Students can Download Class 10 Maths Chapter 10 Quadratic Equations Ex 10.3 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 10 Maths helps you to revise the complete Karnataka State Board Syllabus and to clear all their doubts, score well in final exams.

## Karnataka State Syllabus Class 10 Maths Chapter 10 Quadratic Equations Ex 10.3

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

Answer:

i) 2x^{2} – 7x + 3 = 0 divide by 2

Taking square root

Therefore, Roots of the equation are 3 and \(\frac{1}{2}\)

ii) 2x^{2} + x – 4 = 0 divide by 2

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

Therefore Roots of the equation are –\(\frac{\sqrt{3}}{2}\) and –\(\frac{\sqrt{3}}{2}\)

iv) 2x^{2} + x + 4 = 0 divide by 2

x^{2} + \(\frac{1}{2}\)x + 2 = 0

∴ Roots don’t exist.

Question 2.

Find the roots of the quadratic equations given in Q.1 above by applying the quadratic formula.

Answer:

i) 2x^{2} – 7x + 3 = 0

2x^{2} – 7x + 3 = 0

It is in the form of ax^{2} + bx + c = 0

a = 2, b = – 7 and c = 3

The roots are 3 and \(\frac{1}{2}\)

ii) 2x^{2} + x – 4 = 0

It is in the form ax^{2} + bx + c = 0

a = 2, b = 1 and c = – 4

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

It is in the form ax^{2} + bx + c = 0

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

Therefore Roots are \(\frac{-\sqrt{3}}{2}\) and \(\frac{-\sqrt{3}}{2}\)

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

It is in the form ax^{2} + bx + c = 0

a = 2, b = 1 and c = 4

Therefore Roots does not exist.

Question 3.

Find the roots of the following equations:

(i) x – \(\frac{1}{x}\) = 3, x ≠ 0

(ii) \(\frac{1}{x+4}-\frac{1}{x-7}\) = \(\frac{11}{30}\), x ≠ – 4, 7

Answer:

x^{2} – 1 = 3x

x^{2} – 3x – 1 = 0

It is in the form ax^{2} + bx + c = 0

a = 1, b = – 3 and c = – 1

ii)

-30 = x^{2} – 3x – 28

x^{2} – 3x – 28 + 30 = 0

x^{2} – 3x + 2 = 0

It is in the form ax^{2} + bx + c = 0

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

x = 2 (or) x = 1

The roots are 2 and 1

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.

Answer:

Let the present age of Rehman be ‘x’ years Rehman age 3 years ago is x – 3 and Rehman age 5 years from now is (x + 5) years.

3(2x + 2) = x^{2} + 2x – 15

6x + 6 = x^{2} + 2x – 15

x^{2} + 2x – 6x – 15 – 6 = 0

x^{2} – 4x -21 = 0

x^{2} – 7x + 3x – 21 =0

x (x – 7) + 3 (x – 7) = 0

(x – 7) (x + 3) = 0

x – 7 = 0 and x + 3 = 0

x = 7 and x = – 3

Age is always positive ∴ x = 7

Present age of Rehman’s is 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.

Answer:

Let the marks scored by Shefali in Maths be ‘x’ and marks scored in English by Shefali will be (30 – x).

(x + 2) (30 – x – 3) = 210

(x + 2)(27 – x) = 210

27x – x^{2} + 54 – 2x = 210

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

-x^{2} + 25x + 54 – 210 = 0

-x^{2} + 25x – 156 = 0

x^{2} – 25x + 156 = 0

x^{2} – 13x – 12x + 156 = 0

x(x – 13) – 12 (x- 13) = 0

(x – 13) (x – 12) = 0

If x – 13 = 0, then x = 13

If x – 12 = 0, then x = 12

∴ x = 13 OR x = 12

∴ Shefali scores in Maths, x = 13

Shefali’s marks in English = 30 – x

= 30 – 13

= 17.

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.

Answer:

Let the shortest side of rectangular field is x.

The diagonal of rectangular field is x + 60

The longest side of a rectangular field be x + 30.

AC^{2} = AB^{2} + BC^{2} [By Pythagoras theorem]

(x + 60)^{2} = (x)^{2} + (x + 30)^{2}

x^{2} + 120x + 3600

= x^{2} + x^{2} + 60x + 900

x^{2} + 120x + 3600

= 2x^{2} + 60x + 900

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

x^{2} – 60x – 2700 = 0

x^{2} – 90x + 30x – 2700 = 0

x (x – 90) + 30 (x – 90) = 0

(x – 90) (x + 30) = 0

x – 90 = 0 (or) x + 30 = 0

x = 90 (or) x = – 30

Length is always Positive.

∴ Shorter side of Rectangular field is 90 m

∴ Longest side of Rectangular field = x + 30 = 90 + 30 = 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.

Answer:

Let x and y be the larger and smaller numbers

(x)^{2} – y^{2} = 180 → (1)

y^{2} = 8x → (2)

Put equation (2) in equation (1)

x^{2} – 8x = 180

x^{2} – 8x – 180 = 0

x^{2} – 18x+ 10x- 180 = 0

x(x – 18) + 10 (x – 18) = 0

(x – 18) (x + 10) = 0

x – 18 = 0 and x + 10 = 0

x = 18 (or) x = – 10

∴ Larger number is 18

Smaller number y^{2} = 8x

y^{2} = 8 × 18

y^{2} = 144

y^{2} = \(\sqrt{144}\)

y = ± 12

The pair numbers be 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.

Answer:

Let the uniform speed of a train be x km/hr. Then, time taken to cover 360 km = \(\frac{360}{x}\)

New increased speed = (x + 5) km/hr.

time taken to cover 360 km = \(\frac{360}{x+5}\)

difference in time = 1

360x + 1800 – 360x = x^{2} + 5x

x^{2} + 5x – 1800 = 0

x^{2} + 45x – 40x- 1800 = 0

x (x + 45) – 40 (x + 45) = 0

(x + 45) (x – 40) = 0

x + 45 = 0 (or) x – 40 = 0

x = – 45 (or) x = 40

Speed of fife train is 40 km/hr.

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.

Answer:

Let the time taken by the smaller tap to fill the tank be x hrs

The time taken by the larger tap to fill the tank be x – 10 hrs.

Part of water filled by smaller tap in

1hr = \(\frac{1}{x}\) and part of water filled by larger tap in 1 hr = \(\frac{1}{x-10}\)

Part of water filled by two taps in 9\(\frac{3}{8}\)hrs = \(\frac{75}{8}\)

Part of water filled by two taps in 1 hr is \(\frac{75}{8}\) hrs

75 (2x – 10) = 8 (x^{2} – 10x)

150x – 750 = 8x^{2} – 80x

8x^{2} – 80x – 150x + 750 = 0

8x^{2} – 230x + 750 = 0 divide by 2

4x^{2} – 115x + 375 = 0

4x^{2} – 100x – 15x + 375 = 0

4x (x – 25) – 15 (x – 25) = 0

(x – 25) (4x – 15) = 0

x – 25 = 0 & 4x – 15 = 0

x = 25 and 4x = 15

x = \(\frac{15}{4}\)

∴ x = 25

∴ Smaller tap takes 25 hours

Larger tap takes = x – 10 = 25 – 10

= 15 hours

∴ time should not get negative

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.

Answer:

Let the speed of passenger train be x km/hr and speed of express train be (x + 11) km/hr

Distance between two stations d = 132 km

Time taken by passenger train

t_{1} = \(\frac{d}{s}\)

t_{1} = \(\frac{132}{x}\)

Time taken by express train t_{2} = \(\frac{132}{x+11}\)

difference in time is 1 hr

132x + 1452 – 132x = x^{2} + 10x

x^{2} + 10x – 1452 = 0

x^{2} + 44x – 33x – 1452 = 0

x(x + 44) – 33 (x + 44) = 0

(x + 44) (x – 33) = 0

x + 44 = 0 (or) x – 33 = 0

x = – 44 (or) x = 33

x = 33

Speed of Passenger train = 33 km/hr

Speed of express train = x + 11

= 33 + 11

= 44 km/hr

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.

Answer:

Let ‘x’ be the length of the side of first square and ‘y’ be the length of side of second square.

∴ x^{2} + y^{2} = 468 → (i)

then ‘x’ be the length of bigger square

4x – 4y = 24 divide by 4

x – y = 6

x = 6 + y → (ii)

Put equation (ii) in equation (i)

(6 + y)^{2} + y^{2} = 468

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

2y^{2} + 12y + 36 – 468 = 0

2y^{2} + 12y – 432 = 0 divide by 2

y^{2} + 6y – 216 = 0

y^{2} + 18y – 12y – 216 = 0

y (y + 18) – 12 (y + 18) = 0

(y + 18) (y – 12) = 0

y + 18 = 0 (or) y – 12 = 0

y = – 18 (or) y = 12

y = 12 ∴ sides can’t be negative.

Therefore x = y + 6 = 12 + 6 = 18

sides of a square are 12m and 18m.