NCERT Solutions for Exercise 4.1 Class 10 Maths Chapter 4

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.

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

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.

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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 $x^2 -8x-1280 = 0$

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.

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