NCERT Solutions for Class 10 Maths Chapter 4 Exercise 4.1
NCERT Solutions for Class 10 Maths Chapter 4 - Quadratic Equations Exercise 4.1
In this page we have NCERT Solutions for Class 10th Maths: Chapter 4 - Quadratic Equations for
EXERCISE 4.1 on page number 73 and 74. Hope you like them and do not forget to like , social_share
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Question 1
Check whether the following are quadratic equations
(x+1)^{2}=2(x-3)
x^{2} -2x=(-2)(3-x)
(x-2)(x+1)=(x-1)(x+3)
(x – 3)(2x +1) = x(x + 5)
(2x – 1)(x – 3) = (x + 5)(x – 1)
x^{2} + 3x + 1 = (x – 2)^{2}
(x + 2)^{3} = 2x (x^{2} – 1)
x^{3} – 4x^{2} – x + 1 = (x – 2)^{3}
Solution 1
we know that Quadratic equation
ax^{2} +bx+c =0 where a≠0
(x+1)^{2}=2(x-3)
We know that
(a+b)^{2}=a^{2}+b^{2}+2ab
⇒x^{2} + 2x+1=2x-6
Simplifying it
⇒;x^{2} +7=0
Since it is of a quadratic form : ax^{2} +bx+c =0 where a≠0
with b=0
So it is a quadratic equation
x^{2} -2x=(-2)(3-x)
Simplifying it
x^{2} -2x=-6+2x
⇒x^{2} -4x+6=0
Since it is a quadratic form
ax^{2} +bx+c =0 where a≠0
So it is a quadratic equation
x^{2} -2x=(-2)(3-x)
(x-2)(x+1)=(x-1)(x+3)
Multiplying both the factors
⇒x^{2} -2x+2+x= x^{2} +3x-x-3
Simplifying
-3x+1=0
It is not of the quadratic form
ax^{2} +bx+c =0 where a≠0
So it is not a quadratic equation
(x – 3)(2x +1) = x(x + 5)
Multiplying both the factors
2x^{2}+x-6x-3=x^{2}+5x
Simplifying
x^{2} -10x-3=0
Since it is a quadratic form
ax^{2} +bx+c =0 where a≠0
⇒So it is a quadratic equation
(2x – 1)(x – 3) = (x + 5)(x – 1)
Multiplying both the factors on both sides
2x^{2} -6x-x+3=x^{2} –x+5x-5
⇒x^{2} -11x +8=0
Since it is a quadratic form
ax^{2} +bx+c =0 where a≠0
So it is a quadratic equation
x^{2} + 3x + 1 = (x – 2)^{2}
We know that
(a+b)^{2}=a^{2}+b^{2}+2ab
⇒x^{2} + 3x + 1 =x^{2}-4x+4
⇒7x-3=0
Since it is not of quadratic form
ax^{2} +bx+c =0 where a≠0
So it is a not quadratic equation
(x + 2)^{3} = 2x (x^{2} – 1)
Important formula you must have remembered in old classes
(a+b)^{3}= a^{3} +b^{3}+3ab^{2}+3a^{2}b
⇒x^{3} +8+6x^{2}+12x=2x^{3} -2x
Simplifying
x^{3}-6x^{2} -14x-8=0
Since it is not of quadratic form
ax^{2} +bx+c =0 where a≠0
So it is a not quadratic equation
x^{3} – 4x^{2} – x + 1 = (x – 2)^{3}
Important formula you must have remembered in old classes
(a-b)^{3}= a^{3} -b^{3}+3ab^{2}-3a^{2}b
x^{3} – 4x^{2} – x + 1 =x^{3}-8-6x^{2}+12x
Simplifying
2x^{2} -13x+9=0
Since it is a quadratic form
ax^{2} +bx+c =0 where a≠0
So it is a quadratic equation
Question 2
Represent the following situations in the form of quadratic equations :
The area of a rectangular plot is 528 m^{2}. 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.
The product of two consecutive positive integers is 306. We need to find the Integers.
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.
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 the breath of the plot= x m
As per given condition in the question
Length =2x+1
Now we know that Area is given by
A=LB
A=528 m^{2}
So
528=(2x+1)x
⇒ x^{2}+2x-528=0
Which is a quadratic equation
let the two consecutive positive integers are x and x+1
The product of these would be
x(x+1)
It is given that product is 306
So
⇒ x(x+1)=306
⇒ x^{2}+x-306=0
⇒ Which is a quadratic equation
Let Rohan present age=x year
Then Rohan Mother present age would =x+26
After 3 year,
Rohan age would be =x+3
Rohan mother’s age would be =x+26+3=x+29
According to question, The product of their ages (in years) =360
Then
(x+3)(x+29)= 360
Simplifying
⇒ x^{2} +29x +3x+87=360
⇒ x^{2} +32x -273=0
Which is a quadratic equation
Let the speed of the train is x km/hr
Now distance travelled by the train=480 km
Few important formula here
Speed=Distance/time
Or Time= Distance /Speed Case I
Time taken to travel 480 km by train will be =480/x Case II
Now the speed of the train is reduced by 8 km/hr,
So speed would (x-8)
Now Time taken to travel 480 km will be =480/x-8
Now as per the question
480/(x-8) - 480/x =3
⇒ [480x-480(x-8)]/x(x-8) =3
⇒ 480x-480x+3840=3x(x-8)
⇒ 3x^{2} -24x-3840=0
⇒ Which is a quadratic equation
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