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) 2x^{2} – 7x + 3 = 0

(ii) 2x^{2} + x – 4 = 0

(iii) 4x^{2} + 4\(\sqrt {3}\) x + 3 = 0

(iv) 2x^{2} + x + 4 = 0

Solution:

(i) According to the question the quadratic equation is

2x^{2} – 7x + 3 = 0

or 2x^{2} – 7x = – 3

or 2(x^{2} – \(\frac{7}{2}\)x) = -3

or x^{2} – \(\frac{7}{2}\)x = \(\frac{-3}{2}\)

(since to make a perfect square the coefficient of x^{2} 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

2x^{2} + x – 4 = 0

or 2x^{2} + x = 4

or x^{2} + \(\frac{1}{2}\)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

2x^{2} – 7x + 3 = 0

Comparing it with ax^{2} + bx + c = 0

a = 2, b = – 7, c = 3

Now

b^{2} – 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

2x^{2} + x – 4 = 0

comparing it with ax^{2} + bx + c = 0

a = 1, b = l, c = – 4 Now,

b^{2} – 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

4x^{2} + 4\(\sqrt {3}\)x + 3 = 0

Comparing it with ax^{2} + bx + c = 0,

a = 4, b = 4\(\sqrt {3}\) , c = 3

Now, b^{2} – 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

2x^{2} + x + 4 = 0

Comparing it with ax^{2} + bx + c = 0,

a = 2, b = 1, c = 4

Now, b^{2} – 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 x^{2} – 1 = 3x

or x^{2} – 3x – 1 = 0

Comparing it with ax^{2} + bx + c = 0

a = 1, b = – 3, c = – 1

Now, b^{2} – 4ac = (-3)^{2} – 4 . 1 . (- 1)

= 9 + 4= 13 > 0

(ii) According to the question

or -11 × 30 = 11(x^{2} – 3x – 28)

or -30 = x^{2} – 3x – 28

or x^{2} – 3x – 28 + 30 = 0

or x^{2} – 3x + 2 = 0

Comparing it with ax^{2} + bx + c = 0

a = 1, b = -3, c = 2

Now, b^{2} – 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 = x^{2} + 2x – 15

or x^{2} + 2x – 15 – 6x – 6 = 0

or x^{2} – 4x – 21 =0, which is quadratic in A.

Therefore comparing it with ax^{2} + bx + c = 0,

a = 1, b = – 4, c = – 21

Now, b^{2} – 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 – x^{2} + 54 – 2x

= -x^{2} + 25x + 54

According to the second condition of the problem

-x^{2} + 25x + 54 = 210

or -x^{2} + 25x + 54 – 210 = 0

or -x^{2} + 25x – 156 = 0

or x^{2} – 25x + 156 = 0

Comparing it with ax^{2} + bx + c = 0,

a = 1, b = – 25, c = 156

Now, b^{2} – 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 x^{2} + 3600 + 120x = x^{2} + x^{2} + 900 + 60x

or x^{2} + 3600 + 120x – 2x – 900 – 60x = 0

or -x^{2} + 60x + 2700 = 0

or x^{2} – 60x – 2700 = 0

Comparing it with ax^{2} + bx + c = 0,

a = 1, b = -60, c = – 2700

∴ b^{2} – 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,

x^{2} – y^{2} = 180 ….(1)

According to the second condition of the problem,

y^{2} = 8x ….(2)

From (1) and (2),

x^{2} – 8x = 180

or x^{2} – 8x – 180 = 0

Comparing it with ax^{2} + bx + c = 0,

a = 1, b = – 8, c = – 180

∴ b^{2} – 4ac = (-8)^{2} – 4 × 1 × (-180)

= 64 + 720

= 784 > 0

= 18 and – 10

When x = – 10, then from (2),

y^{2} = 8 (-10) = -80, which is not possible.

Therefore leaving x = – 10

When x = 18, then from (2),

y^{2} = 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 = x^{2} + 5x

or x^{2} + 5x – 1800 = 0

Comparing it with ax^{2} + bx + c = 0

a = 1, b = 5, c = -1800

∴ b^{2} – 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(x^{2} + 10x)

or 150x + 750 = 8x^{2} + 80x

or 8x^{2} + 80x – 150x – 750 = 0

or 8x^{2} – 70x – 750 = 0

or 4x^{2} – 35x – 375 = 0

Comparing it with ax^{2} + bx + c = 0,

a = 4, b = – 35, c = – 375

∴ b^{2} – 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 = x^{2} + 11x

or x^{2} + 11x – 1452 = 0

Comparing it with ax^{2} + bx + c = 0,

a = 1, b = 11, c = – 1452

∴ b^{2} – 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 m^{2}. 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 = x^{2} m^{2}

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 = y^{2} m^{2}

and, perimeter of the square = 4y m

According to the first condition,

x^{2} + y^{2} = 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} + y^{2} = 468

or 36 + y^{2} + 12y + y^{2} = 468

or 2y^{2} + 12y + 36 – 468 = 0

or 2y^{2} + 12y – 432 = 0

or y^{2} + 6y – 216 = 0

Comparing it with ay^{2} + by + c = 0,

a = 1, b = 6, c = -216

∴ b^{2} – 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.

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