Quadratic Equations

Let's Solve the Questions of

Class X

4. Quadratic Equations

Exercise 4.3 

Subject Mathematics 

Publication: NCERT

Exercise 4.3

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) 2x² - 7x + 3 = 0

Dividing both sides by 2.

x² - 7x/2 + 3/2 = 0

Now add 49/16 both sides.

x² - 7x/2 + 49/16 + 3/2 = 49/16

or, (x - 7/4)² = 49/16 - 3/2

or, (x - 7/4)² = (49 - 24)/16

or, (x - 7/4)² = 25/16

or, x - 7/4 = ±5/4

or, x = ±5/4 + 7/4

or, x = 3 or ½ Ans

(ii) 2x² + x – 4 = 0

Solution:

Dividing both sides by 2.

x² + x/2 - 2 = 0

Now, add 1/16 both sides

x² + x/2 + 1/16 - 2 = 1/16

or, (x + 1/4)² = 1/16 + 2

or, (x + ¼)² = 33/16

or, x + ¼ = ±√33/4

or, x = ±√33/4 - ¼ Ans.

(iii)

Solution:

Dividing both sides by 4.

x² + √3x + ¾ = 0

(x + √3/2)² = 0

or, x + √3/2 = 0

∴ x = -√3/2 Ans.

(iv)

Solution:

Dividing both sides by 2.

x² + x/2 + 2 = 0

Now add 1/16 both sides.

x² + x/2 + 1/16 + 2 = 1/16

or, (x + ¼)²  = 1/16 - 2

or, (x + ¼)² = (1 - 32)/2

or, x + ¼ =√ -31/2

The square root of negative integer is not possible.

So, the roots of the given quadratic equation is not exist.

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

(i) 2x² – 7x + 3 = 0

Solution:

From question it is clear that

a = 2

b = -7

c = 3

We know that

x = (-b ±√b² -4ac)/2a

Now putting the value of a, b and c

x = (7±√(-7)² - 4×2×3)/2×2

x = (7±√49 – 24)/4

x = (7±√25)/4

x = (7±5)/4

x = 3 or ½ Ans.

Similarly, we can solve rest other questions of this section.

3. Find the roots of the following equations:

(i) 

x – 1/x = 3, x ≠0

Solution:

(x² – 1)÷x = 3

x² – 3x – 1 = 0

a = 1

b = –3

c = –1

D = b² – 4ac

D = 9 + 4

13

∴ x = (–b ±√D)/2a

Now, putting the values

x = (3 ±√13)/2 Ans.

(ii) 

1/(x + 4) – 1/(x –7) = 11/30

x≠–4, 7.

Solution:

(x –7 + x + 4)/(x + 4)(x –7)

or, (2x –3)/(x²–3x –28) = 11/30

or, 60x –90 = 11x² –33x –308

or, 11x²–93x–218 = 0

a = 11

b = –93

c = –218

D = b² –4ac

or, D = –93² –4×11×–218

or, D = 8649 + 9592

or, D = 18241

x ={ –(–b) ± √D}/2a

or,x = {93 ±√18241}/2×11

or, x = {93±135.05}/22

x = 228.05/22 or 42.05/22

∴ x = 10.37 or 1.91 Ans.

4. The sum of the reciprocals of Rehman’s ages, (in years) 3 years ago and 5 years from now is 1/3

Find his present age.

Solution:

Let the present age of Rehman be x.

Now, from question.

Age of Rehman before 3 yrs. = x - 3.

Age of Rehan after 5 yrs from now = x + 5.

Now, from question

1/(x - 3) + 1/(x + 5) = ⅓

or, (x + 5 + x - 3)/(x - 3)(x + 5) = ⅓

or, (2x + 2)/(x² + 2x - 15) = ⅓

or, x² + 2x - 15 = 6x + 6

or, x² - 4x - 21 = 0

or, (x - 2)² - 4 - 21 = 0

or, (x - 2)² = 25

or, x - 2 = ±5

or, x = ±5 + 2

∴ x = -3 or 7 Ans

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 that be got by Shefali in Mathematics be x.

∴ In English = 30 – x

According to question

(x + 2)(30 – x – 3) = 210

or, (2 + x)(27 – x)

or, (54 + 25x – x²) = 210

or, x² – 25x + 156) = 0

∴ a = 1

b = –25

c = 156

D = b² – 4ac

or, D = –25² – 4×1×156

or, D = 625 – 624

or, D = 1

x = (–b±√D)/2a

x ={–(–25) ±√1}/2×1

x = (25 ±1)/2×1

x = 26/2 or 24/2

x = 13 or 12

When Marks obtained in Mathematics = 13

Marks obtained in English = 17

When Marks obtained in Mathematics = 12

Marks obtained in English = 18.

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 be x m.

‌‌∴ Longer side = x + 30 m.

∴ Diagonal = x + 60 m.

We know that

shorter side² + longer side² = diagonal²

x² + (x + 30)² = (x + 60)²

or, 2x² + 60x + 900 = x² + 120x + 3600

or, 2x² - 60x - 2700 = 0

or, x² - 30x - 1350 = 0

a = 1, b = -30, c = -1350

D = b² - 4ac

D = (-30)² - 4×1×-1350

D = 900 + 5400

D = 6300

x = (-b±√D)/2a

x = (-(-30)±√6300)/2×1

x = (30±79.37)/2

Side of the rectangle never be in negative.

x = 54.69 m

Longer side = 54.69+30 =84.69 m.

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 be x.

Square of smaller number = 8x

Now from question

Share of larger number - Square of smaller number = 180

or, x² - 8x = 180

x² - 8x - 180 = 0

a = 1, b = -8, c = -180

D = b² - 4ac

or, D = (-8)² -4×1×-180

or, D = 64 + 720

or, D = 784

x = (-b ±√D)/2a

or, x = (8 ±√784)/2

or, x = (8 ±28)/2

or, x = 18 or -10

If larger number is 18

Smaller number = √8×18 = √144 = 12

So, required numbers are 18 & 12 Ans.

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 speed of the train be x km/h

Now from question

360/x  - 360/x+5 = 1

or, 360(x + 5 - x)/(x(x + 5)) = 1

or, 360×5 = x(x + 5)

or, 1800 = x² + 5x

or, x² + 5x - 1800 = 0

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

D = b² - 4×a×c

or, D = 5² - 4×1×-1800

or, D = 25 + 7200

or, D = 7225

x = (-b ±√D)/2×a

or, x = (-5 ±√7225)/2

or, x = (-5 ±85)/2

or, x = 80/2 or -90/2

or, x = 40 or -45

Speed of train never be in negetive.

So, Speed of train = 40 km/h Ans.

9. Two water taps together can fill a tank in 9-3/8

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 the smaller tap be take x hours to fill the tank.

∴ Larger will take x -10 hours.

Both take 75/8 hours.

From question

Both taps fill the tank in 1 hours = 1/x + 1/x -10

Both taps in 1 hour = 8/75 

∴ 1/x + 1/(x -10) = 8/75

(x -10 + x)/x(x -10) = 8/75

or, 75(2x -10) = 8(x² -10x)

or, 150x -750 = 8x² -80x

or, 8x² -230x -750 = 0

or, 4x² -115x -375 = 0

a = 4, b = -115, c = -375

D = b² -4ac

or, D = -115² -4×4×-375

or, D = 13225 + 6000

or, D = 19225

x = (-b ±√D)/2a

or, x = (-(-115) ±√19225)/2×4

or, x = (115 ±138.65)/8

or, x = 253.65/8 or -23.65/8

or, x = 31.71 or -2.96

Time never be in negative.

Time taken by smaller tap = 31.71 hours

Time taken by larger tap = 21.71 hours

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 11km/h more than that of the passenger train, find the average speed of the two trains.

Solution:

The average speed of the passenger train be x km/h

∴ Speed of the express train = x +11 km/h

Time taken by passenger train to cover 132 km = 132÷x hours

Time taken by express train to cover 132 km = 132÷(x+11)

Now, from question

(132÷x) -{132÷(x+11)}= 1

or, 132×(x+11-x) ÷x×(x+11) = 1

or, 1452 ÷(x²+11x) = 1

or, x²+11x-1452 = 0

a = 1, b = 11, c =-1452

D = b² -4ac

or, D = 11² -4×1×-1452

or, D = 121 + 5808

or, D = 5929

x = (-b ±√D)÷2×a

or, x = (-11 ±√5929)÷2

or, x = (-11 ±77)÷2

or, x = 66÷2 or -88÷2

or, x = 33 or -44

Speed never be in negative.

So, x = 33 km/h

Speed of express train = 44 km/h Ans.

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 side of smaller square be x m.

Perimeter of the small square = 4x

Perimeter of the larger square 24 + 4x

Side of larger square = (24 + 4x) ÷ 4

or, Side of larger square = 6 + x

Now, from question

(6 + x)² + x² = 468

or, 36 +12x + x² + x² = 468

or, 2x² +12x + 36 -468 = 0

or, 2x² +12x -432 = 0

or, x² + 6x -216 = 0

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

D = b² -4×a×c

or, D =6² -4×1×-216

or, D = 36 + 864

or, D = 900

x = (-b ±√D) ÷ 2×a

x = (-6 ±30) ÷ 2

or, x = 24 ÷ 2 or -36 ÷ 2

or, x = 12 or -18

Side never be in negative.

∴ x = 12

Side of larger square = 6 + 12 = 18

Required Answers

12 m & 18 m Ans.

This Exercise become over.

Quadratic Equations

Quadratic Equations Class X Exercise 4.3

Exercise 4.4

Areas Related to Circles

12. Areas Related to Circles

Circle

The collection of points which is located at a certain distance from a certain point is said to be circle.

The certain distance is known as radius of the circle.

The certain point is known as centre of the circle.

The collection of points is known as circumference of the circle.

It is a geometrical figure in which infinite sides.

Area

The region which is surrounded by a closed geometrical figure is said to be area.

Calculation of area of a circle

When circle is divided into 16 equal parts from the centre of the circle. It is converted into a rectangle of length πr and breadth of r.

Important formulae of the circle

Area of cirle = πr²

Circumference of the circle = 2πr

Area of sector of a circle = πr²θ/360⁰

Area of minor segment = πr²2θ/360⁰ - r²sin2θ/2

Exercise 12.1

1. The radii of two circles are 19 cm and 9 cm respectively. Find the radius of the circle which has circumference equal to the sum of the circumferences of the two circles.

Solution:

Let the radius of the circle be R.

We know that

Circumference of the circle = C = 2πr

From question it is clear that

r₁ = 19 cm

r₂ = 9 cm

R = ?

Therefore,

C = C₁ + C₂

Therefore,

R = 2π(r₁ + r₂)/2π

Therefore,

R = 19 + 9

R = 28 cm Ans.

2. The radii of two circles are 8 cm and 6 cm respectively. Find the radius of the circle having area equal to the sum of the areas of the two circles.

Solution

Let the radius be r.

From question it is clear that

r =?

r₁ = 8 cm

r₂ = 6 cm

Therefore,

Radius of the circle = r² = π(r₁² + r₂²)/π

r² = 8² + 6²

r² = 64 + 36

r² = 100

r = 10 cm. Ans.

द्विघात समीकरण

द्विघात समीकरण को सामान्यतः

ax² + bx + c = 0

से निरूपित किया जाता है ,जहां a, b एवं c वास्तविक संख्या है तथा a का मान शून्य नही है।

द्विघात समीकरण में चर (x) का मान ज्ञात किया जाता है जिसे द्विघात समीकरण का मूल कहा जाता है।

किसी द्विघात समीकरण का दो और केवल दो ही मूल होता है जिसे क्रमशः α एवं β से निरूपित किया जाता है।

द्विघात समीकरण में a, b एवं c एक दूसरे से समन्धित होते है जो इस प्रकार है।

D = b² - 4ac

D को द्विघात समीकरण का विवेचक या विवित्तकर कहा जाता है।

D के मान से मूल की प्रकृति ज्ञात की जा सकती है।

1. मूल वास्तविक एवं आसमान होता है जब D का मान शून्य से बड़ा होता है।

2. मूल वास्तविक एवं समान होता है जब D का मान शून्य होता है

3. मूल वास्तविक नही होता है जब D का मान शून्य से छोटा होता है।

आगे जारी है ...........

9. Some Applications of Trigonometry

Class X

Exercise 9.1

NCERT Textbook Solution

Learning Points

• Angle of Elevation
• Angle of Depression

Angle of Elevation

The angle between line of sight and horizontal level when the observer is below the object is said to be angle of elevation.

Angle of Depression

The angle between line of sight and horizontal level when the observer is above the object is said to be the angle of depression.

Exercise 9.1

Q.No.1. 

 A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is 30° (see Fig. 9.11).

Fig. 9.11

Solution:

From question it is clear that

Angle of Elevation = Ө = 30°

Height of the pole = AB = ?

Now applying trigonometric ratio sin𝞱

From figure it is clear that

sin𝞱 = AB/AC

Now putting the value of 

1/2 = AB/20

AB = 10 m Ans.