# NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations (Ex 4.1) Exercise 4.1

## NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations (Ex 4.1) Exercise 4.1

Before talking about the NCERT solutions for Class 10 Maths Chapter 4 exercise 4.1, let’s do a quick recapitulation- from chapter Polynomials, we already know what quadratic polynomials are. Isn’t it? When a polynomial of quadratic order of the form as  ax2 + bx + c = 0 (a ≠ 0) becomes zero, we termed it as a quadratic equation. Now, ax2 + bx + c = 0 (a ≠ 0) is known as the general form of the quadratic equation where ‘a’ and ‘b’ are called variables of the first order and c is a constant. Maths class 10 chapter 4 exercise 4.1 puts students into the practice of solving quadratic equations. And class 10 maths chapter 4 exercise 4.1 solutions lend them a helping hand and make them understand the right method of solving those problems. Students can now download free Chapter 4 Exercise 4.1 NCERT Book Solutions PDF for Class 10 to boost their exam preparation. Vedantu provides NCERT Solutions for Class 10 Science as well which will help students to prepare for their science exam.

NCERT Solutions for Class 10 Maths Chapter 4 Exercise 4.1 is the best guide for the students preparing for this chapter. NCERT Solutions Class 10 Maths Chapter 4 - Quadratic Equations Exercise 4.1 is available in a pdf format and have all the solutions to the questions provided in the exercise. The subject experts at Vedantu have provided accurate stepwise solutions to the questions.

The students are advised to practice all the questions given in all the exercises of each chapter repeatedly to excel in board and competitive examinations. NCERT Solutions Class 10 Maths Chapter 4 - Quadratic Equations should be understood thoroughly as it carries maximum marks and has different concepts to understand. The more you practice the questions the better you become.

NCERT Solutions are the best way to prepare for an examination.  So it is suggested to download NCERT Solutions for Class 10 Maths Chapter 4 Exercise 4.1 through the pdf link given below.

### Topics Covered

• Introduction

• Standard Form of Quadratic Equation

### What is a Quadratic Equation?

The name Quadratic is derived from the word "quad" meaning square. It is because the variable gets squared ( x2). It is also known as the "Equation of Polynomial Degree 2" (because of the power "2" on the x).

### Standard Form of Quadratic Equation

The Standard Form of a Quadratic Equation is given as:

ax2 + bx + c = 0

Here, a, b and c are known values and the value of ‘a’ can't be 0.

Also, "x" is the variable or unknown value.

7x2 + 5x + 21 = 0

Here, the value of a = 7, b = 5, and c = 21

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## Access NCERT Solutions for Maths Chapter 4 – Quadratic Equations

Exercice No: 4.1

1. Check whether the following are quadratic equations:

(i) ${{\left( \text{x + 1} \right)}^{\text{2}}}\text{ = 2}\left( \text{x - 3} \right)$

Ans: We are given an equation: ${{\left( \text{x + 1} \right)}^{\text{2}}}\text{ = 2}\left( \text{x - 3} \right)$

We will simplify the given equation.

${{\left( \text{x + 1} \right)}^{\text{2}}}\text{ = 2}\left( \text{x - 3} \right)$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{ + 2x + 1 = 2x - 6}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{ + 1 = - 6}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{ + 7 = 0}$

The equation obtained after simplifying is of the form $\text{a}{{\text{x}}^{\text{2}}}\text{ + bx + c = 0}$.

Therefore, the given equation is a quadratic equation.

(ii) ${{\text{x}}^{\text{2}}}\text{ - 2x = }\left( \text{-2} \right)\left( \text{3 - x} \right)$

Ans: We are given an equation: ${{\text{x}}^{\text{2}}}\text{ - 2x = }\left( \text{-2} \right)\left( \text{3 - x} \right)$

We will simplify the given equation.

${{\text{x}}^{\text{2}}}\text{ - 2x = }\left( \text{-2} \right)\left( \text{3 - x} \right)$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{ - 2x = - 6 + 2x}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{ - 4x + 6 =0}$

The equation obtained after simplifying is of the form $\text{a}{{\text{x}}^{\text{2}}}\text{ + bx + c = 0}$.

Therefore, the given equation is a quadratic equation.

(iii) $\left( \text{x - 2} \right)\left( \text{x + 1} \right)\text{ = }\left( \text{x - 1} \right)\left( \text{x + 3} \right)$

Ans: We are given an equation: $\left( \text{x - 2} \right)\left( \text{x + 1} \right)\text{ = }\left( \text{x - 1} \right)\left( \text{x + 3} \right)$

We will simplify the given equation.

$\left( \text{x - 2} \right)\left( \text{x + 1} \right)\text{ = }\left( \text{x - 1} \right)\left( \text{x + 3} \right)$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{ - 2x + x - 2 = }{{\text{x}}^{\text{2}}}\text{ - x + 3x - 3}$

$\Rightarrow \text{- 2x + x - 2 = - x + 3x - 3}$

$\Rightarrow \text{- x - 2 = 2x - 3}$

$\Rightarrow \text{3x - 1 = 0}$

This equation obtained is not of the form $\text{a}{{\text{x}}^{\text{2}}}\text{ + bx + c = 0}$.

Therefore, the given equation is not a quadratic equation.

(iv) $\left( \text{x - 3} \right)\left( \text{2x + 1} \right)\text{ = x}\left( \text{x + 5} \right)$

Ans: We are given an equation: $\left( \text{x - 3} \right)\left( \text{2x + 1} \right)\text{ = x}\left( \text{x + 5} \right)$

We will simplify the given equation.

$\left( \text{x - 3} \right)\left( \text{2x + 1} \right)\text{ = x}\left( \text{x + 5} \right)$

$\Rightarrow \text{2}{{\text{x}}^{\text{2}}}\text{- 6x + x - 3 = }{{\text{x}}^{\text{2}}}\text{ + 5x}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{- 10x - 3 = 0}$

The equation obtained after simplifying is of the form $\text{a}{{\text{x}}^{\text{2}}}\text{ + bx + c = 0}$.

Therefore, the given equation is a quadratic equation.

(v) $\left( \text{2x - 1} \right)\left( \text{x - 3} \right)\text{ = }\left( \text{x + 5} \right)\left( \text{x - 1} \right)$

Ans: We are given an equation: $\left( \text{2x - 1} \right)\left( \text{x - 3} \right)\text{ = }\left( \text{x + 5} \right)\left( \text{x - 1} \right)$

We will simplify the given equation.

$\Rightarrow \text{2}{{\text{x}}^{\text{2}}}\text{ - x - 6x + 3 = }{{\text{x}}^{\text{2}}}\text{ + 5x - x - 5}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{ - 7x + 3 = 4x - 5}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{ - 11x + 8 = 0}$

The equation obtained after simplifying is of the form $\text{a}{{\text{x}}^{\text{2}}}\text{ + bx + c = 0}$.

Therefore, the given equation is a quadratic equation.

(vi) ${{\text{x}}^{\text{2}}}\text{ + 3x + 1 = }{{\left( \text{x - 2} \right)}^{\text{2}}}$

Ans: We are given an equation: ${{\text{x}}^{\text{2}}}\text{ + 3x + 1 = }{{\left( \text{x - 2} \right)}^{\text{2}}}$

We will simplify the given equation.

${{\text{x}}^{\text{2}}}\text{ + 3x + 1 = }{{\left( \text{x - 2} \right)}^{\text{2}}}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{ + 3x + 1 = }{{\text{x}}^{\text{2}}}\text{ - 4x + 4}$

$\Rightarrow \text{3x + 1 = - 4x + 4}$

$\Rightarrow \text{7x - 3 = 0}$

This equation obtained is not of the form $\text{a}{{\text{x}}^{\text{2}}}\text{ + bx + c = 0}$.

Therefore, the given equation is not a quadratic equation.

(vii) ${{\left( \text{x + 2} \right)}^{\text{3}}}\text{ = 2x}\left( {{\text{x}}^{\text{2}}}\text{ - 1} \right)$

Ans: We are given an equation: ${{\left( \text{x + 2} \right)}^{\text{3}}}\text{ = 2x}\left( {{\text{x}}^{\text{2}}}\text{ - 1} \right)$

We will simplify the given equation.

${{\left( \text{x + 2} \right)}^{\text{3}}}\text{ = 2x}\left( {{\text{x}}^{\text{2}}}\text{ - 1} \right)$

$\Rightarrow {{\text{x}}^{\text{3}}}\text{ + 6}{{\text{x}}^{\text{2}}}\text{ + 12x + 8 = 2}{{\text{x}}^{\text{3}}}\text{ - 2x}$

$\Rightarrow \text{6}{{\text{x}}^{\text{2}}}\text{ + 14x + 8 = }{{\text{x}}^{\text{3}}}$

$\Rightarrow {{\text{x}}^{\text{3}}}\text{ - 6}{{\text{x}}^{\text{2}}}\text{ - 14x - 8 = 0}$

This equation obtained is not of the form $\text{a}{{\text{x}}^{\text{2}}}\text{ + bx + c = 0}$.

Therefore, the given equation is not a quadratic equation.

(viii) ${{\text{x}}^{\text{3}}}\text{ - 4}{{\text{x}}^{\text{2}}}\text{ - x + 1 = }{{\left( \text{x - 2} \right)}^{\text{3}}}$

Ans: We are given an equation: ${{\text{x}}^{\text{3}}}\text{ - 4}{{\text{x}}^{\text{2}}}\text{ - x + 1 = }{{\left( \text{x - 2} \right)}^{\text{3}}}$

We will simplify the given equation.

$\Rightarrow {{\text{x}}^{\text{3}}}\text{ - 4}{{\text{x}}^{\text{2}}}\text{ - x + 1 = }{{\text{x}}^{\text{3}}}\text{ - 6}{{\text{x}}^{\text{2}}}\text{ + 12x - 8}$

$\Rightarrow \text{- 4}{{\text{x}}^{\text{2}}}\text{ - x + 1 = - 6}{{\text{x}}^{\text{2}}}\text{ + 12x - 8}$

$\Rightarrow \text{2}{{\text{x}}^{\text{2}}}\text{ - 13x + 9 = 0}$

This equation obtained is of the form $\text{a}{{\text{x}}^{\text{2}}}\text{ + bx + c = 0}$.

Therefore, the given equation is a quadratic equation.

2. Represent the following situations in the form of quadratic equations.

(i) The area of a rectangular plot is $\text{528 }{{\text{m}}^{\text{2}}}$. 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.

Ans: It is given that the area of rectangular plot is $\text{528 }{{\text{m}}^{\text{2}}}$.

Let us assume the breadth of the plot be $\text{b = x m}$.

It is given that the length is one more than twice the breadth.

So, the length will be $\text{l = }\left( \text{2x + 1} \right)\text{ m}$.

The area of the rectangle will be  $\text{A = l }\!\!\times\!\!\text{ b}$

$\therefore \text{l }\!\!\times\!\!\text{ b = 528}$

$\Rightarrow \text{x}\left( \text{2x + 1} \right)\text{ = 528}$

$\Rightarrow \text{2}{{\text{x}}^{\text{2}}}\text{ + x - 528 = 0}$

Therefore, the required quadratic equation is $\text{2}{{\text{x}}^{\text{2}}}\text{ + x - 528 = 0}$.

(ii) The product of two consecutive positive integers is $\text{306}$. We need to find the integers.

Ans: It is given that the product of two consecutive positive integers is $\text{306}$.

If we assume the first integer to be $\text{x}$, then the next consecutive integer will be $\left( \text{x+1} \right)$.

So, we can write $\text{x}\left( \text{x + 1} \right)\text{ = 306}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{ + x - 306 = 0}$

Therefore, the required quadratic equation is ${{\text{x}}^{\text{2}}}\text{ + x - 306 = 0}$.

(iii) Rohan’s mother is $\text{26}$ years older than him. The product of their ages (in years) $\text{3}$ years from now will be $\text{360}$. We would like to find Rohan’s present age.

Ans: Let us assume Rohan’s age be $\text{x}$.

His mother is $\text{26}$ years older than him.

So, his mother’s age will be $\left( x+26 \right)$

After $\text{3}$ years, Rohan’s age $\text{= x + 3}$

His mother’s age $\text{= x + 26 + 3 = x + 29}$

It is given that after $\text{3}$ years, the product of their ages is $\text{360}$.

$\therefore \left( \text{x + 3} \right)\left( \text{x + 29} \right)\text{ = 360}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{ + 32x + 87 = 360}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{ + 32x - 273 = 0}$

Therefore, the required quadratic equation is ${{\text{x}}^{\text{2}}}\text{ + 32x - 273 = 0}$.

(iv) A train travels a distance of $\text{480}$ $\text{km}$ at a uniform speed. If the speed had been $\text{8}$ $\text{km/h}$ less, then it would have taken $\text{3}$ hours more to cover the same distance. We need to find the speed of the train.

Ans: Let us assume the speed of the train be $\text{x km/h}$.

The train travels a distance of $\text{480 km}$ at a uniform speed.

So, the time taken to travel the given distance $\text{= }\frac{\text{480}}{\text{x}}\text{ hrs}$.

It is given that the train would take $\text{3}$ hours more if the

speed had been $\text{8 km/h}$ less.

Now, the new speed of the train $\text{= }\left( \text{x - 8} \right)\text{ km/h}$.

The new time taken to travel the same distance $\text{= }\left( \frac{\text{480}}{\text{x}}\text{ + 3} \right)\text{ hr}$.

We know that $\text{speed }\!\!\times\!\!\text{ time = distance}$

$\therefore \left( \text{x - 8} \right)\left( \frac{\text{480}}{\text{x}}\text{ + 3} \right)\text{ = 480}$

$\Rightarrow \text{480 + 3x - }\frac{\text{3840}}{\text{x}}\text{ - 24 = 480}$

$\Rightarrow \text{3x - }\frac{\text{3840}}{\text{x}}\text{ = 24}$

$\Rightarrow \text{3}{{\text{x}}^{\text{2}}}\text{ - 24x - 3840 = 0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{ - 8x - 1280 = 0}$

Therefore, the required quadratic equation is ${{\text{x}}^{\text{2}}}\text{ - 8x - 1280 = 0}$.

## Ex 4.1 Class 10 - History of Quadratic Equation

The history of the solution of quadratic equations extends across the world for more than 4000 years as class 10 Math chapter 4 exercise 4.1 will tell you. The earliest known records are Babylonian clay tablets from about 1600 BCE where the diagonal of a unit square is given to five decimal places of accuracy. But, as you will know from class 10 quadratic equation exercise 4.1, the use of algebraic symbols only began in the 15th century. Exercise 4.1 Maths class 10 will also inform you that Euclid, a famous Greek mathematician invented the quadratic equations and its solutions while calculating dimensions such as length; breadth and height via geometry. Later Sridharacharya made the calculations of these quadratic equations by completing the square methodology which became more convenient; better known as now as Sridharacharya Formula of Quadratic Equation.

### Class 10 Maths Chapter 4 Exercise 4.1 - Concept of Quadratic Equation

The root of the quadratic equation ax2 + bx + c = 0 (a ≠ 0) is given in NCERT class 10 Maths chapter 4 exercise 4.1 by the following formula:

$x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$

This formula is known as Sridharacharya formula, as you will know from exercise 4.1 class 10 Maths.

Here, the quantity   b2 −4ac   is called the discriminant of the polynomial.

According to Sridharacharya, as mentioned in NCERT Maths class 10 chapter 4 exercise 4.1, every quadratic equation has two roots and the types depend on the value of this discriminant.

If b2−4ac  > 0, the roots of the quadratic equation are real and distinct in nature.

If b2−4ac  < 0, the roots of the quadratic equation are unreal i.e. has a complex solution.

If b2−4ac   = 0, the roots of the quadratic equation are real and equal.

Example 1:

Examine the nature of the roots of the following quadratic equation.

3x2 + 8x + 4  =  0.

Solution:

The given quadratic equation is in the general form:

ax2 + bx + c  =  0

Then, we have a  =  3, b  =  8 and c  =  4.

Therefore:

b2 - 4ac  =  82 - 4(3)(4)

b2 - 4ac  =  64 - 48

b2 - 4ac  =  16

Here, b2 - 4ac > 0 and also a perfect square.

So, the roots are real and distinct.

Example 2:

Examine the nature of the roots of the following quadratic equation.

5x2 - 4x + 2  =  0

Solution:

The given quadratic equation is in the general form:

ax2 + bx + c  =  0

Then, we have a  =  5, b  =  -4 and c  =  2.

Therefore:

b2 - 4ac  =  (-4)2 - 4(5)(2)

b2 - 4ac  =  16 - 40

b2 - 4ac  =  -24

Here, b2 - 4ac < 0.

So, the roots are imaginary.

### Ex 4.1 Class 10 - Solving Techniques of Quadratic Equations

A quadratic equation may be solved either by factorizing the left side( when the right side is zero) or by completing a square on the left side.

Example 1: Solve  6x2 + 11x - 10 = 0

Solution: 6x2 + 11x - 10 = 0

⇒ 6x2 + 15x - 4x - 10 = 0

⇒ 3x(2x + 5) - 2(2x + 5) = 0

⇒ (2x + 5)(3x - 5) = 0

Therefore, either 2x + 5 = 0 or, 3x - 5 = 0

Hence, x = -5/2 or x = ⅔.

### Class 10 Chapter 4 Exercise 4.1: Conversion of Non-Quadratic to Quadratic Equation

Sometimes the equations given are not in quadratic order and hence we need to convert it into a quadratic form to solve it.

Example 1: Solve $2^{2x + 3} + 2^{x + 3} = 1 + 2^{x}$

Solution: $2^{2x + 3} + 2^{x + 3} = 1 + 2^{x}$

or, $2^{2x} . 2^{3} + 2^{x} . 2^{3} = 1 + 2^{x}$

or, $8 . 2^{2x} + 8 . 2^{x} = 1 + 2^{x}$…(1)

Now, let $2^{x} = y$ Then, $y^{2} = (2^{x})^{2} = 2^{2x}$

∴ from (1) we get,

$8y^{2} + 8y = 1 + y$

or, $8y^{2} + 7y - 1 = 0$

or, $8y^{2} - 8y - y - 1 = 0$

or, $8y(y + 1) - 1(y + 1) = 0$

or, $(y + 1)(8y - 1) = 0$

∴ y = -1

or ⅛

If, y = -1, then, $2^{x} = -1$, which is not possible.

If, y = ⅛, then $2^{x} = \frac{1}{8}$

or, $2^{x} = \frac{1}{2^{3}} = 2^{-3}$

∴ x = -3.

Example 2: Solve $\sqrt{\frac{x}{1 - x}} + \sqrt{\frac{1 - x}{x}} = \frac{13}{6}$

Solution: Let us put $y = \sqrt{\frac{1}{1-x}}$, then $\frac{1}{y} = \sqrt{\frac{1 - x}{x}}$

$\therefore y + \frac{1}{y} = \frac{13}{6} or, \frac{y^{2} + 1}{y} = \frac{13}{6}$

⇒ 6y2 + 6 = 13y

⇒ 6y2 - 13y + 6 = 0

⇒ 6y2 - 4y - 9y + 6 = 0

⇒ (3y - 2)(2y - 3) = 0

∴ y = ⅔ or, y = 3/2

If y = ⅔ , then $\sqrt{\frac{x}{1-x}} = \frac{2}{3}$

$\Rightarrow \therefore \frac{x}{1 - x} = \frac{4}{9}$

⇒ 9x = 4 - 4x

⇒ 13x = 4

∴ x = 4/13

If y = 3/2, then $\sqrt{\frac{x}{1-x}} = \frac{3}{2}$

$\Rightarrow \therefore \frac{x}{1 - x} = \frac{9}{4}$

⇒ 4x = 9 - 9x

⇒ 13x = 9

∴ x = 9/13

### Class 10 Maths Ex 4.1 - Sum and Product of Roots of a Quadratic Equation

If α and β be the roots of the quadratic equation ax2 + bx + c = 0 (a ≠ 0) then,

$\alpha + \beta = -\frac{b}{a}$ is called the sum of roots and $\alpha \beta = \frac{c}{a}$ is called products of roots of the quadratic equations.

Example 1: If the roots of the equation x2 + px + 7 = 0 are denoted by α and β and α2 + β2 = 22, find the value of p.

Solution: $\alpha + \beta = \frac{-p}{1} = (-p)$ and $\alpha \beta = \frac{7}{1} = 7$

Now, α2 + β2 = 22 (given)

or, (α + β)2 - 2αβ = 22

or, (-p)2 - 2(7) = 22

or, p2 = 22 + 14 = 36

p = ±6

### Class 10 Maths Ch 4 Ex 4.1- Problems based on Quadratic Equation

Class 10th Maths Chapter 4 exercise 4.1 has several problems for you to solve. Following are a few of them.

Example 1: In a class test, the sum of Rekha’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: with the help of class 10 Maths exercise 4.1, let us marks in Maths as xx,

Then as per the problem,

Marks in English would be = 30 − x30 − x.

Now if Rekha got 2 marks less, then Maths marks would = (x − 2)(x − 2)

and if she got 3 marks less in English, then English marks would = (30 − x − 3) = (27 − x)(30 − x − 3) = (27 − x)

As per problem, the product is equal to 210 so

(x − 2)(27 − x) = 120(x − 2)(27 − x) = 120

−x2 + 25x + 54 = 210 − x2 + 25x + 54 = 210

or

x2 − 25 x + 156 = 0 x 2 − 25 x +156 = 0

x2 − 90x + 30x − 2700 = 0 x 2 − 90x + 30x − 2700 = 0

or x = 12 or 13

So Rekha’ marks in Maths is 12 or 13 and the English marks are 18 or 17.

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### NCERT Solutions for Class 10 Maths Chapter 4 Exercises

 Chapter 4 Quadratic Equations All Exercises in PDF Format Exercise 4.2 6 Questions and Solutions Exercise 4.3 11 Questions and Solutions Exercise 4.4 5 Questions and Solutions

1.The Altitude of a Right Triangle is 7 cm Less than its Base. If the Hypotenuse is 13 cm, Find the Other Two Sides.

You can find the following in class 10 Maths ex 4.1 solutions:

Let x be the base, the altitude will be x - 7

Now as per Pythagoras theorem,

x2 + (x − 7)2 = 132

= x2 + x2 - 14x + 49 = 169

= 2x2 -14x = 120

= x2 - 7x = 60

= x2 - 12x + 5x - 60 = 0

= x(x - 12) + 5(x - 12) = 0

= (x - 12) (x + 5) = 0

x = 12 or -5

Since x cannot be negative x = 12.

So base is 12 cm and altitude is (12 - 7) = 5 cm.

2. If Zeba were Younger by 5 years than what she really is then the Square of her age Would have been 11 more than five times her actual age. What is her age now?

The solution according to exercise 4.1 class 10 Maths NCERT solutions is as follows:

Let age of Zeba be x years
As per the question:

(x − 5)2 = 11 + 5x(x − 5)2 = 11 + 5x
=> x2 + 25 − 10x = 11 + 5x(x)2 + 25 − 10x = 11 + 5x
=> x2 − 15x + 14 = 0 x 2 − 15x + 14 = 0 x 2 − 14x − x + 14 = 0 x 2 − 14x − x + 14 = 0
=> (x - 1)(x - 14)

Therefore, age of Zeba is 14 years.

3. What is a Quadratic Equation, and what is its general form?

When a quadratic order polynomial is of the type ax2 + bx + c = 0 (a ≠  0), we call it a Quadratic Equation. The general version of the Quadratic Equation is now ax2 + bx + c = 0 (a ≠ 0), where ‘a', ‘b' and ‘c’ are constants i.e. numbers. The constant ‘b’ and ‘c’ can also be equal to zero but ‘a’ cannot be zero.

4. How many questions are there in Chapter 4 Maths Class 10?

There are a total of four exercises. In the first exercise, there are two questions. In the second, there are six; in the third, there are 11; and in the fourth exercise, there are five. So, all in all, there are 23 questions and some of these questions have some sub-parts too. In theNCERT Solutions for Class 4, Maths Chapter 4 there are solved examples as well that will help in preparation for the board exams.

5. How to find the nature of roots?

The term b2 – 4ac is called the discriminant of a quadratic equation since b2 – 4ac determines whether the quadratic equation ax2 + bx + c = 0 has real roots or not. So, a Quadratic Equation ax2 + bx + c = 0 has

(i) two distinct real roots, if b2 – 4ac > 0,

(ii) two equal real roots, if b2 – 4ac = 0,

(iii) no real roots, if b2 – 4ac < 0.

6. How can I top in Class 10 Maths?

In order to score the best marks in Maths, regular practice is required. You must practice and revise all the solved examples and questions given in the NCERT book. You must all be thorough with all the concepts, theorems and formulae of all the chapters. You can visit the page NCERT Solutions for Class 10 Maths, on the official website of Vedantu for a complete guide.

7. Where can I get the NCERT Solution for Class 10 Chapter 4 Maths?

You can easily access the NCERT Solutions for Class 10 Maths, on the Vedantu official website (vedantu.com) and on the Vedantu app absolutely free of cost. You can also get revision notes and other study material free of cost on the Vedantu website (vedantu.com). The answers are provided comprehensively and are very easy to understand; all the formulae used in the questions are also provided along with the answers to facilitate a quick revision.