NCERT Solutions For Class 10 Maths Chapter 4 Quadratic Equations

Q: What are the important concepts learned in the NCERT Solutions for chapter 4 maths class 10 Quadratic Equations?

Class 10 maths chapter 4 solutions required multiple concepts including what is an equation, how to get the solution of an equation, methods of quadratic equation solution, the factoring method for solving a quadratic equation, the quadratic formula for solving a quadratic equation, the nature of roots of a quadratic equation, graphs of quadratic equations, and many.

Q: What are the important topics covered in the NCERT Solutions for Class 10 Maths Chapter 4?

Chapter 4 quadratic equation class 10 ncert solutions contain multiple topics like Quadratic equations, Solutions of a quadratic equation, Discriminant of a quadratic equation, Factoring method for solving a quadratic equation, Quadratic formula for solving a quadratic equation, Nature of roots of a quadratic equation, and Graphs of quadratic equations.

Q: How do you download the Class 10 Maths Quadratic Equations NCERT Textbooks PDF?

Students can visit the NCERT official website. There they can find options to select class, subject, and book title. Using these inputs students can download NCERT chapter 4 maths class 10 which is quadratic equations class 10 pdf. Students can also find solutions to these NCERTs from Careers360's official website

Q: How do you solve a quadratic equation in Class 10?

There are multiple methods to solve a quadratic equation including the factoring method, quadratic formula method, graphical method, and many others. These methods depend on the type of quadratic equation so first to know what is quadratic polynomial type and then apply the above methods to find a quadratic polynomial solution.

Q: Which of the following is a quadratic equation?

To determine whether an equation is a quadratic equation, you can look for two conditions: firstly, the highest power of the variable is 2 and secondly, the coefficient of the x^2 term (that is, the "a" in the equation) is not equal to zero. These two are some examples of quadratic equations 2x^2+3x+5 and x^2+5x+6.

Edited By Saumya.Srivastava | Updated on Feb 8, 2023 - 12:51 p.m. IST | #CBSE Class 10th

NCERT Solutions For Class 10 Maths Chapter 4 Quadratic Equations will help students understand the basic concepts in a better way. When we equate the quadratic polynomial to zero, then we get a quadratic equation. With the selp of NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations, students will be able to strategise their preparation. Here students will get NCERT solutions from Class 6 to 12 for Science and Maths. They must complete the NCERT Class 10 Maths syllabus to the earliest so that they can revise in a strategic way. NCERT solutions for class 10 Maths chapter 4 Quadratic Equations is providing an in-depth and step by step solution to each question. Each exercises of Quadratic Equation Class 10 are solved here.

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NCERT solutions for Class 10 Maths chapter 4 Quadratic Equations 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 $x^2 -8x-1280 = 0$

Quadratic Equation Class 10 Excercise: 4.2

Q1 (i) Find the roots of the following quadratic equations by factorization: $x^2 - 3x - 10 =0$

Answer:

Given the quadratic equation: $x^2 - 3x - 10 =0$

Factorization gives, $x^2 - 5x+2x - 10 =0$

$\Rightarrow x^2 - 5x+2x - 10 =0$

$\Rightarrow x(x-5) +2(x-5) =0$

$\Rightarrow (x-5)(x+2) =0$

$\Rightarrow x= 5\ or\ -2$

Hence, the roots of the given quadratic equation are $5\ and\ -2$ .

Q1 (ii) Find the roots of the following quadratic equations by factorization: $2x^2 + x - 6 = 0$

Answer:

Given the quadratic equation: $2x^2 + x - 6 = 0$

Factorisation gives, $2x^2 +4x-3x - 6 = 0$

$\Rightarrow 2x(x+2) -3(x+2) =0$

$\Rightarrow (x+2)(2x-3) = 0$

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

Hence, the roots of the given quadratic equation are

$-2\ and\ \frac{3}{2}$

Q1 (iii) Find the roots of the following quadratic equations by factorization: $\sqrt2x^2 + 7x + 5\sqrt2 = 0$

Answer:

Given the quadratic equation: $\sqrt2x^2 + 7x + 5\sqrt2 = 0$

Factorization gives, $\sqrt2x^2 + 5x+2x + 5\sqrt2 = 0$

$\Rightarrow x(\sqrt2 x +5) +\sqrt2 (\sqrt 2 x +5)= 0$

$\Rightarrow (\sqrt2 x +5)(x+\sqrt{2}) = 0$

$\Rightarrow x=\frac{-5}{\sqrt 2 }\ or\ -\sqrt 2$

Hence, the roots of the given quadratic equation are

$\frac{-5}{\sqrt 2 }\ and\ -\sqrt 2$

Q1 (iv) Find the roots of the following quadratic equations by factorization: $2x^2 -x + \frac{1}{8} = 0$

Answer:

Given the quadratic equation: $2x^2 -x + \frac{1}{8} = 0$

Solving the quadratic equations, we get

$16x^2-8x+1 = 0$

Factorization gives, $\Rightarrow 16x^2-4x-4x+1 = 0$

$\Rightarrow 4x(4x-1)-1(4x-1) = 0$

$\Rightarrow (4x-1)(4x-1) = 0$

$\Rightarrow x=\frac{1}{4}\ or\ \frac{1}{4}$

Hence, the roots of the given quadratic equation are

$\frac{1}{4}\ and\ \frac{1}{4}$

Q1 (v) Find the roots of the following quadratic equations by factorization: $100x^2 -20x +1 = 0$

Answer:

Given the quadratic equation: $100x^2 -20x +1 = 0$

Factorization gives, $100x^2 -10x-10x +1 = 0$

$\Rightarrow 10x(10x-1)-10(10x-1) = 0$

$\Rightarrow (10x-1)(10x-1) = 0$

$\Rightarrow x=\frac{1}{10}\ or\ \frac{1}{10}$

Hence, the roots of the given quadratic equation are

$\frac{1}{10}\ and\ \frac{1}{10}$ .

Q2 Solve the problems given in Example 1. (i) $x^2-45x+324 = 0$ (ii) $x^2-55x+750 = 0$

Answer:

From Example 1 we get:

Equations:

(i) $x^2-45x+324 = 0$

Solving by factorization method:

Given the quadratic equation: $x^2-45x+324 = 0$

Factorization gives, $x^2-36x-9x+324 = 0$

$\Rightarrow x(x-36) - 9(x-36) = 0$

$\Rightarrow (x-9)(x-36) = 0$

$\Rightarrow x=9\ or\ 36$

Hence, the roots of the given quadratic equation are $x=9\ and \ 36$ .

Therefore, John and Jivanti have 36 and 9 marbles respectively in the beginning.

(ii) $x^2-55x+750 = 0$

Solving by factorization method:

Given the quadratic equation: $x^2-55x+750 = 0$

Factorization gives, $x^2-30x-25x+750 = 0$

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

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

$\Rightarrow x=25\ or\ 30$

Hence, the roots of the given quadratic equation are $x=25\ and \ 30$ .

Therefore, the number of toys on that day was $30\ or\ 25.$

Q3 Find two numbers whose sum is 27 and the product is 182.

Answer:

Let two numbers be x and y .

Then, their sum will be equal to 27 and the product equals 182.

$x+y = 27$ ...............................(1)

$xy =182$ .................................(2)

From equation (2) we have:

$y = \frac{182}{x}$

Then putting the value of y in equation (1), we get

$x+\frac{182}{x} = 27$

Solving this equation:

$\Rightarrow x^2-27x+182 = 0$

$\Rightarrow x^2-13x-14x+182 = 0$

$\Rightarrow x(x-13)-14(x-13) = 0$

$\Rightarrow (x-14)(x-13) = 0$

$\Rightarrow x = 13\ or\ 14$

Hence, the two required numbers are $13\ and \ 14$ .

Q4 Find two consecutive positive integers, the sum of whose squares is 365.

Answer:

Let the two consecutive integers be $'x'\ and\ 'x+1'.$

Then the sum of the squares is 365.

. $x^2+ (x+1)^2 = 365$

$\Rightarrow x^2+x^2+1+2x = 365$

$\Rightarrow x^2+x-182 = 0$

$\Rightarrow x^2 - 13x+14x+182 = 0$

$\Rightarrow x(x-13)+14(x-13) = 0$

$\Rightarrow (x-13)(x-14) = 0$

$\Rightarrow x =13\ or\ 14$

Hence, the two consecutive integers are $13\ and\ 14$ .

Q5 The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.

Answer:

Let the length of the base of the triangle be $b\ cm$ .

Then, the altitude length will be: $b-7\ cm$ .

Given if hypotenuse is $13\ cm$ .

Applying the Pythagoras theorem; we get

$Hypotenuse^2 = Perpendicular^2 + Base^2$

So, $(13)^2 = (b-7)^2 +b^2$

$\Rightarrow 169 = 2b^2+49-14b$

$\Rightarrow 2b^2-14b-120 = 0$ Or $b^2-7b-60 = 0$

$\Rightarrow b^2-12b+5b-60 = 0$

$\Rightarrow b(b-12) + 5(b-12) = 0$

$\Rightarrow (b-12)(b+5) = 0$

$\Rightarrow b= 12\ or\ -5$

But, the length of the base cannot be negative.

Hence the base length will be $12\ cm$ .

Therefore, we have

Altitude length $= 12cm -7cm = 5cm$ and Base length $= 12\ cm$

Q6 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 Rs 90, find the number of articles produced and the cost of each article.

Answer:

Let the number of articles produced in a day $= x$

The cost of production of each article will be $=2x+3$

Given the total production on that day was $Rs.90$ .

Hence we have the equation;

$x(2x+3) = 90$

$2x^2+3x-90 = 0$

$\Rightarrow 2x^2+15x-12x-90 = 0$

$\Rightarrow x(2x+15) - 6(2x+15) = 0$

$\Rightarrow (2x+15)(x-6) = 0$

$\Rightarrow x =-\frac{15}{2}\ or\ 6$

But, x cannot be negative as it is the number of articles.

Therefore, $x=6$ and the cost of each article $= 2x+3 = 2(6)+3 = 15$

Hence, the number of articles is 6 and the cost of each article is Rs.15.

Class 10 Maths Chapter 4 Quadratic Equations Excercise: 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 $(12\ m+6\ m) = 18\ m$ .

Quadratic Equation Class 10 Excercise: 4.4

Q1 (i) Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:

$2x^2 - 3x +5 = 0$

Answer:

For a quadratic equation, $ax^2+bx+c = 0$ the value of discriminant determines the nature of roots and is equal to:

$D = b^2-4ac$

If D>0 then roots are distinct and real.

If D<0 then no real roots.

If D= 0 then there exists two equal real roots.

Given the quadratic equation, $2x^2 - 3x +5 = 0$ .

Comparing with general to get the values of a,b,c.

$a = 2, b =-3,\ c= 5$

Finding the discriminant:

$D= (-3)^2 - 4(2)(5) = 9-40 = -31$

$\because D<0$

Here D is negative hence there are no real roots possible for the given equation.

Q1 (ii) Find the nature of the roots of the following quadratic equations. If the real roots exist, find them: $3x^2 - 4\sqrt3x + 4 = 0$

Answer:

$b^2-4ac=(-4\sqrt{3})^2-(4\times4\times3)=48-48=0$

Here the value of discriminant =0, which implies that roots exist and the roots are equal.

The roots are given by the formula

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{4\sqrt{3}\pm\sqrt{0}}{2\times3}=\frac{2}{\sqrt{3}}$

So the roots are

$\frac{2}{\sqrt{3}},\ \frac{2}{\sqrt{3}}$

Q1 (iii) Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:

$2x^2 - 6x + 3 = 0$

Answer:

The value of the discriminant

$b^2-4ac=(-6)^2-4\times2\times3=12$

The discriminant > 0. Therefore the given quadratic equation has two distinct real root

roots are

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-6\pm\sqrt{12}}{2\times2}=\frac{3}{2}\pm\frac{\sqrt{3}}{2}$

So the roots are

$\frac{3}{2}+\frac{\sqrt{3}}{2}, \frac{3}{2}-\frac{\sqrt{3}}{2}$

Q2 (i) Find the values of k for each of the following quadratic equations so that they have two equal roots.

$2x^2 + kx + 3 = 0$

Answer:

For two equal roots for the quadratic equation: $ax^2+bx+c =0$

The value of the discriminant $D= 0$ .

Given equation: $2x^2 + kx + 3 = 0$

Comparing and getting the values of a,b, and, c.

$a = 2, \ b = k,\ c = 3$

The value of $D = b^2-4ac = (k)^2 - 4(2)(3)$

$\Rightarrow (k)^2 = 24$

Or, $\Rightarrow k=\pm \sqrt{24} = \pm 2\sqrt{6}$

Q2 (ii) Find the values of k for each of the following quadratic equations so that they have two equal roots

$kx(x-2) + 6 = 0$

Answer:

For two equal roots for the quadratic equation: $ax^2+bx+c =0$

The value of the discriminant $D= 0$ .

Given equation: $kx(x-2) + 6 = 0$

Can be written as: $kx^2-2kx+6 = 0$

Comparing and getting the values of a,b, and, c.

$a = k, \ b = -2k,\ c = 6$

The value of $D = b^2-4ac = (-2k)^2 - 4(k)(6) = 0$

$\Rightarrow 4k^2 - 24k = 0$

$\Rightarrow 4k(k-6) = 0$

$\Rightarrow k= 0\ or\ 6$

But $k= 0$ is NOT possible because it will not satisfy the given equation.

Hence the only value of $k$ is 6 to get two equal roots.

Q3 Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m ² ? If so, find its length and breadth.

Answer:

Let the breadth of mango grove be $'b'$ .

Then the length of mango grove will be $'2b'$ .

And the area will be:

$Area = (2b)(b) = 2b^2$

Which will be equal to $800m^2$ according to question.

$\Rightarrow 2b^2 = 800m^2$

$\Rightarrow b^2 - 400 = 0$

Comparing to get the values of $a,b,c$ .

$a=1, \ b= 0 , \ c = -400$

Finding the discriminant value:

$D = b^2-4ac$

$\Rightarrow 0^2-4(1)(-400) = 1600$

Here, $D>0$

Therefore, the equation will have real roots.

And hence finding the dimensions:

$\Rightarrow b^2 - 400 = 0$

$\Rightarrow b = \pm 20$

As negative value is not possible, hence the value of breadth of mango grove will be 20m.

And the length of mango grove will be: $= 2\times10 = 40m$

Q4 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.

Answer:

Let the age of one friend be $x\ years.$

and the age of another friend will be: $(20-x)\ years.$

4 years ago, their ages were, $x-4\ years$ and $20-x-4 \ years$ .

According to the question, the product of their ages in years was 48.

$\therefore (x-4)(20-x-4) = 48$

$\Rightarrow 16x-64-x^2+4x= 48$

$\Rightarrow -x^2+20x-112 = 0$ or $\Rightarrow x^2-20x+112 = 0$

Now, comparing to get the values of $a,\ b,\ c$ .

$a = 1,\ b= -20,\ c =112$

Discriminant value $D = b^2-4ac = (-20)^2 -4(1)(112) = 400-448 = -48$

As $D<0$ .

Therefore, there are no real roots possible for this given equation and hence,

This situation is NOT possible.

Q5 Is it possible to design a rectangular park of perimeter 80 m and area 400 m ² ? If so, find its length and breadth.

Answer:

Let us assume the length and breadth of the park be $'l'\ and\ 'b'$ respectively.

Then, the perimeter will be $P = 2(l+b) = 80$

$\Rightarrow l+b = 40\ or\ b = 40 - l$

The area of the park is:

$Area = l\times b = l(40-l) = 40l - l^2$

Given : $40l - l^2 = 400$

$l^2 - 40l +400 = 0$

Comparing to get the values of a, b and c.

The value of the discriminant $D = b^2-4ac$

$\Rightarrow = b^2-4ac = (-40)^2 - 4(1)(400) = 1600 -1600 = 0$

As $D = 0$ .

Therefore, this equation will have two equal roots.

And hence the roots will be:

$l =\frac{-b}{2a}$

$l =\frac{-40}{2(1)} = \frac{40}{2} =20$

Therefore, the length of the park, $l =20\ m$

and breadth of the park $b = 40-l = 40 -20 = 20\ m$ .

Topics of Class 10 Maths Chapter 4 Quadratic Equations

Representation of situation in a quadratic equation.
Checking if an equation is quadratic or not.
Solving a quadratic equation using factorization/ roots of quadratic solutions.
The solution of the quadratic equation using completing the square method.
Solving a quadratic equation using the Sridharacharya formula.
Product of roots in a quadratic equation.
Sum of roots in a quadratic equation.

There are 4 exercise in Class 10 Maths chapter 4. Get the exercise wise solutions from the following links.

NCERT solutions for class 10 Maths chapter wise

Chapter No.	Chapter Name
Chapter 1	Real Numbers
Chapter 2	Polynomials
Chapter 3	Pair of Linear Equations in Two Variables
Chapter 4	Quadratic Equations
Chapter 5	Arithmetic Progressions
Chapter 6	Triangles
Chapter 7	Coordinate Geometry
Chapter 8	Introduction to Trigonometry
Chapter 9	Some Applications of Trigonometry
Chapter 10	Circles
Chapter 11	Constructions
Chapter 12	Areas Related to Circles
Chapter 13	Surface Areas and Volumes
Chapter 14	Statistics
Chapter 15	Probability

How To use NCERT Solutions For Class 10 Maths Chapter 4 Quadratic Equations?

First of all list down all the questions in which you need assistance and go through the NCERT Solutions For Class 10 Maths Chapter 4 Quadratic Equations of that particular question.
When you complete the first step then your next target should be previous papers. You can pick past year papers and practice them thoroughly.
Once you complete NCERTs and previous year papers, try to solve the questions of that particular chapter from different state board books.

Subject-wise NCERT Solutions of Class 10

Subject wise NCERT Exemplar solutions

Also Check NCERT Books and NCERT Syllabus here

Solutions

Frequently Asked Question (FAQs) - NCERT Solutions For Class 10 Maths Chapter 4 Quadratic Equations

Question: Which of the following is a quadratic equation?

Answer:

To determine whether an equation is a quadratic equation, you can look for two conditions: firstly, the highest power of the variable is 2 and secondly, the coefficient of the x^2 term (that is, the "a" in the equation) is not equal to zero. These two are some examples of quadratic equations 2x^2+3x+5 and x^2+5x+6.

Question: How do you solve a quadratic equation in Class 10?

Answer:

There are multiple methods to solve a quadratic equation including the factoring method, quadratic formula method, graphical method, and many others. These methods depend on the type of quadratic equation so first to know what is quadratic polynomial type and then apply the above methods to find a quadratic polynomial solution.

Question: How do you download the Class 10 Maths Quadratic Equations NCERT Textbooks PDF?

Answer:

Students can visit the NCERT official website. There they can find options to select class, subject, and book title. Using these inputs students can download NCERT chapter 4 maths class 10 which is quadratic equations class 10 pdf. Students can also find solutions to these NCERTs from Careers360's official website

Question: What are the important topics covered in the NCERT Solutions for Class 10 Maths Chapter 4?

Answer:

Chapter 4 quadratic equation class 10 ncert solutions contain multiple topics like Quadratic equations, Solutions of a quadratic equation, Discriminant of a quadratic equation, Factoring method for solving a quadratic equation, Quadratic formula for solving a quadratic equation, Nature of roots of a quadratic equation, and Graphs of quadratic equations.

Question: What are the important concepts learned in the NCERT Solutions for chapter 4 maths class 10 Quadratic Equations?

Answer:

Class 10 maths chapter 4 solutions required multiple concepts including what is an equation, how to get the solution of an equation, methods of quadratic equation solution, the factoring method for solving a quadratic equation, the quadratic formula for solving a quadratic equation, the nature of roots of a quadratic equation, graphs of quadratic equations, and many.

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Questions related to CBSE Class 10th

Showing 71 out of 71 Questions

394 Views

my sons dob is 13/02/2009 is hi eligible for 10th board exam in 2024,,from cbse. olz reply me

Shubhangi 30th Jul, 2022

The eligibility age criteria for class 10th CBSE is 14 years of age. Since your son will be 15 years of age in 2024, he will be eligible to give the exam.

1410 Views

can Hindi marks be replace with IT 402(additional subject) in CBSE class 10th board

Aditi Singh 22nd Jul, 2022

That totally depends on what you are aiming for. The replacement of marks of additional subjects and the main subject is not like you will get the marks of IT on your Hindi section. It runs like when you calculate your total percentage you have got, you can replace your lowest marks of the main subjects from the marks of the additional subject since CBSE schools goes for the best five marks for the calculation of final percentage of the students.

However, for the admission procedures in different schools after 10th, it depends on the schools to consider the percentage of main five subjects or the best five subjects to admit the student in their schools.

125 Views

i got less marks in maths can it get replaced by additional hindi in cbse

Aditi Singh 22nd Jul, 2022

Replacement of marks of additional subjects and your main course subject is not like they will swap the marks you got in Hindi and Mathematics on your marksheet. It works like if you calculate the total percentage you got in your class, you can take the marks of your additional subject to consider in place of the subject you have got the least marks in. CBSE schools consider this method only to release the merit list of their students.

If you're in 10th and aiming to get admission in 11th in any school, it depends on the schools if they consider your main five subject marks or the best five subject marks for the admission.

If you're in 12th and are trying for different colleges to get in, it depends on your course which you wish to study further and also some criterias and eligibility of the colleges. In most cases for general courses, colleges take the best three or best four subjects marks consideration for the admission process.

So, good luck.

91 Views

when will cbse term 2 result will declare please tale

Shubham Dhane 21st Jul, 2022

Hello,

Central Board of Secondary Education will declare term 2 CBSE exam result 2022 for Class 10 in the third week of July. Please keep an eye on the official website or the link below for the latest information. Students need to enter their CBSE board roll number and other details to check term 2 cbseresults.nic.in 2022 Class 10 results. The board had announced the CBSE Class 10 results 2022 for term 1 on March 11, 2022.

https://school.careers360.com/boards/cbse/cbse-result

51 Views

cbse 10 claas bord result check now

Jyoti Eyasmin 20th Jul, 2022

Hello Aspirant,

Class 10 cbse results are not out yet. It is expected to be declared on 23rd of July. Keep checking the official website of cbse.

Good luck.

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NCERT Solutions For Class 10 Maths Chapter 4 Quadratic Equations

NCERT solutions for Class 10 Maths chapter 4 Quadratic Equations Excercise: 4.1

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Quadratic Equation Class 10 Excercise: 4.2

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