Chapter 4 of the Class 10 NCERT Mathematics textbook covers Quadratic Equations a fundamental topic in algebra. This chapter helps students understand how to solve quadratic equations using various methods and apply these solutions to real-world problems. Exercise 4.1 focuses on solving basic quadratic equations reinforcing the concepts introduced in the chapter.
What is a Quadratic Equation?
A quadratic equation is a polynomial equation of degree 2 which can be written in the standard form:
ax2+bx+c=0
where a, b, and c are constants, and πβ 0. The solutions to the quadratic equation are the values of the x that satisfy the equation. These solutions can be found using various methods including factoring completing the square and using the quadratic formula:
Question 1. Check whether the following are quadratic equations :
(i) (x + 1)2 = 2(x β 3)
Solution:
Here,
LHS = (x + 1)2
= x2 + 2x + 1 (Using identity (a+b)2 = a2 + 2ab + b2)
and, RHS = 2(xβ3)
= 2x β 6
As, LHS = RHS
x2 + 2x + 1 = 2x β 6
x2 + 7 = 0 .............(I)
As, the eqn. (I) is in the form of ax2 + bx + c = 0 where (aβ 0).
Hence, the equation is QUADRATIC equation because highest power of x is 2.
(ii) x2 β 2x = (β2) (3 β x)
Solution:
Here,
LHS = x2 β 2x
and, RHS = (β2) (3 β x)
= 2xβ6
As, LHS = RHS
x2 β 2x = 2x β 6
x2 β 4x + 6 = 0 .............(I)
As, the eqn. (I) is in the form of ax2 + bx + c = 0 where (aβ 0).
Hence, the equation is QUADRATIC equation because highest power of x is 2.
(iii) (x β 2)(x + 1) = (x β 1)(x + 3)
Solution:
Here,
LHS = (x β 2)(x + 1)
= x2 + (β2+1)x + (β2)(1) (Using identity (x+a) (x+b) = x2 + (a+b)x + ab)
= x2 - x - 2
and, RHS = (x β 1)(x + 3)
= x2 + (β1+3)x + (β1)(3) (Using identity (x+a) (x+b) = x2 + (a+b)x + ab)
= x2 + 2x - 3
As, LHS = RHS
x2 β x β 2 = x2 + 2x - 3
3x β 1 = 0 .............(I)
As, the eqn. (I) is not in the form of ax2 + bx + c = 0 because (a=0).
Hence, the equation is NOT QUADRATIC equation because highest power of x is 1.
(iv) (x β 3)(2x +1) = x(x + 5)
Solution:
Here,
LHS = (x β 3)(2x +1)
= 2x2 + x +(β3)(2x) + (β3)(1)
= 2x2 β 5x - 3
and, RHS = x(x + 5)
= x2 + 5x
As, LHS = RHS
2x2 β 5x - 3 = x2 + 5x
x2 β 10x - 3 = 0 .............(I)
As, the eqn. (I) is in the form of ax2 + bx + c = 0 where (aβ 0).
Hence, the equation is QUADRATIC equation because highest power of x is 2.
(v) (2x β 1)(x β 3) = (x + 5)(x β 1)
Solution:
Here,
LHS = (2x β 1)(x β 3)
= 2x2 + (2x)(β3) +(β1)(x) + (β1)(β3)
= 2x2 β 7x + 3
and, RHS = (x + 5)(x β 1)
= x2 + 5(x) + (β1)(x) + (5)(β1) (Using identity (x+a) (x+b) = x2 + (a+b)x + ab)
= x2 + 4(x) β 5
As, LHS = RHS
2x2 β 7x + 3 = x2 + 4(x) β 5
x2 β 11x + 8 = 0 .............(I)
As, the eqn. (I) is in the form of ax2 + bx + c = 0 where (aβ 0).
Hence, the equation is QUADRATIC equation because highest power of x is 2.
(vi) x2 + 3x + 1 = (x β 2)2
Solution:
Here,
LHS = x2 + 3x + 1
and, RHS = (x β 2)2
= x2 β 4x + 4 (Using identity (aβb)2 = a2 β 2ab + b2)
As, LHS = RHS
x2 + 3x + 1 = x2 β 4x + 4
7x β 3 = 0 .............(I)
As, the eqn. (I) is not in the form of ax2 + bx + c = 0 because (a=0).
Hence, the equation is NOT QUADRATIC equation because highest power of x is 1.
(vii) (x + 2)3 = 2x (x2 β 1)
Solution:
Here,
LHS = (x + 2)3
= x3 + 23 + 3x(2)(x+2) (Using identity (x+y)3 = x3 + y3 + 3xy(x+y))
= x3 + 8 + 6x2 +12x
and, RHS = 2x (x2 β 1)
= 2x3 β 2x
As, LHS = RHS
x3 + 8 + 6x2 +12x = 2x3 β 2x
x3 β 6x2 β 14x β 8 = 0 .............(I)
As, the eqn. (I) is not in the form of ax2 + bx + c = 0 where (aβ 0).
Hence, the equation is NOT QUADRATIC equation because highest power of x is 3.
(viii) x3 β 4x2 β x + 1 = (x β 2)3
Solution:
Here,
LHS = x3 β 4x2 β x + 1
and, RHS = (x β 2)3
= x3 β 23 β 3x(2)(xβ2) (Using identity (xβy)3 = x3 β y3 β 3xy(x-y))
= x3 β 8 β 6x2 +12x
As, LHS = RHS
x3 β 4x2 β x + 1 = x3 β 8 β 6x2 +12x
2x2 β 13x + 9 = 0 .............(I)
As, the eqn. (I) is in the form of ax2 + bx + c = 0 where (aβ 0).
Hence, the equation is QUADRATIC equation because highest power of x is 2.
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 meters) is one more than twice its breadth. We need to find the length and breadth of the plot.
Solution:
Let's consider,
Breadth of the rectangular plot = b m
Then, length of the plot = (2b + 1) m.
As, Area of rectangle = length Γ breadth
528 m2 = (2x + 1) Γ x
2x2 + x =528
2x2 + x β 528 = 0 ...................(I)
As, the eqn. (I) is in the form of ax2 + bx + c = 0 where (aβ 0).
Hence, the equation is QUADRATIC equation because highest power of x is 2.
Therefore, the length and breadth of plot, satisfies the quadratic equation, 2x2 + x β 528 = 0, which is the required representation of the problem mathematically.
(ii) The product of two consecutive positive integers is 306. We need to find the integers.
Solution:
Let's consider,
The first integer number = x
Next consecutive positive integer will be = x + 1
Product of two consecutive integers = x Γ (x +1)
x2 + x = 306
x2 + x β 306 = 0 ...................(I)
As, the eqn. (I) is in the form of ax2 + bx + c = 0 where (aβ 0).
Hence, the equation is QUADRATIC equation because highest power of x is 2.
Therefore, the two integers x and x+1, satisfies the quadratic equation, x2 + x β 306 = 0, which is the required representation of the problem mathematically.
(iii) Rohanβs 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 Rohanβs present age.
Solution:
Let's consider,
Age of Rohanβs = x years
So, Rohanβs motherβs age = x + 26
After 3 years,
Age of Rohanβs = x + 3
Age of Rohanβs mother will be = x + 26 + 3 = x + 29
The product of their ages after 3 years will be equal to 360, such that
(x + 3)(x + 29) = 360
x2 + 29x + 3x + 87 = 360
x2 + 32x + 87 β 360 = 0
x2 + 32x β 273 = 0 ...................(I)
As, the eqn. (I) is in the form of ax2 + bx + c = 0 where (aβ 0).
Hence, the equation is QUADRATIC equation because highest power of x is 2.
Therefore, the age of Rohan and his mother, satisfies the quadratic equation, x2 + 32x β 273 = 0, which is the required representation of the problem mathematically.
(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.
Solution:
Let's consider,
The speed of train = x km/hr
Time taken to travel 480 km = (480/x) hr
Time = Distance / Speed
Here, According to the given condition,
The speed of train = (x β 8) km/hr
As, the train will take 3 hours more to cover the same distance.
Hence, Time taken to travel 480 km = 480/(x+3) km/hr
As,
Speed Γ Time = Distance
(x β 8)(480/(x + 3) = 480
480 + 3x β 3840/x β 24 = 480
3x β 3840/x = 24
3x2 β 8x β 1280 = 0 ...................(I)
As, the eqn. (I) is in the form of ax2 + bx + c = 0 where (aβ 0).
Hence, the equation is QUADRATIC equation because highest power of x is 2.
Therefore, the speed of the train, satisfies the quadratic equation, 3x2 β 8x β 1280 = 0, which is the required representation of the problem mathematically.
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Summary
Exercise 4.1 focuses on identifying quadratic equations and finding their solutions. It covers the standard form of quadratic equations (axΒ² + bx + c = 0, where a β 0), distinguishing between quadratic and non-quadratic equations, and solving simple quadratic equations by factorization. Students learn to recognize the coefficients a, b, and c in quadratic equations and practice solving equations where one solution is provided to find the other solution.