NCERT Solutions for Exercise 4.3 Class 10 Maths Chapter 4

NCERT Solutions for Exercise 4.3 Class 10 Maths Chapter 4 - Quadratic Equations

Edited By Ramraj Saini | Updated on Nov 16, 2023 11:44 AM IST | #CBSE Class 10th

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NCERT Solutions For Class 10 Maths Chapter 4 Exercise 4.3

NCERT Solutions for Exercise 4.3 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 4.3 class 10 deals concept of finding the roots of quadratic equations by completing squares and applying the quadratic formula. Exercise 4.3 Class 10 Maths, quadratic equations help us to explore the ideas of ways of finding quadratic equations in depth. Completing the square is nothing but finding the value that makes a quadratic equation a square trinomial. The square trinomial can then be solved easily by using algebraic identities and formulas. NCERT solutions for Class 10 Maths chapter 4 exercise 4.3 consists of 11 simple problems with word problems that are easy to solve and also explore the concepts of solving quadratic equations by using quadratic formulas.

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NCERT Solutions for Exercise 4.3 Class 10 Maths Chapter 4 - Quadratic Equations

NCERT solutions for Class 10 Maths exercise 4.3, 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 4.2 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 Exercise: 4.3

Q1 (i) Find the roots of the following quadratic equations, if they exist, by the method of completing the square $2x^2 - 7x +3 = 0$

Answer:

Given equation: $2x^2 - 7x +3 = 0$

On dividing both sides of the equation by 2, we obtain

$\Rightarrow x^2-\frac{7}{2}x+\frac{3}{2} = 0$

$\Rightarrow (x-\frac{7}{4})^2 + \frac{3}{2} - \frac{49}{16} = 0$

$\Rightarrow (x-\frac{7}{4})^2 = \frac{49}{16} - \frac{3}{2}$

$\Rightarrow (x-\frac{7}{4})^2 =\frac{25}{16}$

$\Rightarrow (x-\frac{7}{4}) =\pm \frac{5}{4}$

$\Rightarrow x =\frac{7}{4}\pm \frac{5}{4}$

$\Rightarrow x = \frac{7}{4}+\frac{5}{4}\ or\ x = \frac{7}{4} - \frac{5}{4}$

$\Rightarrow x = 3\ or\ \frac{1}{2}$

Q1 (ii) Find the roots of the following quadratic equations, if they exist, by the method of completing the square $2x^2 + x -4 = 0$

Answer:

Given equation: $2x^2 + x -4 = 0$

On dividing both sides of the equation by 2, we obtain

$\Rightarrow x^2+\frac{1}{2}x-2 = 0$

Adding and subtracting $\frac{1}{16}$ in the equation, we get

$\Rightarrow (x+\frac{1}{4})^2 -2 - \frac{1}{16} = 0$

$\Rightarrow (x+\frac{1}{4})^2 =2+\frac{1}{16}$

$\Rightarrow (x+\frac{1}{4})^2 = \frac{33}{16}$

$\Rightarrow (x+\frac{1}{4}) =\pm \frac{\sqrt{33}}{4}$

$\Rightarrow x =\pm \frac{\sqrt{33}}{4} -\frac{1}{4}$

$\Rightarrow x = \frac{\pm \sqrt{33} - 1}{4}$

$\Rightarrow x = \frac{ \sqrt{33} - 1}{4}\ or\ x = \frac{ -\sqrt{33} - 1}{4}$

Q1 (iii) Find the roots of the following quadratic equations, if they exist, by the method of completing the square $4x^2 + 4\sqrt3 + 3 = 0$

Answer:

Given equation: $4x^2 + 4\sqrt3 + 3 = 0$

On dividing both sides of the equation by 4, we obtain

$\Rightarrow x^2+\sqrt3x+\frac{3}{4} = 0$

Adding and subtracting $(\frac{\sqrt3}{2})^2$ in the equation, we get

$\Rightarrow (x+\frac{\sqrt3}{2})^2 +\frac{3}{4} - (\frac{\sqrt3}{2})^2 = 0$

$\Rightarrow (x+\frac{\sqrt3}{2})^2 = \frac{3}{4} - \frac{3}{4} = 0$

$\Rightarrow (x+\frac{\sqrt3}{2}) = 0\ or\ (x+\frac{\sqrt3}{2}) = 0$

Hence there are the same roots and equal:

$\Rightarrow x = \frac{-\sqrt3}{2}\ or\ \frac{-\sqrt3}{2}$

Q2 (iv) Find the roots of the following quadratic equations, if they exist, by the method of completing the square $2x^2 + x + 4 = 0$

Answer:

Given equation: $2x^2 + x + 4 = 0$

On dividing both sides of the equation by 2, we obtain

$\Rightarrow x^2+\frac{x}{2}+2 = 0$

Adding and subtracting $(\frac{1}{4})^2$ in the equation, we get

$\Rightarrow (x+\frac{1}{4})^2 +2- (\frac{1}{4})^2 = 0$

$\Rightarrow (x+\frac{1}{4})^2 = \frac{1}{16} -2 = \frac{-31}{16}$

$\Rightarrow (x+\frac{1}{4}) = \pm \frac{\sqrt{-31}}{4}$

$\Rightarrow x = \pm \frac{\sqrt{-31}}{4} - \frac{1}{4}$

$\Rightarrow x = \frac{\sqrt{-31}-1}{4} \ or\ x = \frac{-\sqrt{-31}-1}{4}$

Here the real roots do not exist (in the higher studies we will study how to find the root of such equations).

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

Answer:

(i) $2x^2-7x+3 = 0$

The general form of a quadratic equation is : $ax^2+bx+c = 0$ , where a, b, and c are arbitrary constants.

Hence on comparing the given equation with the general form, we get

$a = 2,\ b = -7\ c = 3$

And the quadratic formula for finding the roots is:

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

Substituting the values in the quadratic formula, we obtain

$\Rightarrow x= \frac{7 \pm \sqrt{49-24}}{4}$

$\Rightarrow x= \frac{7 \pm 5}{4}$

$\Rightarrow x= \frac{7 + 5}{4} = 3\ or\ x= \frac{7 - 5}{4} = \frac{1}{2}$

Therefore, the real roots are: $x =3,\ \frac{1}{2}$

(ii) $2x^2+x-4 = 0$

The general form of a quadratic equation is : $ax^2+bx+c = 0$ , where a, b, and c are arbitrary constants.

Hence on comparing the given equation with the general form, we get

$a = 2,\ b = 1\ c =-4$

And the quadratic formula for finding the roots is:

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

Substituting the values in the quadratic formula, we obtain

$\Rightarrow x= \frac{-1 \pm \sqrt{1+32}}{4}$

$\Rightarrow x= \frac{-1 \pm \sqrt{33}}{4}$

$\Rightarrow x= \frac{-1 + \sqrt{33}}{4} \ or\ x= \frac{-1 - \sqrt{33}}{4}$

Therefore, the real roots are: $x = \frac{-1+\sqrt{33}}{4}\ or\ \frac{-1-\sqrt{33}}{4}$

(iii) $4x^2+4\sqrt3x+3 = 0$

The general form of a quadratic equation is : $ax^2+bx+c = 0$ , where a, b, and c are arbitrary constants.

Hence on comparing the given equation with the general form, we get

$a = 4,\ b = 4\sqrt{3}\ c =3$

And the quadratic formula for finding the roots is:

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

Substituting the values in the quadratic formula, we obtain

$\Rightarrow x= \frac{-4\sqrt{3} \pm \sqrt{48-48}}{8}$

$\Rightarrow x= \frac{-4\sqrt{3} \pm 0}{8}$

Therefore, the real roots are: $x = \frac{-\sqrt{3}}{2}\ or\ \frac{-\sqrt{3}}{2}$

(iv) $2x^2+x+4 = 0$

The general form of a quadratic equation is : $ax^2+bx+c = 0$ , where a, b, and c are arbitrary constants.

Hence on comparing the given equation with the general form, we get

$a = 2,\ b = 1,\ c =4$

And the quadratic formula for finding the roots is:

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

Substituting the values in the quadratic formula, we obtain

$\Rightarrow x= \frac{-1 \pm \sqrt{1-32}}{4}$

$\Rightarrow x= \frac{-1 \pm \sqrt{-31}}{4}$

Here the term inside the root is negative

Therefore there are no real roots for the given equation.

Q3 (i) Find the roots of the following equations: $x - \frac{1}{x} = 3, x\neq 0$

Answer:

Given equation: $x - \frac{1}{x} = 3, x\neq 0$

So, simplifying it,

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

$\Rightarrow x^2-3x-1 = 0$

Comparing with the general form of the quadratic equation: $ax^2+bx+c = 0$ , we get

$a=1,\ b=-3,\ c=-1$

Now, applying the quadratic formula to find the roots:

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

$\Rightarrow x= \frac{3 \pm \sqrt{9+4}}{2}$

$\Rightarrow x= \frac{3 \pm \sqrt{13}}{2}$

Therefore, the roots are

$\Rightarrow x = \frac{3+\sqrt{13}}{2}\ or\ \frac{3 - \sqrt{13}}{2}$

Q3 (ii) Find the roots of the following equations: $\frac{1}{x+4} - \frac{1}{x- 7} = \frac{11}{30},\ x\neq -4,7$

Answer:

Given equation: $\frac{1}{x+4} - \frac{1}{x- 7} = \frac{11}{30},\ x\neq -4,7$

So, simplifying it,

$\Rightarrow \frac{x-7-x-4}{(x+4)(x-7)} = \frac{11}{30}$

$\Rightarrow \frac{-11}{(x+4)(x-7)} = \frac{11}{30}$

$\Rightarrow (x+4)(x-7) = -30$

$\Rightarrow x^2-3x-28 = -30$ or $\Rightarrow x^2-3x+2 = 0$

Can be written as:

$\Rightarrow x^2-x-2x+2 = 0$

$\Rightarrow x(x-1) -2(x-1) = 0$

$\Rightarrow (x-2)(x-1) = 0$

Hence the roots of the given equation are:

$\Rightarrow x = 1\ or\ 2$

Q4 The sum of the reciprocals of Rehman’s ages, (in years) 3 years ago and 5 years from now is $\frac{1}{3}$ . Find the present age.

Answer:

Let the present age of Rehman be $'x'$ years.

Then, 3 years ago, his age was $(x-3)$ years.

and 5 years later, his age will be $(x+5)$ years.

Then according to the question we have,

$\frac{1}{(x-3)}+\frac{1}{(x+5)} = \frac{1}{3}$

Simplifying it to get the quadratic equation:

$\Rightarrow \frac{x+5+x-3}{(x-3)(x+5)} = \frac{1}{3}$

$\Rightarrow \frac{2x+2}{(x-3)(x+5)} = \frac{1}{3}$

$\Rightarrow 3(2x+2)= (x-3)(x+5)$

$\Rightarrow 6x+6 = x^2+2x-15$

$\Rightarrow x^2-4x-21 = 0$

$\Rightarrow x^2-7x+3x-21 = 0$

$\Rightarrow x(x-7)+3(x-7) = 0$

$\Rightarrow (x-7)(x+3) = 0$

Hence the roots are: $\Rightarrow x = 7,\ -3$

However, age cannot be negative

Therefore, Rehman is 7 years old in the present.

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.

Answer:

Let the marks obtained in Mathematics be 'm' then, the marks obtain in English will be '30-m'.

Then according to the question:

$(m+2)(30-m-3) = 210$

Simplifying to get the quadratic equation:

$\Rightarrow m^2-25m+156 = 0$

Solving by the factorizing method:

$\Rightarrow m^2-12m-13m+156 = 0$

$\Rightarrow m(m-12)-13(m-12) = 0$

$\Rightarrow (m-12)(m-13) = 0$

$\Rightarrow m = 12,\ 13$

We have two situations when,

The marks obtained in Mathematics is 12, then marks in English will be 30-12 = 18.

Or,

The marks obtained in Mathematics is 13, then marks in English will be 30-13 = 17.

Q6 The diagonal of a rectangular field is 60 meters more than the shorter side. If the longer side is 30 meters more than the shorter side, find the sides of the field.

Answer:

Let the shorter side of the rectangle be x m.

Then, the larger side of the rectangle wil be $= (x+30)\ m$ .

Diagonal of the rectangle:

$= \sqrt{x^2+(x+30)^2}\ m$

It is given that the diagonal of the rectangle is 60m more than the shorter side.

Therefore,

$\sqrt{x^2+(x+30)^2} = x+60$

$\Rightarrow x^2+(x+30)^2 = (x+60)^2$

$\Rightarrow x^2+x^2+900+60x = x^2+3600+120x$

$\Rightarrow x^2-60x-2700 = 0$

Solving by the factorizing method:

$\Rightarrow x^2-90x+30x-2700 = 0$

$\Rightarrow x(x-90)+30(x-90)= 0$

$\Rightarrow (x+30)(x-90) = 0$

Hence, the roots are: $x = 90,\ -30$

But the side cannot be negative.

Hence the length of the shorter side will be: 90 m

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

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.

Answer:

Given the difference of squares of two numbers is 180.

Let the larger number be 'x' and the smaller number be 'y'.

Then, according to the question:

$x^2-y^2 = 180$ and $y^2 = 8x$

On solving these two equations:

$\Rightarrow x^2-8x =180$

$\Rightarrow x^2-8x -180 = 0$

Solving by the factorizing method:

$\Rightarrow x^2-18x+10x -180 = 0$

$\Rightarrow x(x-18)+10(x-18) = 0$

$\Rightarrow (x-18)(x+10) = 0$

$\Rightarrow x=18,\ -10$

As the negative value of x is not satisfied in the equation: $y^2 = 8x$

Hence, the larger number will be 18 and a smaller number can be found by,

$y^2 = 8x$ putting x = 18, we obtain

$y^2 = 144\ or\ y = \pm 12$ .

Therefore, the numbers are $18\ and\ 12$ or $18\ and\ -12$ .

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.

Answer:

Let the speed of the train be $x\ km/hr.$

Then, time taken to cover $360km$ will be:

$=\frac{360}{x}\ hr$

According to the question,

$\Rightarrow (x+5)\left ( \frac{360}{x}-1 \right ) = 360$

$\Rightarrow 360-x+\frac{1800}{x} - 5 = 360$

Making it a quadratic equation.

$\Rightarrow x^2+5x-1800 = 0$

Now, solving by the factorizing method:

$\Rightarrow x^2+45x-40x-1800 = 0$

$\Rightarrow x(x+45)-40(x+45) = 0$

$\Rightarrow (x-40)(x+45) = 0$

$\Rightarrow x = 40,\ -45$

However, the speed cannot be negative hence,

The speed of the train is $40\ km/hr$ .

Q9 Two water taps together can fill a tank in $9\frac{3}{8}$ 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.

Answer:

Let the time taken by the smaller pipe to fill the tank be $x\ hr.$

Then, the time taken by the larger pipe will be: $(x-10)\ hr$ .

The fraction of the tank filled by a smaller pipe in 1 hour:

$= \frac{1}{x}$

The fraction of the tank filled by the larger pipe in 1 hour.

$= \frac{1}{x-10}$
Given that two water taps together can fill a tank in $9\frac{3}{8} = \frac{75}{8}$ hours.

Therefore,

$\Rightarrow \frac{1}{x}+\frac{1}{x-10} = \frac{8}{75}$

$\Rightarrow \frac{x-10+x}{x(x-10)} = \frac{8}{75}$

$\Rightarrow \frac{2x-10}{x(x-10)} = \frac{8}{75}$

Making it a quadratic equation:

$\Rightarrow 150x-750 = 8x^2-80x$

$\Rightarrow 8x^2-230x+750 = 0$

$\Rightarrow 8x^2-200x-30x+750 = 0$

$\Rightarrow 8x(x-25) - 30(x-25) = 0$

$\Rightarrow (x-25)(8x+30) = 0$

Hence the roots are $\Rightarrow x = 25,\ \frac{-30}{8}$

As time is taken cannot be negative:

Therefore, time is taken individually by the smaller pipe and the larger pipe will be $25$ and $25-10 =15$ hours respectively.

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

Answer:

Let the average speed of the passenger train be $x\ km/hr$ .

Given the average speed of the express train $= (x+11)\ km/hr$

also given that the time taken by the express train to cover 132 km is 1 hour less than the passenger train to cover the same distance.

Therefore,

$\Rightarrow \frac{132}{x} - \frac{132}{x+11} = 1$

$\Rightarrow 132\left [ \frac{x+11-x}{x(x+11)} \right ] = 1$

$\Rightarrow \frac{132\times11}{x(x+11)} = 1$

Can be written as quadratic form:

$\Rightarrow x^2+11x-1452 = 0$

$\Rightarrow x^2+44x-33x-1452 = 0$

$\Rightarrow x(x+44)-33(x+44)= 0$

$\Rightarrow (x+44)(x-33) = 0$

Roots are: $\Rightarrow x = -44,\ 33$

As the speed cannot be negative.

Therefore, the speed of the passenger train will be $33\ km/hr$ and

The speed of the express train will be $33+11 = 44\ km/hr$ .

Q11 Sum of the areas of two squares is 468 m ² . If the difference of their perimeters is 24 m, find the sides of the two squares.

Answer:

Let the sides of the squares be $'x'\ and\ 'y'$ . (NOTE: length are in meters)

And the perimeters will be: $4x\ and\ 4y$ respectively.

Areas $x^2\ and\ y^2$ respectively.

It is given that,

$x^2 + y^2 = 468\ m^2$ .................................(1)

$4x-4y = 24\ m$ .................................(2)

Solving both equations:

$x-y = 6$ or $x= y+6$ putting in equation (1), we obtain

$(y+6)^2 +y^2 = 468$

$\Rightarrow 2y^2+36+12y = 468$

$\Rightarrow y^2+6y - 216 = 0$

Solving by the factorizing method:

$\Rightarrow y^2+18y -12y-216 = 0$

$\Rightarrow y(y+18) -12(y+18) = 0$

$\Rightarrow (y+18)(y-12)= 0$

Here the roots are: $\Rightarrow y = -18,\ 12$

As the sides of a square cannot be negative.

Therefore, the sides of the squares are $12m$ and $(12\ m+6\ m) = 18\ m$ .

Benefits of NCERT Solutions for Class 10 Maths Exercise 4.3 :

NCERT solutions for Class 10 Maths exercise 4.3 is recommended for students who want to do well in the CBSE Term exams.
In NCERT book exercise 4.3 Class 10 Maths, all the minute concepts of solution of a Quadratic Equation by Completing the Square are also covered to help students face other competitive exams more confidently.
By solving the NCERT syllabus Class 10th Maths chapter 4 exercise 4.3 exercises, students are able to answer all the questions based on quadratic equations and also students can attain a good score not only on the final answer but also on each step.

Also see-

NCERT Solutions of Class 10 Subject Wise

Subject Wise NCERT Exemplar Solutions

Frequently Asked Questions (FAQs)

1. State true /false : A quadratic equation can be solved using the method of completing square.

Yes, the given statement is true. A quadratic equation can be solved using the method of completing squares.

2. What is the easiest and the fastest method to calculate the roots of a quadratic equation?

One of the fastest methods to calculate the roots of a quadratic equation is using the quadratic formula.

3. What is completing the square according to NCERT solutions for Class 10 Maths chapter 4 exercise 4.3 ?

Completing the square is nothing but finding the value that makes a quadratic equation a square trinomial

4. In NCERT solutions for Class 10 Maths chapter 4 exercise 4.3 , how many questions and what types of questions are covered?

The questions are centred on the concept of finding the roots of quadratic equations by completing squares, and the NCERT solutions for Class 10 Maths chapter 4 exercise 4.3 consists of 11 problems with subsections

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Final Destination:
- Once you reach Ramgart Road, Sadhu Ashram should be easy to spot. Look for any signage or ask nearby people to guide you to the exact location.

If you need detailed directions from a specific location or more information about Sadhu Ashram, feel free to ask

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sai raja 01 September,2024

show the previous year exams ?

Hello Aspirant, Hope your doing great, your question was incomplete and regarding what exam your asking.

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SWETCHCHA MISKA 10 July,2024

as an icse student in class 10 is above 80 percent good enough to get into commerce in 11th cbse

Yes, scoring above 80% in ICSE Class 10 exams typically meets the requirements to get into the Commerce stream in Class 11th under the CBSE board . Admission criteria can vary between schools, so it is advisable to check the specific requirements of the intended CBSE school. Generally, a good academic record with a score above 80% in ICSE 10th result is considered strong for such transitions.

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Manasvini Gupta 23 July,2024

i had cleared 10th from cbse in 2022 2 years ago now i wish to apply for 12th as private candidate in 2025. can i do so?

hello Zaid,

Yes, you can apply for 12th grade as a private candidate .You will need to follow the registration process and fulfill the eligibility criteria set by CBSE for private candidates.If you haven't given the 11th grade exam ,you would be able to appear for the 12th exam directly without having passed 11th grade. you will need to give certain tests in the school you are getting addmission to prove your eligibilty.

best of luck!

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Ashvadha 03 July,2024

i had cleared 10th in 2022 from cbse can i apply for 12th as private candidate this year?

According to cbse norms candidates who have completed class 10th, class 11th, have a gap year or have failed class 12th can appear for admission in 12th class.for admission in cbse board you need to clear your 11th class first and you must have studied from CBSE board or any other recognized and equivalent board/school.

You are not eligible for cbse board but you can still do 12th from nios which allow candidates to take admission in 12th class as a private student without completing 11th.

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sadafreen356 03 July,2024

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Central Board of Secondary Education Class 10th Examination

Applications Open

Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

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) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

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 .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

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.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

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

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

NCERT Solutions for Exercise 4.3 Class 10 Maths Chapter 4 - Quadratic Equations

NCERT Solutions For Class 10 Maths Chapter 4 Exercise 4.3

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