NCERT solutions for class 10 Maths chapter 4 Quadratic Equations (Exercise 4.3)

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NCERT solutions for class 10 Maths


Chapter 4


Quadratic Equations


Exercise 4.3


Ex 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

Solutions:

(i) 2x2 – 7x + 3 = 0

⇒ 2x2 – 7= – 3

Dividing both sides by 2 ,

⇒ x2 -7x/2 = -3/2

⇒ x-2 × x ×7/4 = -3/2

On adding (7/4)2 to both sides ,

⇒ (x)2-2×x×7/4 +(7/4)2 = (7/4)2-3/2

⇒ (x-7/4)2 = (49/16) – (3/2)

⇒(x-7/4)= 25/16

⇒(x-7/4)2 = ±5/4

⇒ x = 7/4 ± 5/4

⇒ x = 7/4 + 5/4 or x = 7/4 – 5/4

⇒ x = 12/4 or x = 2/4

⇒ x = 3 or x = 1/2

(ii) 2x2 + x – 4 = 0

⇒ 2x2 + x = 4

Dividing both sides by 2,

⇒ x2 +x/2 = 2

Adding (1/4)to both sides,

⇒ (x)+ 2 × x × 1/4 + (1/4)2 = 2 + (1/4)2

⇒ (x + 1/4)2 = 33/16

⇒ x + 1/4 = ± √33/4

⇒ x = ± √33/4 – 1/4

⇒ x = ± √33-1/4

Either x = √33-1/4 or x = -√33-1/4

(iii) 4x2 + 4√3x + 3 = 0

Converting the equation into a2+2ab+bform,

⇒ (2x)2 + 2 × 2x × √3 + (√3)2 = 0

⇒ (2x + √3)2 = 0

⇒ (2x + √3) = 0 and (2x + √3) = 0

Either x = -√3/2 or x = -√3/2.

(iv) 2x2 + x + 4 = 0

⇒ 2x2 + x = -4

Dividing both sides by 2,

⇒ x2 + 1/2x = 2

⇒ x2 + 2 × x × 1/4 = -2

Adding (1/4)to both sides ,

⇒ (x)+ 2 × x × 1/4 + (1/4)2 = (1/4)– 2

⇒ (x + 1/4)2 = 1/16 – 2

⇒ (x + 1/4)2 = -31/16 [The square of numbers cannot be negative.]

Hence, this quadratic equation has no real roots.


Ex 4.3 Question 2.


Find the roots of the quadratic equations given in Q.1 above 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) 2x2 – 7x + 3 = 0

a = 2, b = -7 and c = 3

Using quadratic formula

⇒ x = (7±√(49 – 24))/4

⇒ x = (7±√25)/4

⇒ x = (7±5)/4

⇒ x = (7+5)/4 or x = (7-5)/4

⇒ x = 12/4 or 2/4

∴  x = 3 or 1/2

(ii) 2x2 + x – 4 = 0

a = 2, b = 1 and c = -4

Using quadratic formula,

⇒x = -1±√1+32/4

⇒x = -1±√33/4

∴ x = -1+√33/4 or x = -1-√33/4

(iii) 4x2 + 4√3x + 3 = 0

a = 4, b = 4√3 and c = 3

Using quadratic formula,

⇒ x = -4√3±√48-48/8

⇒ x = -4√3±0/8

∴ x = -√3/2 or x = -√3/2

(iv) 2x2 + x + 4 = 0

a = 2, b = 1 and c = 4

Using quadratic formula,

⇒ x = -1±√1-32/4

⇒ x = -1±√-31/4

The square of numbers cannot be negative. Hence, this quadratic equation has no real roots.


Ex 4.3 Question 3.


Find the roots of the following equations:

(i) x-1/x = 3, x ≠ 0
(ii) 1/x+4 – 1/x-7 = 11/30, x = -4, 7

Solution:

(i) x-1/x = 3

⇒ x2 – 3x -1 = 0

a = 1, b = -3 and c = -1

Using quadratic formula,

⇒ x = 3±√9+4/2

⇒ x = 3±√13/2

∴ x = 3+√13/2 or x = 3-√13/2

(ii) 1/x+4 – 1/x-7 = 11/30

⇒ x-7-x-4/(x+4)(x-7) = 11/30

⇒ -11/(x+4)(x-7) = 11/30

⇒ (x+4)(x-7) = -30

⇒ x2 – 3x – 28 = 30

⇒ x2 – 3x + 2 = 0

Using factorization method,

⇒ x2 – 2x – x + 2 = 0

⇒ x(x – 2) – 1(x – 2) = 0

⇒ (x – 2)(x – 1) = 0

⇒ x = 1 or 2


Ex 4.3 Question 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 current age of Rahman = x years.

Three years ago, Rehman’s age = (x – 3) years.

Five years after, Rehman’s age = (x + 5) years.

∴ 1/x-3 + 1/x-5 = 1/3

(x+5+x-3)/(x-3)(x+5) = 1/3

(2x+2)/(x-3)(x+5) = 1/3

⇒ 3(2x + 2) = (x-3)(x+5)

⇒ 6x + 6 = x2 + 2x – 15

⇒ x2 – 4x – 21 = 0

⇒ x2 – 7x + 3x – 21 = 0

⇒ x(x – 7) + 3(x – 7) = 0

⇒ (x – 7)(x + 3) = 0

⇒ x = 7, -3 [Age of a person cannot be negative.]

∴Rahman’s present age = 7 years.


Ex 4.3 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 Shefali’s marks in Maths = x.

Then, Shefali’s marks in English = 30 – x.

If she got 2 marks more  in maths and 3 less marks in english

Marks in maths = x + 2

Marks in english = 30 – x – 3

(x + 2)(30 – x – 3) = 210

(x + 2)(27 – x) = 210

⇒ -x2 + 25x + 54 = 210

⇒ x2 – 25x + 156 = 0

⇒ x2 – 12x – 13x + 156 = 0

⇒ x(x – 12) -13(x – 12) = 0

⇒ (x – 12)(x – 13) = 0

Either x = 12 or x = 13

If x = 12, then

Marks in maths = 12 and marks in English will be 30 – 12 = 18

If x = 13

Marks in Maths are 13,  and marks in English will be 30 – 13 = 17.


Ex 4.3 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 smaller side = x m.

Then, hypotenuse = x + 60 m

Longer side = x + 30 m

⇒ x2 + (x + 30)2 = (x + 60)2

⇒ x2 + x2 + 900 + 60x = x2 + 3600 + 120x

⇒ x2 – 60x – 2700 = 0

⇒ x2 – 90x + 30x – 2700 = 0

⇒ x(– 90) + 30(x -90) = 0

⇒ (– 90)(x + 30) = 0

Either = 90 or x = -30  [side of the field cannot be negative.]

Therefore, x= 90 m.Hence , the smaller side = 90 m.

Longer side = (90 + 30) m = 120 m.


Ex 4.3 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

Let , the smaller = y

y2 = 8x

According to the question ,

x– y2 = 180 and y2 = 8x

⇒ x– 8x = 180

⇒ x– 8x – 180 = 0

⇒ x– 18x + 10x – 180 = 0

⇒ x(x – 18) +10(x – 18) = 0

⇒ (x – 18)(x + 10) = 0

Either x = 18 or x = -10

However, the larger number cannot considered as negative number, as 8 times of the larger number will be negative and hence, the square of the smaller number will be negative which is not possible.

Therefore, the larger number will be 18 only.

x = 18

∴ y2 = 8x = 8 × 18 = 144

⇒ y = ±√144 = ±12

∴ Smaller number = ±12

Therefore, the numbers are 18 and 12 or 18 and -12.


Ex 4.3 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 speed of the train = x km/hr.

Distance = 360 km

Time taken  = 360/x hr.          [time = distance / speed]

According to the question

⇒ (x + 5)(360-1/x) = 360

⇒ 360 – x + 1800-5/x = 360

⇒ x+ 5x + 10x – 1800 = 0

⇒ x(x + 45) -40(x + 45) = 0

⇒ (x + 45)(x – 40) = 0

⇒ x = 40, -45  [The speed of train cannot be negative.]

Therefore, the speed of train = 40 km/h.

 


Ex 4.3 Question 9.


Two water taps together can fill a tank in
NCERT Solutions for Class 10 Chapter 4- 1 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 time taken by the smaller tap to fill the tank = x hr.

Time taken by the larger tap y2 = 8x = (x – 10) hr

Part of tank filled by smaller tap in 1 hour = 1/x

Part of tank filled by larger tap in 1 hour = 1/(– 10)

1/x + 1/x-10 = 8/75

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

⇒ 2x-10/x(x-10) = 8/75

⇒ 75(2x – 10) = 8x2 – 80x

⇒ 150x – 750 = 8x2 – 80x

⇒ 8x2 – 230x +750 = 0

⇒ 8x2 – 200x – 30x + 750 = 0

⇒ 8x(x – 25) -30(x – 25) = 0

⇒ (x – 25)(8x -30) = 0

⇒ x = 25 or x = 30/8 = 15/4

Therefore, the smaller tap tkaes 25 hours  and the larger tap take 25 and 25 – 10 =15 hours respectively.

 


Ex 4.3 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 speeds 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 passenger train =  x km/h.

Speed of express train = (x + 11) km/h

Distance travelled = 132 km

Time taken by passenger train = 132/x

Time taken by express train = 132/(x+11)

(132/x) – (132/(x+11)) = 1

132(x+11-x)/(x(x+11)) = 1

132 × 11 /(x(x+11)) = 1

⇒ 132 × 11 = x(x + 11)

⇒ x2 + 11x – 1452 = 0

⇒ x2 +  44x -33x -1452 = 0

⇒ x(x + 44) -33(x + 44) = 0

⇒ (x + 44)(x – 33) = 0

⇒ x = – 44 or x = 33  [Speed cannot be negative.]

∴The speed of the passenger train = 33 km/h.

The speed of the express train = 33 + 11 = 44 km/h.

 


Ex 4.3 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 side of the larger square = x m and y m.

Let the side of the smaller square = y m

Therefore, the perimeter of the larger square =  4x

The perimeter of the larger square = 4y

And area of the larger squares =  x2

Area of the larger squares = y2

4x – 4y = 24

x – y = 6

x = y + 6

Also, x+ y2 = 468

⇒ (6 + y2) + y2 = 468

⇒ 36 + y2 + 12y + y2 = 468

⇒ 2y2 + 12y + 432 = 0

⇒ y2 + 6y – 216 = 0

⇒ y2 + 18y – 12y – 216 = 0

⇒ y(+18) -12(y + 18) = 0

⇒ (y + 18)(y – 12) = 0

⇒ y = -18 or x = 12 [The side of a square cannot be negative.]

Hence, the sides of the  samller square = 12 m

The sides of the  samller square = (12 + 6) m = 18 m.