NCERT Solutions for Exercise 4.1 Class 10 Maths Chapter 4 - Quadratic Equations

NCERT Solutions for Exercise 4.1 Class 10 Maths Chapter 4 - Quadratic Equations

Edited By Ramraj Saini | Updated on Nov 15, 2023 02:37 PM IST

NCERT Solutions For Class 10 Maths Chapter 4 Exercise 4.1

NCERT Solutions for Exercise 4.1 Class 10 Maths Chapter 4 Quadratic Equations are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. This ex 4.1 class 10 deals with the concept of quadratic equations which is nothing but a polynomial with degree 2. The general form of the quadratic equation in the variable x is ax2+ bx + c = 0 where a, b, c are real numbers. Class 10 maths ex 4.1 consists of 2 simple problems with subsections that are easy to solve and also explore the concepts of the relationship between roots/zeroes. In exercise 4.1 Class 10 Maths, there are three main methods to solve the quadratic equation i.e. Factorizations, Completing the square and Quadratic Formula.

NCERT solutions for Class 10 Maths exercise 4.1, focused on the concepts of solving quadratic equations and understanding more about the relationship between the roots of the equation and the nature of the equation. These class 10 maths ex 4.1 solutions are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise together using the link provided below.

Download Free Pdf of NCERT Solutions for Class 10 Maths chapter 4 exercise 4.1

Download PDF

Assess NCERT Solutions for Class 10 Maths chapter 4 exercise 4.1

Quadratic Equations class 10 chapter 4 Excercise: 4.1

Q1 (i) Check whether the following are quadratic equations : (x+1)^2 = 2(x-3)

Answer:

We have L.H.S. (x+1)^2 = x^2+2x+1

Therefore, (x+1)^2 = 2(x-3) can be written as:

\Rightarrow x^2+2x+1 = 2x-6

i.e., x^2+7 = 0

Or x^2+0x+7 = 0

This equation is of type: ax^2+bx+c = 0 .

Hence, the given equation is a quadratic equation.

Q1 (ii) Check whether the following are quadratic equations : x^2 - 2x = (-2)(3-x)

Answer:

Given equation x^2 - 2x = (-2)(3-x) can be written as:

\Rightarrow x^2 -2x = -6+2x

i.e., x^2-4x+6 = 0

This equation is of type: ax^2+bx+c = 0 .

Hence, the given equation is a quadratic equation.

Q1 (iii) Check whether the following are quadratic equations : (x-2)(x+1) = (x-1)(x+3)

Answer:

L.H.S. (x-2)(x+1) can be written as:

= x^2+x-2x-2 = x^2-x-2

and R.H.S (x-1)(x+3) can be written as:

= x^2+3x-x-3 = x^2+2x-3

\Rightarrow x^2-x-2 = x^2+2x-3

i.e., 3x-1 = 0

The equation is of the type: ax^2+bx+c = 0,a\neq0 .

Hence, the given equation is not a quadratic equation since a=0.

Q1 (iv) Check whether the following are quadratic equations : (x-3)(2x+1) = x(x+5)

Answer:

L.H.S. (x-3)(2x+1) can be written as:

= 2x^2+x-6x-3 = 2x^2-5x-3

and R.H.S (x)(x+5) can be written as:

= x^2+5x

\Rightarrow 2x^2-5x-3 = x^2+5x

i.e., x^2-10x-3 = 0

This equation is of type: ax^2+bx+c = 0,a\neq0 .

Hence, the given equation is a quadratic equation.

Q1 (v) Check whether the following are quadratic equations : (2x -1)(x-3) = (x+5)(x-1)

Answer:

L.H.S. (2x-1)(x-3) can be written as:

= 2x^2-6x-x+3 = 2x^2-7x+3

and R.H.S (x+5)(x-1) can be written as:

=x^2-x+5x-5 = x^2+4x-5

\Rightarrow 2x^2-7x+3 = x^2+4x-5

i.e., x^2-11x+8 = 0

This equation is of type: ax^2+bx+c = 0,a \neq 0 .

Hence, the given equation is a quadratic equation.

Q1 (vi) Check whether the following are quadratic equations : x^2 +3x +1 = (x-2)^2

Answer:

L.H.S. x^2+3x+1

and R.H.S (x-2)^2 can be written as:

= x^2-4x+4

\Rightarrow x^2+3x+1 = x^2- 4x+4

i.e., 7x-3 = 0

This equation is NOT of type: ax^2+bx+c = 0 , a\neq0 .

Here a=0, hence, the given equation is not a quadratic equation.

Q1 (vii) Check whether the following are quadratic equations : (x+2)^3 = 2x(x^2 -1)

Answer:

L.H.S. (x+2)^3 can be written as:

= x^3+8+6x(x+2) =x^3+6x^2+12x+8

and R.H.S 2x(x^2-1) can be written as:

= 2x^3-2x

\Rightarrow x^3+6x^2+12x+8 = 2x^3-2x

i.e., x^3-6x^2-14x-8 = 0

This equation is NOT of type: ax^2+bx+c = 0 .

Hence, the given equation is not a quadratic equation.

Q1 (viii) Check whether the following are quadratic equations : x^3 -4x^2 -x +1 = (x -2)^3

Answer:

L.H.S. x^3 -4x^2 -x +1 ,

and R.H.S (x-2)^3 can be written as:

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

\Rightarrow x^3-4x^2-x+1 = x^3-6x^2+12x-8

i.e., 2x^2-13x+9=0

This equation is of type: ax^2+bx+c = 0 .

Hence, the given equation is a quadratic equation.

Q2 (i) Represent the following situations in the form of quadratic equations : The area of a rectangular plot is 528m^2 . 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.

Answer:

Given the area of a rectangular plot is 528m^2 .

Let the breadth of the plot be 'b' .

Then, the length of the plot will be: = 2b +1 .

Therefore the area will be:

=b(2b+1)\ m^2 which is equal to the given plot area 528m^2 .

\Rightarrow 2b^2+b = 528

\Rightarrow 2b^2+b - 528 = 0

Hence, the length and breadth of the plot will satisfy the equation 2b^2+b - 528 = 0

Q2 (ii) Represent the following situations in the form of quadratic equations : The product of two consecutive positive integers is 306. We need to find the integers.

Answer:

Given the product of two consecutive integers is 306.

Let two consecutive integers be 'x' and 'x+1' .

Then, their product will be:

x(x+1) = 306

Or x^2+x- 306 = 0 .

Hence, the two consecutive integers will satisfy this quadratic equation x^2+x- 306 = 0 .

Q2 (iii) Represent the following situations in the form of quadratic equations: 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 the age of Rohan be 'x' years.

Then his mother age will be: 'x+26' years.

After three years,

Rohan's age will be 'x+3' years and his mother age will be 'x+29' years.

Then according to question,

The product of their ages 3 years from now will be:

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

\Rightarrow x^2+3x+29x+87 = 360 Or

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

Hence, the age of Rohan satisfies the quadratic equation x^2+32x-273 = 0 .

Q2 (iv) Represent the following situations in the form of quadratic equations : 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 the train be 's' km/h.

The distance to be covered by the train is 480\ km .

\therefore The time taken will be

=\frac{480}{s}\ hours

If the speed had been 8\ km/h less, the time taken would be: \frac{480}{s-8}\ hours .

Now, according to question

\frac{480}{s-8} - \frac{480}{s} = 3

\Rightarrow \frac{480x - 480(x-8)}{(x-8)x} = 3

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

\Rightarrow 3840 = 3x^2-24x

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

Dividing by 3 on both the side

x^2 -8x-1280 = 0

Hence, the speed of the train satisfies the quadratic equation x^2 -8x-1280 = 0

More About NCERT Solutions for Class 10 Maths Exercise 4.1:

Exercise 4.1 Class 10 Maths consists of a question based on checking whether the given equations are quadratic equations. Mathematically, the roots of the equation can also be found by the Graphical method. To solve a linear system of equations graphically, we first need to draw a graph of the equations. The solution to the linear equation is where the lines intersect graphically. The solutions of simultaneous linear equations are given by the method (x, y). The NCERT solutions for Class 10 Maths exercise 4.1 mainly focused on the concepts of solving quadratic equations. Two questions related to checking whether the given equation is a quadratic equation and solving quadratic equations are given in NCERT syllabus exercise 4.1 Class 10 Maths. Also students can get access of Quadratic Equations Class 10 Notes to revise all the concepts quickly.

Benefits of NCERT Solutions for Class 10 Maths Exercise 4.1:

  • NCERT Solutions for Class 10 Maths exercise 4.1 will assist us in resolving and validating all troubles on this exercise, as properly as higher apprehend quadratic equations.
  • If you go through the NCERT solution for Class 10th Maths chapter 4 exercise 4.1, you can also be used in calculating areas or rooms, determining a product’s profit in business or finding the speed of an object.
  • Exercise 4.1 Class 10 Maths, is based on solving the quadratic equations solutions and their roots which are important concepts of the chapter.

Also see-

NCERT Solutions of Class 10 Subject Wise

Subject Wise NCERT Exemplar Solutions

Frequently Asked Questions (FAQs)

1. Check whether the polynomial 7x^2+6x+34=0 a quadratic polynomial.

The maximum index here is 2 . Also it is in the form of ax^2+bx+c=0 . 

Therefore 7x^2+6x+34=0 is a quadratic polynomial. 

2. Factorize the polynomial y^2-10y+24

y^2-10y+24=y2-4y-6y+24 

=y(y-4)-6(y-4) 

=(y-4)(y-6) 

3. Solve and check whether the equation n(n-1)=3 is a quadratic equation?

n(n-1)=3

n^2-n=3 

n^2-n-3=0 

The maximum index here is 2 . Also it is in the form of ax^2+bx+c=0 .

Therefore n(n-1)=3 is a quadratic polynomial. 

4. Write the coefficient of x in x^2-8x+130=0

The coefficient of x is -8 

5. What are the method used to solve the quadratic equations?

There are three main methods to solve the quadratic equation. 

They are, 

  • Factorisation 

  • Completing the square

  •  Quadratic Formula. 

6. How can quadratic equations be applied in our real life?

Quadratic equations can be used in calculating areas or rooms, determining a product’s profit in business or finding the speed of an object.

7. Define quadratic equation as per NCERT solutions for Class 10 Maths chapter 4 exercise 4.1 ?

A polynomial of degree two is alluded to as a quadratic polynomial. A quadratic polynomial has the overall structure ax^2+bx+c=0, where a, b and c  are real numbers.

8. What number of inquiries are tended to in the NCERT solutions for Class 10 Maths chapter 4 exercise 4.1, what kinds of inquiries right?

The inquiries are focused on the idea of making and settling quadratic conditions, and the  NCERT solutions for Class 10 Maths chapter 4 exercise 4.1 comprises of two questions with subsections.

Articles

Upcoming School Exams

Application Date:07 October,2024 - 22 November,2024

Application Date:07 October,2024 - 22 November,2024

Application Correction Date:08 October,2024 - 27 November,2024

View All School Exams

Explore Top Universities Across Globe

Get answers from students and experts

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

Back to top