Class 10 - Chapter 4 Archives – Mathemafia http://sh016.global.temp.domains/~mathemaf/category/class-10-chapter-4/ Mathematics is like music - Enjoy it!!! Wed, 07 May 2025 07:57:23 +0000 en-US hourly 1 https://wordpress.org/?v=7.0 https://mathemafia.com/wp-content/uploads/2020/08/cropped-M__Logo__SM-1-32x32.png Class 10 - Chapter 4 Archives – Mathemafia http://sh016.global.temp.domains/~mathemaf/category/class-10-chapter-4/ 32 32 Exercise 4.2 (NCERT) – Quadratic Equations, Class 10 – All Solutions https://mathemafia.com/exercise-4-2-ncert-quadratic-equations-class-10-all-solutions/ Mon, 05 May 2025 08:31:29 +0000 https://mathemafia.com/?p=18385 Exercise 4.2, chapter 4, Quadratic Equations, Class 10 - All questions are solved using Mathematical symbols and scalable vector graphics. The exercise is primarily about solving the quadratic equations by factorisation.

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Exercise 4.2 (NCERT) – Quadratic Equations, Class 10

Chapter 4 Exercise 4.2 Class 10

Below are the Quick links for all questions of Exercise 4.2, Quadratic Equations, Class 10. Click a link to view the solution of corresponding question.

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Below are the solutions of every question of exercise 4.2 of chapter 4, Quadratic Equations, NCERT of class 10. YouTube video tutorial of every question is also embedded along with the written solution.  

Question 1

Find the roots of the following quadratic equations by factorisation:

(i) x2 – 3x – 10 = 0

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Splitting the middle term

Answer: Roots: -2, 5

Important: To verify that your answer is correct, always put the values of the roots in the given equation and verify that LHS = RHS. 

(ii) 2x2 + x – 6 = 0

Splitting the middle term

Answer: Roots: 3/2, -2

(iii) √2x2 + 7x + 5√2 = 0 

Splitting the middle term

Answer: Roots: -√2, -5/√2

(iv) 2x2 – x + 1/8 = 0 

Answer: Roots: 1/4, 1/4

(v) 100 x2 – 20x + 1 = 0

Splitting the middle term

Answer: Roots: 1/10, 1/10

Question 2

Solve the problems given in Example 1.

(i) John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 124. We would like to find out how many marbles they had to start with.

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Let John’s marble count = x 
Then Jivanti’s marble count = 45 – x 

Condition
Both lost 5 marbles, this implies

John’s marble count = x – 5
Jivanti’s marble count = 45 – x – 5 = 40 – x

product of the remanding marbles = 124

Splitting the middle term

Taking x = 9

No. of marbles John has = x = 9

No. of marbles Jivanti has = 45 – 9 = 36

Taking x = 36

No. of marbles John has = x = 36

No. of marbles Jivanti has = 45 – 36 = 9

Answer: 9, 36

(ii) A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production was Rs. 750. We would like to find out the number of toys produced on that day.

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Let the number of toys produced = x 
Then cost per toy = 55 – x 

Condition

Total cost = 750

=> Total no. of toys x Cost of one toy = 750

=> x(55 – x) = 750

Answer: 25, 30

Question 3

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

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Let one number be = x 

Condition 1:

Sum of numbers = 27

=> Second number = 27 – x

Condition 2:

Product of numbers = 182

Taking x = 13

First number = x = 13
Second number = 27 – 13 = 14

Taking x = 14

First number = x = 14
Second number = 27 – 14 = 13

In both cases the two numbers are 13, 14

Answer: 13, 14

Important: You can solve this question using two variables as well. Eventually, one variable will be eliminated and you will get a quadratic equation. Watch the video to learn this method.

Question 4

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

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Let smaller positive integer be = x
Then, next integer = x + 1

Condition

Sum of squares of these numbers = 365

Reject x = -14

(because it is not a positive integer)

⇒ First number = x = 13

=> Second number = x + 1 = 14

Answer: 13, 14

Important: There is another way to solve this question. Watch the video to learn it.

Question 5

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

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Let the base = x cm
Then the altitude = x – 7 cm
Hypotenuse = 13 cm

Applying Pythagoras Theorem 

Reject x = -5

(becasue length of a side cannot be -ve)

⇒ Base = x = 12 cm

=> Altitude = x – 7 = 5 cm

Answer: Base: 12 cm, Altitude: 5 cm

Question 6

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.

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Let the number of articles produced = x 
Cost per article = 2x + 3

Condition

Total cost = 90

=> Total no. of articles x Cost per article = 90

=> x (2x + 3) = 90

Reject -ve value as number of articles cannot be -ve

⇒ Total no. of articles produced = x = 6

=> Cost per article = 2x + 3 = 12 + 3 = 15 Rs.

Answer: Number of articles = 6, Cost per article = Rs. 15

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]]> Exercise 4.1 (NCERT) – Quadratic Equations, Class 10 – All Solutions https://mathemafia.com/exercise-4-1-ncert-quadratic-equations-class-10-all-solutions/ Mon, 05 May 2025 05:15:56 +0000 https://mathemafia.com/?p=18260 Exercise 4.1, chapter 4, Quadratic Equations, Class 10 - All questions are solved using Mathematical symbols and scalable vector graphics. The exercise is primarily about identifying and forming the quadratic equations.

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Exercise 4.1 (NCERT) – Quadratic Equations, Class 10

Chapter 4 Exercise 4.1 Class 10

Below are the Quick links for all questions of Exercise 4.1, Quadratic Equations, Class 10. Click a link to view the solution of corresponding question.

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Below are the solutions of every question of exercise 4.1 of chapter 4, Quadratic Equations, NCERT of class 10. YouTube video tutorial of every question is also embedded along with the written solution.  

Question 1

Check whether the following are quadratic equations :

(i) (x + 1)2 = 2(x – 3)

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Expand the equation

This is in the form ax2 + bx + c = 0 where a = 1, b = 0, c = 7

Answer: Yes, it is a Quadratic Equation.

(ii) x2 – 2x = (–2) (3 – x)

This is in the form ax2 + bx + c = 0 where a = 1, b = -4, c = 6

Answer: Yes, it is a Quadratic Equation.

(iii) (x – 2)(x + 1) = (x – 1)(x + 3)

This is not in the form ax2 + bx + c = 0 because its highest degree is 1.

Answer: No, it is not a Quadratic Equation.

(iv) (x – 3)(2x +1) = x(x + 5)

This is in the form ax2 + bx + c = 0 where a = 1, b = -10, c = -3

Answer: Yes, it is a Quadratic Equation.

(v) (2x – 1)(x – 3) = (x + 5)(x – 1)

This is in the form ax2 + bx + c = 0 where a = 1, b = -11, c = 8

Answer: Yes, it is a Quadratic Equation.

(vi) x2 + 3x + 1 = (x – 2)2

This is not in the form ax2 + bx + c = 0 because its highest degree is 1.

Answer: No, it is not a Quadratic Equation.

(vii) (x + 2)3 = 2x (x2 – 1)

This is not in the form ax2 + bx + c = 0 because its highest degree is 3.

Answer: No, it is not a Quadratic Equation.

(viii) x3 – 4x2– x + 1 = (x – 2)3

This is in the form ax2 + bx + c = 0 where a = 2, b = -13, c = 9

Answer: Yes, it is a Quadratic Equation.

Question 2

Represent the following situations in the form of quadratic equations :

(i) The area of a rectangular plot is 528 m2. The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.

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Let the breadth be x metres
Then, length = 2x + 1 metres

Area = length × breadth = 528

⇒ (2x + 1)(x) = 528
⇒ 2x2 + x = 528
⇒ 2x2 + x − 528 = 0

Answer: 2x2 + x − 528 = 0

(ii) The product of two consecutive positive integers is 306. We need to find the integers.

Let the smaller integer be x
Then the next integer = x + 1

Given: Product of numbers = 306

⇒ x(x + 1) = 306
⇒ x2 + x = 306
⇒ x2 + x − 306 = 0

Answer: x2 + x − 306 = 0

Important: There is one more solution and a different answer of this question. We can form another quadratic equation using the given information. Watch the video to know about it.

(iii) 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.

Let Rohan’s present age be x years
Then his mother’s present age = x + 26 years

After 3 years:
Rohan’s age = x + 3

Mother’s age = x + 26 + 3 = x + 29

Given: Product of their ages = 360

⇒ (x + 3)(x + 29) = 360
⇒ x2 + 32x + 87 = 360
⇒ x2 + 32x − 273 = 0

Answer: x2 + 32x − 273 = 0

(iv) 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.

Case 1: Normal Speed

Let the train’s speed be x km/h

Case 2: Slower Speed

Speed = x – 8 km/h

Time taken at slower speed = 3 Hours more than normal speed

Answer: x2 − 8x − 1280 = 0

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]]> Exercise 4.3 (NCERT) – Quadratic Equations, Class 10 – All Solutions https://mathemafia.com/exercise-4-3-ncert-quadratic-equations-class-10-all-solutions/ Tue, 01 Apr 2025 08:52:15 +0000 https://mathemafia.com/?p=16660 Exercise 4.3, chapter 4, Quadratic Equations, Class 10 - All question solved using proper mathematical symbols. Quadratic Equation Calculator is also available that you can use to solve any QE equation.

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Exercise 4.3 (NCERT) - Quadratic Equations, Class 10

Chapter 4 Exercise 4.3 Class 10

Below are the Quick links for all questions of Exercise 4.3, Quadratic Equations, Class 10.

Click a link to view the solution of corresponding question.

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Quadratic Equation Calculator 

Click a question to view its solution. You can also enter any integer value in the calculator directly to solve the corresponding quadratic equation. 

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NCERT

  • E.g. 7) 2x2 - 4x + 3 = 0
  • Ex: 4.3, Q1) 2x2 - 3x + 5 = 0
  • Ex: 4.3, Q3) 2x2 - 6x + 3 = 0

NCERT Exemplar

  • Ex: 4.2, Q1 (i) x2 - 3x + 4 = 0
  • Ex: 4.2, Q1 (ii) 2x2 + x - 1 = 0
  • Ex: 4.2, Q1 (iv) 3x2 - 4x + 1 = 0
  • Ex: 4.3, Q1 (i) 2x2 - 3x - 5 = 0
  • Ex: 4.3, Q1 (ii) 5x2 + 13x + 8 = 0
  • Ex: 4.3, Q1 (iii) -3x2 + 5x + 12 = 0
  • Ex: 4.3, Q1 (iv) -x2 + 7x - 10 = 0

More Questions

  • 1) 4x2 - 4 = 0
  • 2) x2 = 0
  • 3) x2 - 4 = 0
  • 4) x2 + 4 = 0
  • 5) 4x2 + 4x + 1 = 0
  • 6) 8x2 + 8x + 2 = 0
    Observe this equation has same roots as that of question 5.
  • 7) x2 + 5x + 6 = 0
  • 8) 20x2 - 15x = 0
  • 9) x2 - x + 6 = 0
  • 10) x2 - 10x + 25 = 0

Enter the coefficeints of QE

Coeff. the x2

Coeff. the x

Constant

\( \begin{aligned}\text{Coeff of x}^2\text{ cannot be 0}\end{aligned}\)

Answer

Solution

Comparing the equation with
ax2 + bx + c = 0, we get

Discriminant (D)

Nature of Roots

D > 0 ⇒ Two Real and Distinct Roots

D = 0 ⇒ Two Real and Equal Roots

D < 0 ⇒ Two Complex Roots

Roots

Below are the solutions of every question of exercise 4.3 of chapter 4, Quadratic Equations, NCERT of class 10. YouTube video tutorial of every question is also embedded along with the written solution.  

Question 1

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

(i) 2x2 – 3x + 5 = 0

Comparing the equation with ax2 + bx + c = 0

𝑎=2,𝑏=−3,𝑐=5

Discriminant (D)

𝐷=𝑏2−4𝑎𝑐=(−3)2−4(2)(5)=9−40=−31

Nature of Roots

D < 0 ⇒ No Real Roots

(ii) 3x2 – 4√3x + 4 = 0

Comparing the equation with ax2 + bx + c = 0 

𝑎=3,𝑏=−4√3,𝑐=4

Discriminant (D)

𝐷=𝑏2−4𝑎𝑐=(−43‾√)2−4(3)(4)=(−43‾√)2−4(3)(4)=(16×3)−48=48−48=0

Nature of Roots

D = 0 ⇒ Two Equal Real Roots

Roots

𝑥=−𝑏±𝐷‾‾√2𝑎=−(−43‾√)±0‾√2×3=43‾√+02×3,43‾√−02×3=23‾√3,23‾√3

(iii) 2x2 – 6x + 3 = 0

Comparing the equation with ax2 + bx + c = 0 

𝑎=2,𝑏=−6,𝑐=3

Discriminant (D)

𝐷=𝑏2−4𝑎𝑐=(−6)2−4(2)(3)=36−24=12

Nature of Roots

D > 0 ⇒ Two Distinct Real Roots

Roots

\( \begin{aligned} x &= \frac{-b \pm \sqrt{D}}{2a} \\ &= \frac{-(-6) \pm \sqrt{12}}{2 \times 2}\\ &= \frac{6 \pm 2\sqrt{3} }{4}\\ &= \frac{3 \pm \sqrt{3} }{2}\\ \end{aligned}\)

Question 2

Find the values of k for each of the following quadratic equations, so that they have two equal roots.

(i) 2x2 + kx + 3 = 0

Comparing the equation with ax2 + bx + c = 0 

\(\begin{aligned} &a = 2, b = k, c = 3 \end{aligned}\)

We are given that the equation has two equal roots, this means

Discriminant = 0

\( \begin{aligned} \Rightarrow b^2-4ac &= 0\\ \Rightarrow k^2-4(2)(3) &=0\\ \Rightarrow k^2-24 &=0\\ \Rightarrow k^2 &=24\\ \Rightarrow k &=\pm\sqrt{24}\\ \Rightarrow k &=\pm2\sqrt{6}\\ \end{aligned}\)

(ii) kx (x – 2) + 6 = 0

𝑘𝑥(𝑥–2)+6⇒𝑘𝑥2–2𝑘𝑥+6=0=0

Comparing the equation with ax2 + bx + c = 0 

\(\begin{aligned} &a = k, b = -2k, c = 6 \end{aligned}\)

We are given that the equation has two equal roots, this means

Discriminant = 0

⇒𝑏2−4𝑎𝑐⇒(−2𝑘)2−4(𝑘)(6)⇒4𝑘2−24𝑘⇒4𝑘(𝑘−6)=0=0=0=0
⇒  4𝑘=0  ⇒  𝑘=0  𝑜𝑟  𝑘−6=0𝑜𝑟  𝑘=6

We can clearly observe that the value of k cannot be 0.

Let us put k = 0 in the given equation.

𝑘𝑥(𝑥–2)+6⇒  0×𝑥(𝑥–2)+6⇒  0+6⇒  6=0=0=0=0

This is not possible. Hence the value of k cannot be 0.

Hence k = 6.

Important: Always verify your answer by putting the value of k in the given equation to see if we get valid quadratic equation or not and if time allows, quickly verify if discriminant is 0 or not.

Question 3

Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m2? If so, find its length and breadth.

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For Rectangular Mango Grove

𝐿𝑒𝑡  𝑏𝑟𝑒𝑎𝑑𝑡ℎ  (𝑏)⇒𝐿𝑒𝑛𝑔𝑡ℎ  (𝑙)=𝑥  𝑚𝑒𝑡𝑒𝑟=2𝑥  𝑚𝑒𝑡𝑒𝑟

Area = 800 m2

\( \begin{aligned} \Rightarrow l \times b &= 800\\ \Rightarrow 2x \times x &= 800\\ \Rightarrow 2x^2 &= 800\\ \Rightarrow x^2 &= 400\\ \Rightarrow x &= \pm\sqrt{400}\\ \Rightarrow x &= \pm20\\\\ \Rightarrow breadth = x &= \pm20 \end{aligned}\)

Reject the -ve value because breadth cannot be -ve.

Hence 

𝑏𝑟𝑒𝑎𝑑𝑡ℎ=𝑥𝑙𝑒𝑛𝑔𝑡ℎ=2𝑥=20  𝑚𝑒𝑡𝑒𝑟=2×20=40  𝑚𝑒𝑡𝑒𝑟

Conclusion: Yes, it is possible to design the rectangular mango grove whose length is twice its breadth.

Question 4

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.

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It is recommended to solve such questions by making a table as shown below.

Let x and y are the current (today’s) ages of the two friends.

Today 4 Years Ago
Age (yrs) of Friend 1
x
x – 4
Age (yrs) of Friend 2
y
y – 4
Condition
x + y = 20
(x – 4) * (y – 4) = 48

We got two equations from the given data.

𝑥+𝑦=20  …(1)(𝑥−4)×(𝑦−4)=48  …(2)

Get the value of y in terms of x from equation 1.

(Applying substitution method)

𝑥+𝑦⇒𝑦=20=20−𝑥

Put the value of y in equation 2.

(𝑥−4)×(𝑦−4)(𝑥−4)×(20−𝑥−4)(𝑥−4)×(16−𝑥)16𝑥−64−𝑥2+4𝑥−𝑥2+20𝑥−64−48−𝑥2+20𝑥−112𝑥2−20𝑥+112=48=48=48=48=0=0=0

Comparing the equation with ax2 + bx + c = 0 

\(\begin{aligned} &a = 1, b = -20, c = 112 \end{aligned}\)

Discriminant (D)

\( \begin{aligned} D &= b^2-4ac\\ &= (-20)^2-4(1)(112)\\ &= 400-448\\ &= -48\\ \end{aligned}\)

D < 0 ⇒ No Real Roots

Conclusion: This means the situation provided in the question is not possible.

Question 5

Is it possible to design a rectangular park of perimeter 80 m and area 400 m2? If so, find its length and breadth.

For Rectangular Park

𝐿𝑒𝑡  𝐿𝑒𝑛𝑔𝑡ℎ𝐵𝑟𝑒𝑎𝑑𝑡ℎ=𝑥=𝑦

Perimeter = 80 m

⇒2(𝑥+𝑦)⇒𝑥+𝑦⇒𝑦=80=40=40−𝑥  …(1)

Area = 400 m2

\( \begin{aligned} &\Rightarrow x \times y = 400 \end{aligned}\)
\( \begin{aligned} &\text{Substitute the value of } y \text{ from equation (1)} \end{aligned}\)
⇒𝑥×(40−𝑥)⇒40𝑥−𝑥2⇒𝑥2−40𝑥+400=400=400=0

Comparing the equation with ax2 + bx + c = 0 

\(\begin{aligned} &a = 1, b = -40, c = 400 \end{aligned}\)

Discriminant (D)

𝐷=𝑏2−4𝑎𝑐=(−40)2−4(1)(400)=1600−1600=0

D = 0 ⇒ Two Equal Real Roots

\( \begin{aligned} x &= \frac{-b \pm \sqrt{D}}{2a} \\ &= \frac{-(-40) \pm \sqrt{0}}{2 \times 1}\\ &= \frac{40 \pm 0}{2}\\ &= \frac{40 + 0}{2}, \frac{40 – 0}{2}\\ &= \frac{40 }{2}, \frac{40 }{2}\\ &= 20, 20 \end{aligned}\)

Hence

\( \begin{aligned} Length = x &= 20 m\\ Breadth = 40 – x &= 20 m \end{aligned}\)
\( \begin{aligned} \end{aligned}\)

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\( \begin{aligned} \end{aligned}\)

Quadratic Equation Calculator

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Enter the coefficeints of QE

Coeff. the x2

Coeff. the x

Constant

\( \begin{aligned}\text{Coeff of x}^2\text{ cannot be 0}\end{aligned}\)

Answer

Solution

Comparing the equation with
ax2 + bx + c = 0, we get

Discriminant (D)

Nature of Roots

D > 0 ⇒ Two Real and Distinct Roots

D = 0 ⇒ Two Real and Equal Roots

D < 0 ⇒ Two Complex Roots

Roots

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A polynomial equation of degree 2 is called Quadratic Equation. It’s standard form is

ax2 + bx + c = 0.

Roots of a Quadratic Equation

A quadratic equation has exactly 2 roots. These roots are either both Real or both Complex. It is not possible to have 1 real and 1 complex root for any quadratic equation with real coefficients. 

Click here to read about 1 real and 1 complex root scenario

Nature of Roots

To find the nature of roots, we need to find the discriminant of the quadratic equation. Formula to calculate the discriminant is:

\(\begin{aligned}&D = b^2-4ac\end{aligned}\)
D > 0 ⇒ Two Real and Distinct Roots

D = 0 ⇒ Two Real and Equal Roots

D < 0 ⇒ Two Complex Roots

It is always recommended to find the discriminant first rather than directly solving the equation.

Finding Roots

Once the discriminant is calculated, we can find the roots of the Quadratic Equation using the Quadratic Formula.

\(\begin{aligned}&x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\end{aligned}\)

An easy way to remember this formula is:

\(\begin{aligned}&x = \frac{-b \pm \sqrt{D}}{2a}\end{aligned}\)

Note: The roots of a quadratic equation are the zeros of equivalent quadratic polynomial. 

1 Real Root & 1 Complex Root Scenario

If we have a quadratic equation with imaginary or complex coefficients, then the equation can have 1 real and 1 complex root. For example, the roots of the equation mentioned below are: 3i and 1. 

\(\begin{aligned}&x^2-(3i + 1)x+(3i)=0\end{aligned}\)

An easy way to understand this is –

Form an equation like this:

\(\begin{aligned}&(x-3i) (x-1)=0\\\\&(x-\text{[A complex number]}) (x-\text{[A real number]})=0\end{aligned}\)

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]]> Class 10 | Chapter 4 | All Solutions | Quadratic Equations https://mathemafia.com/class-10-chapter-04-all-solutions-quadratic-equations/ Fri, 17 Dec 2021 07:37:36 +0000 http://sh016.global.temp.domains/~mathemaf/?p=7361 This playlist covers all the questions, concepts and examples of chapter 4 (Quadratic Equations) of class 10, NCERT book. Click a thumbnail to watch the tutorial. Watch complete chapter on YouTube

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This playlist covers all the questions, concepts and examples of chapter 4 (Quadratic Equations) of class 10, NCERT book. Click a thumbnail to watch the tutorial. 

Watch complete chapter on YouTube

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