RBSE Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.3

Rajasthan Board RBSE Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.3 Textbook Exercise Questions and Answers.

RBSE Class 10 Maths Solutions 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\(\sqrt {3}\) x + 3 = 0
(iv) 2x² + x + 4 = 0
Solution:
(i) According to the question the quadratic equation is
2x² – 7x + 3 = 0
or 2x² – 7x = – 3
or 2(x² – \(\frac{7}{2}\)x) = -3
or x² – \(\frac{7}{2}\)x = \(\frac{-3}{2}\)
(since to make a perfect square the coefficient of x² should be unity)
To make a perfect square we add the square of half of the coefficient of x in both

Case II-Taking negative sign

x = \(\frac{2}{4}\) = \(\frac{1}{2}\)
Hence, the roots of the given quadratic equation are 3 and \(\frac{1}{2}\).

(ii) According to the question the quadratic equation is
2x² + x – 4 = 0
or 2x² + x = 4
or x² + \(\frac{1}{2}\)x = \(\frac{4}{2}\)

Case I. Taking positive sign





∵ The square of any real number cannot be negative, therefore, \(\left(x+\frac{1}{4}\right)^{2}\), cannot be negative for any real value of x.
∴ There exists no real value of x which satisfies the given quadratic equation. Hence the given equation has no real roots.
Hence the roots of the equation do not exist.

Question 2.
Find the roots of the quadratic equations given in Q.1 above by applying the quadratic formula.
Solution:
(i) According to the question the quadratic equation is
2x² – 7x + 3 = 0
Comparing it with ax² + bx + c = 0
a = 2, b = – 7, c = 3
Now
b² – 4ac = (-7)² – 4 × 2 × 3
= 49 – 24
= 25 > 0
Using quadratic formula

Hence, 3 and \(\frac{1}{2}\) are the roots of the given quadratic equation.

(ii) According to the question the quadratic equation is
2x² + x – 4 = 0
comparing it with ax² + bx + c = 0
a = 1, b = l, c = – 4 Now,
b² – 4ac = (1)² – 4 × 2 × (- 4)
= 1 + 32
= 33 > 0
Using quadratic formula

Hence, \(\frac{-1+\sqrt{33}}{4}\) and \(\frac{-1-\sqrt{33}}{4}\) are the roots of the given quadratic equation.

(iii) According to the question the quadratic equation is
4x² + 4\(\sqrt {3}\)x + 3 = 0
Comparing it with ax² + bx + c = 0,
a = 4, b = 4\(\sqrt {3}\) , c = 3
Now, b² – 4ac = (4\(\sqrt {3}\))² – 4 × 4 × 3
= 48 – 48 = 0
Using quadratic formula

Hence, \(-\frac{\sqrt{3}}{2}\) and \(\frac{\sqrt{3}}{2}\) are the roots of the given quadratic equation.

(iv) According to the question the quadratic equation is
2x² + x + 4 = 0
Comparing it with ax² + bx + c = 0,
a = 2, b = 1, c = 4
Now, b² – 4ac = (1)² – 4 × 2 × 4
= 1 – 32 = – 31 < 0
∴ x = \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)
Since the square of a real number cannot be negative, therefore there will be no real value of x.
Hence the given equation has no real roots.
Hence the roots of the equation do not exist.

Question 3.
Find the roots of the following equations :
(i) x – \(\frac{1}{x}\) = , x ≠ 0
(ii) \(\frac{1}{x+4}\) – \(\frac{1}{x-7}\) = \(\frac{11}{30}\), x ≠ -4, 7
Solution:
(i) According to the question
x – \(\frac{1}{x}\) = , x ≠ 0
or \(\frac{x^{2}-1}{x}\) = 3
or x² – 1 = 3x
or x² – 3x – 1 = 0
Comparing it with ax² + bx + c = 0
a = 1, b = – 3, c = – 1
Now, b² – 4ac = (-3)² – 4 . 1 . (- 1)
= 9 + 4= 13 > 0

(ii) According to the question

or -11 × 30 = 11(x² – 3x – 28)
or -30 = x² – 3x – 28
or x² – 3x – 28 + 30 = 0
or x² – 3x + 2 = 0
Comparing it with ax² + bx + c = 0
a = 1, b = -3, c = 2
Now, b² – 4ac = (-3)² – 4 × 1 × 2
= 9 – 8
= 1 > 0
Using quadratic formula

Hence, 2 and 1 are the roots of the given quadratic equation.

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 = x years
Then, Age of Rehman 3 years ago = (x – 3) years
Age of Rehman 5 years from now = (x + 5) years
According to the question

or 6x + 6 = x² + 2x – 15
or x² + 2x – 15 – 6x – 6 = 0
or x² – 4x – 21 =0, which is quadratic in A.
Therefore comparing it with ax² + bx + c = 0,
a = 1, b = – 4, c = – 21
Now, b² – 4ac = (-4)² – 4 × 1 × (-21)
= 16 + 84
= 100 > 0

∵ Age cannot be negative therefore leaving
x = – 3,
∴ x = 7
Hence, the present age of Rehman = 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 obtained by Shefali in Mathematics = x
∴ Marks of Shefali in English = 30 – x
According to the first condition
Marks of Shefali in Mathematics = x + 2
and Marks of Shefali in English= 30 – x – 3
= 27 – A
∴ Their product = (x + 2) (27 – x)
= 27x – x² + 54 – 2x
= -x² + 25x + 54
According to the second condition of the problem
-x² + 25x + 54 = 210
or -x² + 25x + 54 – 210 = 0
or -x² + 25x – 156 = 0
or x² – 25x + 156 = 0
Comparing it with ax² + bx + c = 0,
a = 1, b = – 25, c = 156
Now, b² – 4ac = (-25)² – 4 × 1 × 156
= 625 – 624
= 1 > 0

= 13 and 12
Case I.
When x = 13
Then, Marks of Shefali in Mathematics = 13
Marks of Shefali in English = 30 – 13 = 17
Case II.
When x = 12
Then Maiks of Shefali in Mathematics = 12 Marks of Shefali in English = 30 – 12 = 18
Hence, the marks of Shefali in the two subjects are : 13 and 17 or 12 and 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 the shorter side of the rectangular field = AD = x m
Then, longer side of the rectangular field = AB = (x + 30) m
and diagonal of the rectangular field = DB = (x + 60) m
In a rectangle the angle between the length and the breadh is a right angle.
∴ ∠DAB = 90°
Now in right triangle DAB,

By Pythagoras Theorem,
(DB)² = (AD)² + (AB)²
or (x + 60)² – (x)² + (x + 30)²
or x² + 3600 + 120x = x² + x² + 900 + 60x
or x² + 3600 + 120x – 2x – 900 – 60x = 0
or -x² + 60x + 2700 = 0
or x² – 60x – 2700 = 0
Comparing it with ax² + bx + c = 0,
a = 1, b = -60, c = – 2700
∴ b² – 4ac = (-60)² – 4 . 1 . (- 2700)
= 3600 + 10800
= 14400 > 0

= 90 and -30
∵ The length of any side is not negative
Therefore leaving x = – 30
∴ x = 90
Hence, the shorter side of the rectangular field
= 90 m
and, the longer side of the rectangular field
= (90 + 30) m
= 120 m

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 the larger number = x
and the smaller number = y
According to the first condition of the problem,
x² – y² = 180 ….(1)
According to the second condition of the problem,
y² = 8x ….(2)
From (1) and (2),
x² – 8x = 180
or x² – 8x – 180 = 0
Comparing it with ax² + bx + c = 0,
a = 1, b = – 8, c = – 180
∴ b² – 4ac = (-8)² – 4 × 1 × (-180)
= 64 + 720
= 784 > 0

= 18 and – 10
When x = – 10, then from (2),
y² = 8 (-10) = -80, which is not possible.
Therefore leaving x = – 10
When x = 18, then from (2),
y² = 8 (18) = 144
or y = ±\(\sqrt {144}\)
or y = ± 12
Hence the required numbers are 18 and 12 or 18 and – 12.

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:
Let the uniform speed of the train = x km./h
Distance covered by the train= 360 km.
Time taken by the tram = \(\frac{Distance}{Speed}\)
= \(\frac{360}{x}\) hours
Increased speed of the train = (x + 5) km/hour
∴ Time taken by the train with the increased speed
= \(\frac{360}{x+5}\) hours
According to the question,

or 1800 = x² + 5x
or x² + 5x – 1800 = 0
Comparing it with ax² + bx + c = 0
a = 1, b = 5, c = -1800
∴ b² – 4ac = (5)² – 4 × 1 × (-1800)
= 25 + 7200
= 7225 > 0
∴ x = \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\) (quadratic formula)

∵ The speed of any train cannot be negative
Therefore leaving x = – 45
∴ x = 40
Hence the speed of the train =40 km/ hour

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 fdl the tank.
Solution:
Let the time taken by the tap of larger diameter to fill the tank = x hours
Then, the time taken by the tap of smaller diameter to fill the tank = (x + 10) hours
In case of one hour,
The larger tap can fill the tank = \(\frac{1}{x}\)
The smaller tap can fill the tank = \(\frac{1}{x+10}\)
∴ The larger and the smaller tap both can fill the tank
= \(\frac{1}{x}\) + \(\frac{1}{x+10}\) …(1)
But both the taps together take the time in filling the tank
= \(9 \frac{3}{8}\) hours = \(\frac{75}{8}\) hours
Therefore, both the taps together can fill the tank in one hour
= \(\frac{8}{75}\) ….(2)
From equation (1) and (2)

or 75(2x + 10) = 8(x² + 10x)
or 150x + 750 = 8x² + 80x
or 8x² + 80x – 150x – 750 = 0
or 8x² – 70x – 750 = 0
or 4x² – 35x – 375 = 0
Comparing it with ax² + bx + c = 0,
a = 4, b = – 35, c = – 375
∴ b² – 4ac = (-35)² – 4 × 4 × (-375)
= 1225 + 6000
= 7225 > 0

∵ Time cannot be negative
Therefore, leaving x = \(\frac{-25}{4}\)
∴ x = 15
Hence, the time in which the larger tap can fill the tank
= 15 hours
and, the time in which the smaller tap can fill the tank
= (15 + 10) hours
= 25 hours

Question 10.
An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (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 speed of the passenger train = x km/hour
Then, the speed of the express train = (x + 11) km/hour
Distance between Mysore and Bangalore = 132 km.
Time taken by the passenger train = \(\frac{132}{x}\) hours
Time taken by the express train = \(\frac{132}{x+11}\) hours
According to the question,

or 1452 = x² + 11x
or x² + 11x – 1452 = 0
Comparing it with ax² + bx + c = 0,
a = 1, b = 11, c = – 1452
∴ b² – 4ac = (11)² – 4 × 1 × (-1452)
= 121 + 5808
= 5929 > 0

∵ Speed of the train cannot be negative
∴ x = 33
Hence, average speed of the passenger train = 33 km/hour
and the average speed of the express train
= (33 + 11) km./hour
= 44 km./hour

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:
In case of first larger square
Let the length of each side of the square = x m
Then, area of the square = x² m²
and, perimeter of the square = 4x m
In case of second smaller sqaure
Let the length of each side of the square
= y m
Then, area of the square = y² m²
and, perimeter of the square = 4y m
According to the first condition,
x² + y² = 468 ….(1)
According to the second condition,
4x – 4y = 24
or 4 (x – y) = 24
or x – y = 6
or x = 6 + y …(2)
From (1) and (2),
(6 + y)² + y² = 468
or 36 + y² + 12y + y² = 468
or 2y² + 12y + 36 – 468 = 0
or 2y² + 12y – 432 = 0
or y² + 6y – 216 = 0
Comparing it with ay² + by + c = 0,
a = 1, b = 6, c = -216
∴ b² – 4ac = (6)² – 4 × 1 × (-216)
= 36 + 864
= 900 > 0

∵ The length of the side of the square cannot be negative
Therefore leaving y = -18
∴ y = 12
From (2), x = 6 + 12 = 18
Hence, the sides of the two squares are 12 m and 18 m.