These NCERT Solutions for Class 10 Maths Chapter Ex 4.3 Questions and Answers are prepared by our highly skilled subject experts.
NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Exercise 4.3
Question 1.
Find the roots of the following quadratic equations, if they exist, by the method of completing the square:
(i) 2x2 – 7x + 3 = 0
(ii) 2x2 + x – 4 = 0
(iii) 4x2 + 4√3x + 3 = 0
(iv) 2x2 + x + 4 = 0
Solution:
(i) We have,
Adding and subtracting (\(\frac { 7 }{ 4 }\))² in equation (i), we get,
(ii) We have,
Adding and subtracting (\(\frac { 1 }{ 4 }\))² in equation (i), we get,
(iii) We have,
Adding and subtracting \(\left(\frac{\sqrt{3}}{2}\right)^{2}\) in equation (i), we get,
(iv) We have,
Adding and subtracting (\(\frac { 1 }{ 4 }\))² in equation (i), we get,
So, roots do not exist.
Question 2.
Find the roots of the quadratic equations by applying the quadratic formula.
(i) 2x2 – 7x + 3 = 0
(ii) 2x2 + x – 4 = 0
(iii) 4x2 + 4√3x + 3 = 0
(iv) 2x2 + x + 4 = 0
Solution:
(i) We have,
2x2 – 7x + 3 = 0
Here a = 2, b = – 7, and c = 3
∴ By using quadratic formula, we get,
(ii) We have,
2x2 + x – 4 = 0
Here a = 2, b = 1, and c = – 4
∴ By using quadratic formula, we get,
(iii) We have,
4x2 + 4√3x + 3 = 0
Here a = 4, b = 4√3 and c = 3
∴ By using quadratic formula, we get,
(iv) We have,
2x2 + x + 4 = 0
Here a = 2, b = 1, and c = 4
∴ By using quadratic formula, we get,
Here, b² – 4ac < 0
So, the given equation has no any real roots.
Question 3.
Find the roots of the following equations:
(i) x – \(\frac { 1 }{ x }\) = 3 ; x ≠ 0
(ii) \(\frac { 1 }{ x+4 }\) – \(\frac { 1 }{ x-7 }\); x ≠ – 4, 7
Solution:
Question 4.
The sum of the reciprocals of Rehman’s ages, (in years) 3 years ago and 5 years from now is \(\frac { 1 }{ 3 }\) Find his present age.
Solution:
Let the present age of Rehman be x years
3 years ago Rehman’s age was = (x – 3) years
5 years from now Rehman’s age will be = (x + 5) years
∴ According to question
Using the quadratic formula
Since, age cannot be in negative,
so we ignore the value of x = – 3
Therefore, x = 7
So, present age of Rehman = x = 7 years.
Question 5.
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.
Solution:
Let the marks secured by Shefali in Mathematics = x Then,
According to question,
(x + 2) x (30 – x – 3) = 210
or (x + 2) x (27 – x) = 210
or 27x – x² + 54 – 2x – 210
or 25x – x² + 54 = 210
or x² – 25x – 54 + 210 = 0
or x² – 25x + 156 = 0
Here, a = 1, b = – 25 and c = 156
Using the quadratic formula
Therefore, if Shefali’s marks in Mathematics = x = 13
Then, Shefail’s marks in English = 30 – x = 30 = 17
and if Shefail’s marks in Mathematics = x = 12
Then, Shefail’s marks in Mathematics = 30 – x = 30 – 12 = 18
Question 6.
The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 metres more than the shorter side, find the sides of the field.
Solution:
Let ABCD be a rectangular field.
Let AB, its shorter side, be x metres And BC, the longer side be (x + 30) metres
Diagonal AC of the rectangular field be x + 60 metres.
By Pythagoras theorem, we have,
(AC)² = (AB)² + (BC)²
∴ (x – 60)² = x² + (x + 30)²
or x² + 120x + 3600 = x² + x² + 60x + 900
or x² + 120x + 3600 = 2x² + 60x + 900
2x² + 60x + 900 – x² – 120x – 3600 = 0
or x² – 60x – 2700 = 0
Here, a = 1, b = – 60 and c = – 2700
Using the quadratic formula, we get
Since, dimension cannot be in negative,
so we ignore the value of x = – 30
Therefore, x = 90
∴ Length of shorter side of rectangular field = x = 90 metres
and length of longer side of rectangular = x + 30 = 90 + 30 = 120 metres.
Question 7.
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.
Solution:
Let x be the larger number.
So, the square of the smaller number = 8x According to the second condition, the square of the larger number – the square of the smaller number = 180
x² – 8x = 180
⇒ x² – 8x – 180 = 0
⇒ x² – 18x + 10x – 180 = 0
⇒ x (x – 18) + 10 (x – 18) = 0
⇒ (x – 18) (x + 10) = 0
⇒ x = 18 or x = – 10
x = – 10 does not exist, so it is neglected.
So, x = 18 (the larger number)
Smaller number = \(\sqrt{8x}\)
⇒ \(\sqrt{8×18}\) = \(\sqrt{144}\) = ± 12
∴ The smaller and larger numbers are ±12and 18 respectively.
Question 8.
A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.
Solution:
Total distance travelled = 360 km
Let uniform speed be x km/h
Then, increased speed = (x + 5) km/h
According to question,
Since, speed cannot be in negative, so we reject the value of x = – 45
∴ x = 40
So, the usual speed of train = x km/h = 40 km/h.
Question 9.
Two water taps together can fill a tank in 9\(\frac { 3 }{ 8 }\) hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.
Solution:
Let larger pipe fills the tank in x hours and the smaller pipe fills the tank in y hours.
The tank filled by the larger pipe in 1 hour = \(\frac { 1 }{ x }\)
and part of tank filled by larger tap in 1 hour = \(\frac { 1 }{ x-10 }\)
Total = 9\(\frac { 3 }{ 8 }\) hours = \(\frac { 75 }{ 8 }\)
∴ According to questions,
Using the quadratic formula
So, if smaller tap take 3.75 hour to fill the tank then time taken by larger tap = 3.75 = 10 = – 6.25 hour
But we know that time cannot be in negative The value of x = 3.75 is neglected.
Therefore, time taken by the smaller tap to fill the tank = x = 25 hours.
and time taken by the larger tap to fill the bank = x – 10 = 25 – 10 = 15 hours.
Question 10.
An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bengaluru (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 11 km/h more than that of the passenger train, find the average speed of the two trains.
Solution:
Let the average speed of passenger train be x km/h
Then, average speed of express train = x + 11 km/h
Total distance from Mysore to Bangalore is 132 km.
Therefore, time taken by the passenger train to cover the distance of 132 km = \(\frac { 132 }{ x }\)h
According to question,
Using the quadratic formula
But we know that speed of train cannot be in negative.
∴ The value of x = – 44 is neglected.
∴ x = 33
Therefore, the speed of passenger train = x = 33 km/h
and the speed of express train = x + 11 = 33+ 11 = 44 km/h.
Question 11.
Sum of the areas of two squares is 468 m². If the difference of their perimeters is 24 m, find the sides of the two squares.
Solution:
Let the sides of smaller square = x
The sides of larger square = \(\sqrt{468-x^{2}}\)
[∴ Area of square = (side)²]
Now, perimeter of smaller square = 4x
and perimeter of larger square = 4 x \(\sqrt{468-x^{2}}\)
According to question,
4 x \(\sqrt{468-x^{2}}\) = 24
or 4 x \(\sqrt{468-x^{2}}\) = 24 + 4x
or \(\sqrt{468-x^{2}}\) = \(\frac{4(6+x)}{4}\)
or \(\sqrt{468-x^{2}}\) = 6 + x
Squaring both sides, we get,
468 – x² = (6 + x²)
or 468 – x² = 36 + x² + 12x
[∴ (a + b)² = a² + b² – 2ab]
or 36 + x² + 12x + x² – 468 = 0
or 2x² + 12x – 432 = 0
or x² + 6x – 216 = 0
Here, a = 1, b = 6 and c = – 216
Using the quadratic fomula,
But we know that length of the side of square cannot be in negative.
So, the value of x = – 18 is neglected. Therefore, x = 12
So, length of the sides of smaller square = x = 12 m
and length of the sides of larger square = \(\sqrt{468-x^{2}}\)
= \(\sqrt{468-(12)^{2}}\)
= \(\sqrt{468-144}\)
= \(\sqrt{324}\)
= 18 m