# NCERT Solutions for Class 10 Maths Chapter 4

Free download NCERT Solutions for class 10 Maths Chapter 4 Quadratic Equations exercise 4.4, 4.3, 4.2 and 4.1 in Hindi & English medium PDF form. Solutions are as per Latest CBSE syllabus for class 10 2019-2020 MP Board, UP board, Gujrat board & CBSE Board exams. Download Class 10 NCERT Solutions Offline Apps for offline use.

 Class 10: Maths – गणित Chapter 4: Quadratic Equations

## NCERT Solutions for class 10 Maths Chapter 4

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. Buy NCERT books or download NCERT books, revision books and solutions from the following links.

### Class 10 Maths Chapter 4 Solutions – English & Hindi Medium

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.

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### Previous Years CBSE Questions

• Two marks questions
1. Find the roots of the quadratic equation √2 x^2+7x+5√2=0. [CBSE 2017] 2. Find the value of k for which the equation x^2+k(2x+k-1)+2=0 has real and equal roots. [CBSE 2017]
• Three marks questions
1. If the equation (1+m^2 ) x^2+2mcx+c^2-a^2=0 has equal roots then show that c^2=a^2 (1+m^2 ).
• Four marks questions
1. 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]

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.

#### (i) Factor Method

Factorisation method is used when the quadratic equation can be factorised into two linear factors. After factorisation, the quadratic equation is expressed as the product of its two linear factors and this is equated to zero. See the following example to learn the step by step method.

#### (iii) Using the Quadratic Formula

Download assignments with answers. One marks questions assignments contains very short answers question based on logical concepts of the chapter. Two marks questions assignments contains short answers questions based on solution of equations and finding the value of k or other missing constants. Three marks questions assignments contains most of the proving statements between the coefficients of the variable. There are four marks questions contains word problems, like distance – time, age problems, perimeter – area questions, upstream – downstream, work – time etc. First four marks questions assignment includes proving questions, second assignment and third assignments contains word problems. Answers of the assignments age given in the same sheet.

##### 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.

Download Class 10 Maths App for offline use as well as कक्षा 10 गणित App for offline use.

#### गुणनखंड विधि से निम्न द्विघात समीकरण के मूल ज्ञात कीजिए: x^2 – 3x – 10 = 0.

द्विघात समीकरण को सरल करने पर
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।

#### Check whether the following is quadratic equation: (x + 1)^2 = 2(x – 3)

(x + 1)^2 = 2(x – 3)
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.

#### Represent the following situation in the form of quadratic equation: The product of two consecutive positive integers is 306. We need to find the integers.

Let the first integer = x
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.

#### ऐसी दो संख्याएँ ज्ञात कीजिए, जिनका योग 27 हो और गुणनफल 182 हो।

माना पहली संख्या = x
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।

#### दो क्रमागत धनात्मक पूर्णांक ज्ञात कीजिए जिनके वर्गों का योग 365 हो।

माना पहली संख्या = x,
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।

#### 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.

Let, Shefali’s marks in Mathematics = x
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

#### 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.

Let the larger number = x
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

#### 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.

Let the side of larger square = x m
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

#### एक समकोण त्रिभुज की ऊँचाई इसके आधार से 7 cm कम है। यदि कर्ण 13 cm का हो, तो अन्य दो भुजाएँ ज्ञात कीजिए।

माना आधार = x cm
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।

#### क्या एक ऐसी आम की बगिया बनाना संभव है जिसकी लंबाई, चौड़ाई से दुगुनी हो और उसका क्षेत्रफल 800 m^2 हो? यदि है, तो उसकी लंबाई और चौड़ाई ज्ञात कीजिए।

माना बगिया की चौड़ाई = x m
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

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