NCERT Solutions for Class 10 Maths Exercise 4.1 Chapter 4 Quadratic Equations – FREE PDF Download
NCERT Class 10 Maths Ch 4 is one of the most important ones in the NCERT syllabus. Duly following NCERT Solutions for Class 10 Maths Chapter
4 ensures you that there will be no hindrance when you opt for more advanced branches of Maths. This is where CoolGyan comes in. Our free Class 10 Quadratic Equations solutions will help you understand the chapter thoroughly.
NCERT Solutions for Class 10 Maths Chapter 4- Quadratic Equations
1. Check whether the following are Quadratic Equations.
(i) (x+1)2= 2 (x − 3)
(ii) x2-2x= (−2) (3 − x)
(iii) (x − 2) (x + 1) = (x − 1) (x + 3)
(iv) (x − 3) (2x + 1) = x (x + 5)
(v) (2x − 1) (x − 3) = (x + 5) (x − 1)
(vi) x2 + 3x +1 = (x-2)2
(vii) (x+2)3 =2x(x2-1)
(viii) x3-4x2-x+1 = (x-2)3
⇒ 
= 2x – 6
⇒ 
= 0
Here, degree of equation is 2.
Therefore, it is a Quadratic Equation.
(ii) 
= (−2) (3 − x)
⇒ = −6 + 2x
⇒ − 2x + 6 = 0
⇒ 
= 0
Here, degree of equation is 2.
Therefore, it is a Quadratic Equation.
(iii) (x − 2) (x + 1) = (x − 1) (x + 3)
⇒ 
= 
⇒ 
⇒ x − 2x – 2 − 3x + x + 3 = 0
⇒ −3x + 1 = 0
Here, degree of equation is 1.
Therefore, it is not a Quadratic Equation.
(iv) (x − 3) (2x + 1) = x (x + 5)
⇒ 
= 
⇒ 
⇒ 
= 0
Here, degree of equation is 2.
Therefore, it is a quadratic equation.
(v) (2x − 1) (x − 3) = (x + 5) (x − 1)
⇒ 
= 
⇒ 
⇒ x2 − 11x + 8 = 0
Here, degree of Equation is 2.
Therefore, it is a Quadratic Equation.
(vi)
⇒ 
= 
⇒ 
⇒ 7x – 3 = 0
Here, degree of equation is 1.
Therefore, it is not a Quadratic Equation.
(vii)
⇒ 
= 
⇒ 
= 
⇒ 
⇒ 
Here, degree of Equation is 3.
Therefore, it is not a quadratic Equation.
(viii)
⇒ 
= 
(x) (2) (x − 2)
⇒ 
= 
⇒ 
= 0
Here, degree of Equation is 2.
Therefore, it is a Quadratic Equation.
2. Represent the following situations in the form of Quadratic Equations:
(i) The area of rectangular plot is
. 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.
(ii) The product of two consecutive numbers is 306. We need to find the integers.
(iii) Rohan’s mother is 26 years older than him. The product of their ages (in years) after 3 years will be 360. We would like to find Rohan’s present age.
(iv) A train travels a distance of 480 km at uniform speed. If, the speed had been 8km/h less, then it would have taken 3 hours more to cover the same distance. We need to find speed of the train.
. Let breadth of rectangular plot be x metres
Length is one more than twice its breadth.
Therefore, length of rectangular plot is (2x + 1) metres
Area of rectangle = length × breadth
⇒ 528 = x (2x + 1)
⇒ 528 = 
⇒– 528 = 0
This is required Quadratic Equation.
(ii) Let two consecutive numbers be x and (x + 1).
It is given that x (x + 1) = 306
⇒ 
= 306
⇒– 306 = 0
This is the required Quadratic Equation.
(iii) Let present age of Rohan = x years
Let present age of Rohan’s mother = (x + 26) years
Age of Rohan after 3 years = (x + 3) years
Age of Rohan’s mother after 3 years = x + 26 + 3 = (x + 29) years
According to given condition:
(x + 3) (x + 29) = 360
⇒ 
= 360
⇒ 
= 0
This is the required Quadratic Equation.
(iv) Let speed of train be x km/h
Time taken by train to cover 480 km = 480x480xhours
If, speed had been 8km/h less then time taken would be 480x−8480x−8 hours
According to given condition, if speed had been 8km/h less then time taken is 3 hours less.
Therefore, 480x−8−480x=3480x−8−480x=3
⇒ 480 (x – x + 8) = 3x(x−8)3x(x−8)
⇒ 3840=3x2−24x3840=3×2−24x
⇒ 
= 0
Dividing equation by 3, we get
⇒ 
= 0
This is the required Quadratic Equation.