# NCERT Solutions for class 10 Maths Chapter 4

NCERT Solutions for class 10 Maths Chapter 4 Quadratic Equations all exercises in Hindi Medium and English medium PDF format free to download. Uttar Pradesh Board, Prayagraj also implemented NCERT Books for High School Students for the academic session 2020-2021. UP Board students also take the benefits of these solutions and study material. Download here UP Board Solutions for Class 10 Maths Chapter 4 all exercises. 10th Maths Chapter 4 solutions are online as well as download in PDF form for sesion 2020-21. Offline Apps and NCERT Solutions 2020-21 are as per Latest CBSE syllabus for class 10 2020-2021 MP Board, UP board, Gujrat board and CBSE Board exams.

Download Class 10 Maths NCERT Solutions Offline Apps in Hindi or English for offline use. For any educational help, you may contact us without any hesitation. We will try to help you as soon as possible.## NCERT Solutions for class 10 Maths Chapter 4

Class: 10 | Maths (English and Hindi Medium) |

Chaper 4: | Quadratic Equations |

### 10th Maths Chapter 4 Solutions

NCERT Solutions for class 10 Maths Chapter 4 are given below in PDF form or view online. Solutions are in Hindi and English Medium. Uttar Pradesh students also can download UP Board Solutions for Class 10 Maths Chapter 4 here in Hindi Medium. It is very essential to learn quadratic equations, because it have wide applications in other branches of mathematics, physics, in other subjects and also in real life situations. Download NCERT books 2020-21, revision books and solutions from the links given below.

### 10th Maths Chapter 4 Exercise 4.1

### 10th Maths Chapter 4 Exercise 4.2

### 10th Maths Chapter 4 Exercise 4.3

### 10th Maths Chapter 4 Exercise 4.4

#### Class 10 Maths Exercise 4.1, 4.2 Solutions in Video

#### Class 10 Maths Exercise 4.3, 4.4 Solutions in Video

### NCERT Solutions for Class 10 Maths Chapter 4

- NCERT Solutions 10th Maths
- NCERT Solutions 10th Maths
- NCERT Solutions 10th Maths
- NCERT Solutions 10th Maths

##### How a quadratic polynomial is different from a quadratic equation?

A polynomial of degree two is called a quadratic polynomial. When a quadratic polynomial is equated to zero, it is called a quadratic equation. A quadratic equation of the form ax^2 bx + c = 0, a > 0, where a, b, c are constants and x is a variable is called a quadratic equation in the standard form.

##### What is meant by zeros of a quadratic equation?

A zero of a polynomial is that real number, which when substituted for the variable makes the value of the polynomial zero. In case of a quadratic equation, the value of the variable for which LHS and RHS of the equation become equal is called a root or solution of the quadratic equation. There are three algebraic methods for finding the solution of a quadratic equation. These are (i) Factor Method (ii) Completing the square method and (iii) Using the Quadratic Formula.

#### Previous Year’s CBSE Questions

- Two marks questions

Find the roots of the quadratic equation √2 x² + 7x + 5√2 = 0. [CBSE 2017] 2. Find the value of k for which the equation x²+k(2x + k – 1) + 2 = 0 has real and equal roots. [CBSE 2017] - Three marks questions

If the equation (1 + m² ) x² +2mcx + c² – a² = 0 has equal roots then show that c² = a² (1 + m² ). - . Four marks questions

Speed of a boat in still water is 15 km/h. It goes 30 km upstream and returns back at the same point in 4 hours 30 minutes. Find the speed of the stream. [CBSE 2017]

##### HISTORICAL FACTS!

The word quadratic is derived from the Latin word “Quadratum” which means “A square figure”.

Brahmagupta (an ancient Indian Mathematician)(A.D. 598-665) gave an explicit formula to solve a quadratic equation. Later Sridharacharya (A.D. 1025) derived a formula, now known as the quadratic formula, for solving a quadratic equation by the method of completing the square. An Arab mathematician Al-khwarizni(about A.D. 800) also studied quadratic equations of different types. It is believed that Babylonians were the first to solve quadratic equations. Greek mathematician Euclid developed a geometrical approach for finding lengths, which are nothing but solutions of quadratic equations.

### Important Questions on Class 10 Maths Chapter 4

x^2 – 3x – 10 = 0

⇒ x^2 – 5x + 2x + 10 = 0

⇒ x(x – 5) + 2(x – 5) = 0

⇒ (x – 5)(x + 2) = 0

⇒ (x – 5) = 0 या (x + 2) = 0

अर्थात x = 5 या x = -2

अतः दिए गए द्विघात समीकरण के मूल 5 और – 2 हैं।

Simplifying the given equation, we get

(x + 1)^2 = 2(x – 3)

⇒ x^2 + 2x + 1 = 2x – 6

⇒ x^2 + 7 = 0 or x^2 + 0x + 7 = 0

This is an equation of type ax^2 + bx + c = 0.

Hence, the given equation is a quadratic equation.

Therefore, the second integer = x + 1

Hence, the product = x(x + 1)

According to questions, x(x + 1) = 306

⇒ x^2 + x = 306

⇒ x^2 + x – 306 = 0

Hence, the two consecutive integers satisfies the quadratic equation

x^2 + x – 306 = 0.

इसलिए, दूसरी संख्या = 27 – x

प्रश्नानुसार, गुणनफल = x (27 – x) = 182

⇒ 27x – x^2 = 182

⇒ x^2 – 27x + 182 = 0

⇒ x^2 – 13x – 14x + 182 = 0

⇒ x(x – 13) -14(x – 13) = 0

⇒ (x – 13)(x – 14) = 0

⇒ (x – 13) = 0 या (x – 14) = 0

अर्थात x = 13 या x = 14

अतः 13 और 14 अभीष्ठ दो संख्याएँ हैं।

इसलिए, दूसरी संख्या = x + 1

प्रश्नानुसार,

वर्गों का योग = x^2 + (x + 1)^2 = 365

⇒ x^2 + x^2 + 2x + 1 = 365

⇒〖2x〗^2 + 2x – 364 = 0

⇒ x^2 + x – 182 = 0

⇒ x^2 – 13x + 14x + 182 = 0

⇒ x(x -13) + 14(x – 13) = 0

⇒ (x – 13)(x + 14) = 0

⇒ (x – 13) = 0 या (x + 14) = 0

अर्थात x = 13 या x = -14

अतः, 13 और 14 दो अभीष्ठ क्रमागत धनात्मक पूर्णांक हैं।

Therefore, Shefali’s marks in English = 30 – x

If she got 2 marks more in Mathematics and 3 marks less in English,

Marks in Mathematics = x + 2

Marks in English = 30 – x – 3

According to questions, Product

= (x + 2)(27 – x) = 210

⇒ 27x – x^2 + 54 – 2x = 210

⇒〖-x〗^2 + 25x – 156 = 0

⇒ x^2 – 25x + 156 = 0

⇒ x^2 – 12x – 13x + 156 = 0

⇒ x(x – 12) – 13(x – 12) = 0

⇒ (x – 12)(x – 13) = 0

⇒ (x – 12) = 0 or (x – 13) = 0

Either x = 12 or x = 13

If x = 12

then, marks in Maths = 12 and marks in English = 30 – 12 = 18

If x=13

then, marks in Maths = 13 and marks in English = 30 – 13 = 17

Let the smaller number = y

Therefore, y^2=8x

According to question, x^2 – y^2 = 180

⇒ x^2 – 8x = 180 [As y^2 = 8x]

⇒ x^2 – 8x – 180 = 0

⇒ x^2 – 18x + 10x – 180 = 0

⇒ x(x – 18) + 10(x – 18) = 0

⇒ (x – 18)(x + 10) = 0

⇒ (x – 18) = 0 or (x + 10) = 0

Either x = 18 or x = -10

But x ≠ -10 , as x is the larger of two numbers. So, x = 18

Therefore, the larger number = 18

Hence, the smaller number = y = √144 = 12

Let the side of smaller square = y m

According to question, x^2 + y^2 = 468 …(i)

Difference between perimeters, 4x – 4y = 24

⇒ x – y = 6

⇒ x = 6 + y … (ii)

Putting the value of x in equation (i), we get

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

⇒ y^2 + 12y + 36 + y^2 = 468

⇒〖2y〗^2 +12y – 432 = 0

⇒ y^2 + 6y – 216 = 0

⇒ y^2 + 18y – 12y – 216 = 0

⇒ y(y + 18) – 12(y + 18) = 0

⇒ (y + 18)(y – 12) = 0

⇒ (y + 18) = 0 or (y – 12) = 0

Either y = -18 or y = 12

But, y ≠ -18 , as x is the side of square, which can’t be negative.

So, y = 12

Hence, the side of smaller square = 12 m

Putting the value of y in equation (ii), we get

Side of larger square = x = y + 6 = 12 + 6 = 18 m

इसलिए, ऊँचाई = x – 7 cm

दिया है, कर्ण = 13 cm

पाइथागोरस प्रमेय से, x^2 + (x – 7)^2 =〖13〗^2

⇒ x^2 + x^2 – 14x + 49 = 169

⇒〖2x〗^2 – 14x – 120 = 0

⇒ x^2 – 7x – 60 = 0

⇒ x^2 – 12x + 5x – 60 = 0

⇒ x(x – 12) +5(x – 12) = 0

⇒(x – 12)(x + 5) = 0

⇒(x – 12) = 0 या (x + 5) = 0

अर्थात x = 12 या x = -5

लेकिन x ≠ -5 , क्योंकि x त्रिभुज की भुजा है।

इसलिए, x = 12 और दूसरी भुजा x-7 = 12 – 7 = 5

अतः, अन्य दो भुजाएँ 12 cm और 5 cm हैं।

इसलिए, बगिया की लंबाई = 2x m

इसलिए, क्षेत्रफल = x × 2x = 2x^2

प्रश्नानुसार, 2x^2 = 800

⇒x^2 = 400

⇒x= ±20

क्योंकि बगिया की चौड़ाई ऋणात्मक नहीं हो सकती,

अतः बगिया की चौड़ाई = 20 m

इसलिए, बगिया की लंबाई = 2×20 = 40 m