]\frac{x}{2}          [/Tex]= 2

Now on adding[Tex](\frac{1}{4})          [/Tex]2 to both sides of the equation, we get,

(x)2 + 2 × x ×[Tex]\frac{1}{4}          [/Tex]+ ([Tex]\frac{1}{4}          [/Tex])2 = 2 + ([Tex]\frac{1}{4}          [/Tex])2

(x +[Tex]\frac{1}{4}          [/Tex])2 =[Tex]\frac{33}{16}          [/Tex](Using identity: a2 + 2ab + b2 = (a+b)2)

x +[Tex]\frac{1}{4}          [/Tex]= ±[Tex]√(\frac{33}{16}) [/Tex]

x =[Tex]– \frac{1}{4} ± \frac{√33}{4} [/Tex]

x =[Tex]\frac{-1± √33}{4} [/Tex]

Hence, x =[Tex]\frac{-1+ √33}{4}          [/Tex]or x =[Tex]\frac{-1- √33}{4} [/Tex]

(iii) 4x2 + 4√3x + 3 = 0

Solution:

4x2 + 4√3x = -3

Dividing both sides of the equation by 4, we get

x2 + √3x = –[Tex]\frac{3}{4} [/Tex]

Now on adding ([Tex]\frac{√3}{2}          [/Tex])2 to both sides of the equation, we get,

(x)2 + 2 × x ×[Tex]\frac{√3}{2}          [/Tex]+ ([Tex]\frac{√3}{2}          [/Tex])2 = –[Tex]\frac{3}{4}          [/Tex]+ ([Tex]\frac{√3}{2}          [/Tex])2

(x +[Tex]\frac{√3}{2}          [/Tex])2 = 0 (Using identity: a2 + 2ab + b2 = (a+b)2)

Hence, x = –[Tex]\frac{√3}{2}          [/Tex]or x = –[Tex]\frac{√3}{2} [/Tex]

(iv) 2x2+ x + 4 = 0

Solution:

2x2 + x = -4

Dividing both sides of the equation by 2, we get

x2 +[Tex]\frac{x}{2}          [/Tex]= -2

Now on adding ([Tex]\frac{1}{4}          [/Tex])2 to both sides of the equation, we get,

(x)2 + 2 × x ×[Tex]\frac{1}{4}          [/Tex]+ ([Tex]\frac{1}{4}          [/Tex])2 = – 2 + ([Tex]\frac{1}{4}          [/Tex])2

(x +[Tex]\frac{1}{4}          [/Tex])2 =[Tex]\frac{-31}{16}          [/Tex](Using identity: a2 + 2ab + b2 = (a+b)2)

As we know, the square of numbers cannot be negative.

Hence, there is no real root for the given equation, 2x2 + x + 4 = 0.

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

(i) 2x2– 7x + 3 = 0

Solution:

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

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

Then roots of the quadratic equation =[Tex]\frac{-b± √(b^2-4ac)}{2a} [/Tex]

x =[Tex]\frac{-(-7)± √((-7)^2-4(2)(3))}{2(2)} [/Tex]

x =[Tex]\frac{7± √(49-24)}{4} [/Tex]

x =[Tex]\frac{7± √25}{4} [/Tex]

x =[Tex]\frac{7± 5}{4} [/Tex]

x =[Tex]\frac{7+5}{4}          [/Tex]or x =[Tex]\frac{7-5}{4} [/Tex]

x =[Tex]\frac{12}{4}          [/Tex]or[Tex]\frac{2}{4} [/Tex]

x = 3 or[Tex]\frac{1}{2} [/Tex]

(ii) 2x2+ x – 4 = 0

Solution:

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

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

Then roots of the quadratic equation =[Tex]\frac{-b± √(b^2-4ac)}{2a} [/Tex]

x =[Tex]\frac{-1± √(1^2-4(2)(-4))}{2(2)} [/Tex]

x =[Tex]\frac{-1± √(1+32)}{4} [/Tex]

x =[Tex]\frac{-1± √33}{4} [/Tex]

x =[Tex]\frac{-1+√33}{4}          [/Tex]or x =[Tex]\frac{1-√33}{4} [/Tex]

(iii) 4x2 + 4√3x + 3 = 0

Solution:

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

a = 4, b = 4√3 and c = 3

Then roots of the quadratic equation =[Tex]\frac{-b± √(b^2-4ac)}{2a} [/Tex]

x =[Tex]\frac{-(4√3)± √((4√3)^2-4(4)(3))}{2(4)} [/Tex]

x =[Tex]\frac{-(4√3)± √(48-48)}{8} [/Tex]

x =[Tex]\frac{-4√3}{8} [/Tex]

x =[Tex]\frac{-√3}{2}          [/Tex]or x =[Tex]\frac{-√3}{2} [/Tex]

(iv) 2x2+ x + 4 = 0

Solution:

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

a = 2, b = 1 and c = 4

Then roots of the quadratic equation =[Tex]\frac{-b± √(b^2-4ac)}{2a} [/Tex]

x =[Tex]\frac{-1± √(1^2-4(2)(4))}{2(2)} [/Tex]

x =[Tex]\frac{-1± √(1-32)}{4} [/Tex]

x =[Tex]\frac{-1± √(-31)}{4} [/Tex]

As we know, the square of a number can never be negative.

Hence, there is no real solution for the given equation.

Question 3. Find the roots of the following equations:

(i) x2 – 3x -1 = 0, x ≠ 0

Solution:

After rearranging, we get

x2 – 3x -1 = 0

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

a = 1, b = -3 and c = -1

Then roots of the quadratic equation =[Tex]\frac{-b± √(b^2-4ac)}{2a} [/Tex]

x =[Tex]\frac{-(-3)± √((-3)^2-4(1)(-1))}{2(1)} [/Tex]

x =[Tex]\frac{3± √(9+4)}{2} [/Tex]

x =[Tex]\frac{3±√(13)}{2} [/Tex]

x =[Tex]\frac{3+√(13)}{2}          [/Tex]or x =[Tex]\frac{3-√(13)}{2} [/Tex]

(ii) [Tex]\frac{(x-7-x-4)}{(x+4)(x-7)} = \frac{11}{30}       [/Tex], x ≠ -4,7

Solution:

[Tex]\frac{(x-7-x-4)}{(x+4)(x-7)} = \frac{11}{30} [/Tex]

[Tex]\frac{(-11)}{(x+4)(x-7)} = \frac{11}{30} [/Tex]

After rearranging,

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

x2 – 3x – 28 = 30

x2 – 3x + 2 = 0

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

a = 1, b = -3 and c = 2

Then roots of the quadratic equation =[Tex]\frac{-b± √(b^2-4ac)}{2a} [/Tex]

x =[Tex]\frac{-(-3)± √((-3)^2-4(1)(2))}{2(1)} [/Tex]

x =[Tex]\frac{3± √(9-8)}{2} [/Tex]

x =[Tex]\frac{3±√1}{2} [/Tex]

x =[Tex]\frac{3±1}{2} [/Tex]

x =[Tex]\frac{4}{2}          [/Tex]or x =[Tex]\frac{2}{2} [/Tex]

x = 2 or x = 1

Question 4. The sum of the reciprocals of Rehman’s ages, (in years) 3 years ago and 5 years from now is[Tex]\frac{1}{3}          [/Tex]. Find his present age.

Solution:

Let’s take,

Present age of Rehman is x years.

Three years ago, Rehman’s age was (x – 3) years.

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

According to the given condition,

[Tex]\frac{1}{x-3} + \frac{1}{x-5} = \frac{1}{3} [/Tex]

[Tex]\frac{(x+5+x-3)}{(x-3)(x+5)} = \frac{1}{3} [/Tex]

[Tex]\frac{(2x+2)}{(x-3)(x+5)} = \frac{1}{3} [/Tex]

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

6x + 6 = x2 + 2x – 15

x2 – 4x – 21 = 0

x2 – 7x + 3x – 21 = 0 (by factorizing)

x(x – 7) + 3(x – 7) = 0

(x – 7)(x + 3) = 0

x = 7, -3

As, age cannot be negative.

Therefore, Rehman’s present age is 7 years.

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.

Solution:

Let’s take,

The marks of Shefali in Maths be x.

Then, the marks in English will be 30 – x.

According to the given condition,

(x + 2)(30 – x – 3) = 210

(x + 2)(27 – x) = 210

-x2 + 25x + 54 = 210

Multiply the equation by (-1),

x2 – 25x + 156 = 0

x2 – 12x – 13x + 156 = 0

x(x – 12) -13(x – 12) = 0

(x – 12)(x – 13) = 0

x = 12, 13

Hence, if the marks in Maths are 12, then marks in English will be 30 – 12 = 18

and, the marks in Maths are 13, then marks in English will be 30 – 13 = 17.

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.

Solution:

Let’s take,

Breadth = x

Length = x+30

Diagonal = x+60

Diagonal = √(Length2 + Breadth2)

According to the given condition,

√((x+30)2 + (x)2) = x+60

Squaring both sides,

x2 + (x + 30)2 = (x + 60)2

x2 + x2 + 900 + 60x = x2 + 3600 + 120x

x2 – 60x – 2700 = 0

x2 – 90x + 30x – 2700 = 0

x(x – 90) + 30(x -90) = 0

(x – 90)(x + 30) = 0

x = 90, -30

As, side of the field cannot be negative.

Hence, the length of the shorter side will be 90 m.

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

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.

Solution:

Let’s take,

The larger number = x

and, smaller number = y

According to the given condition,

x2 – y2 = 180 and y2 = 8x (It means x has to be positive, because it is obtained by squaring a number)

x2 – 8x = 180

x2 – 8x – 180 = 0

x2 – 18x + 10x – 180 = 0

x(x – 18) +10(x – 18) = 0

(x – 18)(x + 10) = 0

x = 18, -10

As x cannot be negative,

Hence, the larger number will be 18.

x = 18

So, As y2 = 8x

= 8 × 18

= 144

y = ±√144 = ±12

So, Smaller number = ±12

Hence, the numbers are 18 and 12 or 18 and -12.

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.

Solution:

Let’s take

The speed of the train = x km/hr.

As, Speed =[Tex]\frac{Distance}{Time} [/Tex]

Time taken to cover 360 km =[Tex]\frac{360}{x}          [/Tex]hr.

As per the question given,

(x + 5)([Tex]\frac{360}{x}          [/Tex]– 1) = 360

(x + 5)([Tex]\frac{360 – x}{x}          [/Tex]) = 360

(x + 5)(360 – x) = 360x

x2 + 5x -1800 = 0

x2 + 45x – 40x + 1800 = 0

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

(x + 45)(x – 40) = 0

x = 40, -45

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

Hence, the speed of train is 40 km/h.

Question 9. Two water taps together can fill a tank in 9[Tex]\frac{3}{8}          [/Tex]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.

Solution:

Let’s take

The time taken by the smaller pipe to fill the tank = x hr.

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

Part of tank filled by smaller pipe in 1 hour =[Tex]\frac{1}{x} [/Tex]

Part of tank filled by larger pipe in 1 hour =[Tex]\frac{1}{x-10} [/Tex]

According to the given condition,

9[Tex]\frac{3}{8}          [/Tex]hrs taken to fill with both the pipe.

So,

[Tex]\frac{1}{x} + \frac{1}{x-10} = \frac{8}{75} [/Tex]

[Tex]\frac{x-10+x}{x(x-10)} = \frac{8}{75} [/Tex]

[Tex]\frac{2x-10}{x(x-10)} = \frac{8}{75} [/Tex]

75(2x – 10) = 8x2 – 80x

150x – 750 = 8x2 – 80x

8x2 – 230x +750 = 0

8x2 – 200x – 30x + 750 = 0

8x(x – 25) -30(x – 25) = 0

(x – 25)(8x -30) = 0

x = 25,[Tex]\frac{30}{8} [/Tex]

Time taken by the smaller pipe cannot be[Tex]\frac{30}{8}          [/Tex]hours, as the time taken by the larger pipe will become negative.

Hence, time taken by the smaller pipe = 25hours

and, by the larger pipe =15 hours

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 speed of the express train is 11km/h more than that of the passenger train, find the average speed of the two trains.

Solution:

Let’s take

The average speed of passenger train = x km/h.

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

According to the given condition,

[Tex]\frac{132}{x} + \frac{132}{x+11}          [/Tex]= 1

[Tex]\frac{132(x+11-x)}{x(x+11)}          [/Tex]= 1

[Tex]\frac{132 × 11}{x(x+11)}          [/Tex]= 1

132 × 11 = x(x + 11)

x2 + 11x – 1452 = 0

x2 + 44x -33x -1452 = 0

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

(x + 44)(x – 33) = 0

x = – 44, 33

As we know, Speed cannot be negative.

Hence, the speed of the passenger train will be 33 km/h

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

Question 11. 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.

Solution:

Let the sides of the two squares be x and y meter.

Perimeter = 4x and 4y respectively

Area = x2 and y2 respectively.

According to the given condition,

4x – 4y = 24

x – y = 6

x = y + 6 ………………….(I)

and,

x2 + y2 = 468

(6 + y)2 + y2 = 468 (From (I))

36 + y2 + 12y + y2 = 468

2y2 + 12y + 432 = 0

y2 + 6y – 216 = 0

y2 + 18y – 12y – 216 = 0

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

(y + 18)(y – 12) = 0

y = -18, 12

As we know, the side of a square cannot be negative.

Hence, the sides of the squares are 12 m

and, (12 + 6) m = 18 m.

NCERT Solutions for Class 10 Maths Chapter 4 – Quadratic Equation: Exercise 4.4

This exercise contains problems based on the nature of the root of the quadratic equation which can be determined using the quadratic formula. Other than that same word problems are based on real-world scenarios.

Question 1. Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:

(i) 2x2-3x+5=0

Solution:

(i) Given: 2x2-3x+5=0

Here a=2,b=-3 and c=5

[Tex]\therefore         [/Tex] Discriminant, D=b2-4ac

⇒ D = (-3)2– 4 × 2 × 5)

⇒ D = 9-40 = -31 < 0

Hence, the roots are imaginary.

(ii) 3x2-4√3x+4=0

Solution:

(ii) Given: 3x2-4√3x + 4 = 0

Here a=3,b=√3 and c=4

[Tex]\therefore         [/Tex] Discriminant, D=b2-4ac

⇒ D = (-4√3)2 – (4 × 3 × 4)

⇒ D= 48 – 48 = 0

Hence, the roots are real and equal.

Using the formula,

[Tex]x=\frac {-b\pm\sqrt{b^2-4ac}} {2a}                [/Tex], we get 

[Tex]x=\frac {-(-4\sqrt3)\pm\sqrt{(-4\sqrt3)^2-4\times3\times4}} {2\times3} [/Tex]

[Tex]= \frac {4\sqrt3\pm\sqrt{48-48}} {6}=\frac {4\sqrt3} {6}=\frac {2} {\sqrt3} [/Tex]

Hence, the equal roots are [Tex]\frac 2 {\sqrt3}         [/Tex] and [Tex]\frac 2 {\sqrt3}                [/Tex].

(iii) 2x2-6x+3=0

Solution:

(iii) Given: 2x2-6x+3=0

Here, a=2,b=-6 and c=3

[Tex]\therefore         [/Tex] Discriminant, D=b2-4ac

⇒ D = (-6)2 – (4 × 2 × 3)

⇒ D = 36 – 24 = 12 > 0

Hence, the roots are distinct and real.

Using the formula,

[Tex]x=\frac {-b\pm\sqrt{b^2-4ac}} {2a}                [/Tex],we get

[Tex]x=\frac {-(-6)\pm\sqrt{(-6)^2-4\times2\times3}} {2\times2}  [/Tex]

[Tex]x=\frac {6\pm\sqrt{36-24}} {4}  [/Tex]

[Tex]x=\frac {6\pm\sqrt{12}} {4}  [/Tex]

[Tex]x=\frac {6\pm2\sqrt{3}} {4}  [/Tex]

[Tex]x=\frac {3\pm\sqrt{3}} {2}  [/Tex]

Hence, the equal roots are [Tex]\frac {3+\sqrt{3}} {2}         [/Tex]and [Tex]\frac {3-\sqrt{3}} {2}  [/Tex]

Question 2. Find the values of k for each of the following quadratic equations, so that they have two equal roots.

(i) 2x2+kx+3

Solution:

This equation is of the form ax2+bx+x, where a=2, b=k and c=3.

Discriminant, D=b2-4ac

⇒D= k – 4 × 2 × 3

⇒D = k2 -24

For equal roots D=0

⇒ k2-24=0

⇒ k2=24

⇒ k2 = ±24 = ±2√6​
 

(ii) kx(x-2)+6=0

Solution:

⇒  kx2-2kx+6=0

This equation is of the form ax2+bx+c=0, where a=k, b=-2k and c=6.

Discriminant, D=b2-4ac

⇒ D =(-2k)2 – 4 × k × 6

⇒ D =4k2-24k

For equal roots D=0

⇒  4k2-24k=0

⇒  4k(k-24)=0

⇒ k=0 (not possible) or 4k-24=0

⇒  k= 24/4=6

Question 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 rectangular mango grove be x m.

Then, the length of the rectangular mango grove will be 2x m.

The Area of the rectangular mango grove=length × breadth

According to the question, we have

x × 2x= 800

⇒  2x2=800

⇒  x2=400

⇒  x=20

Hence, the rectangular mango grove is possible to design whose length=40 m and breadth=20 m.

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.

Solution:

Let the present age of one friend be x years.

Then, the present age of other friend be (20-x) years.

4 years ago, one friend’s age was (x-4) years

4 years ago, other friend’s age was (20-x-4)=(16-x) years.

According to the question,

(x-4)(16-x)=48

⇒  16x-64-x2+4x=48

⇒  x2-20x+112=0

This equation is of the form ax2+bx+c=0,where a=1, b=-20 and c=112.

Discriminant, D=b2-4ac

⇒ D = (-20)2-4 × 1 × 112 = -48 < 0

Since, there are no real roots.

So the given situation is not possible.

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

Solution:

Let the length of the rectangular park be x.

The perimeter of the rectangular park= 2(length + breadth)

⇒  2(x + breadth)=80

⇒  breadth=40-x

The area of rectangular park= length × breadth

⇒  x(40-x)=400

⇒  40x-x2=400

⇒  x2-40x+400=0

⇒  x2 -20x-20x+400=0

⇒ (x-20)(x-20)=0

⇒ x=20

Hence, the rectangular park is possible to design. So, the length of the park is 20m and the breadth = 40-20=20m.

Important Points to Remember:

  • These NCERT solutions for class 10 Maths Solution are developed by the GFG team, with a focus on students’ benefit.
  • These solutions are entirely accurate and can be used by students to prepare for their board exams.
  • Each solution is presented in a step-by-step format with comprehensive explanations of the intermediate steps. 

Features of NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations

The seasoned faculty at GeeksforGeeks meticulously crafts NCERT Solutions, ensuring precision. These solutions, proven to be 100% accurate, serve as invaluable tools for students gearing up for their CBSE board exams. In addition to aiding board preparations, the comprehensive coverage of minute concepts equips students to tackle various competitive exams with increased confidence. The step-wise answers to exercise questions from the NCERT Textbook further enhance the learning experience, guiding students not only towards correct final answers but also fostering a deep understanding of each procedural step, ultimately leading to higher scores.

Also Check:

FAQs – NCERT Solutions for Class 10 Maths Chapter 4

Q1: Why is it important to learn quadratic equations?

Quadratic Equations are one of the most useful concepts in mathematics. It is used in various real-life applications such as engineering, physics, finance, and computer graphics. It is also useful as a problem-solving tool as many real-world problems can be modelled into a quadratic equation.

Q2: What topics are covered in NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations?

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations covers topics such as the definition of quadratic equations, finding the roots of quadratic equations using different methods, discriminant, nature of roots, and the graphical representation of quadratic equations.

Q3: How can NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations help me?

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations can help you solve the NCERT exercise without any limitations. If you are stuck on a problem you can find its solution in these solutions and free yourself from the frustration of being stuck on some question.

Q4: How many exercises are there in Class 10 Maths Chapter 4 Quadratic Equations?

There are 4 exercises in the Chapter 4 Maths Class 10 Quadratic Equations which covers all the important topics and sub-topics.

Q5: Where can I find NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations?

You can find these NCERT Solutions in this article or you can search for it in the google find it there as well.



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