NCERT Solutions Class 10 Maths Chapter 4 Quadratic Equations Ex 4.3

NCERT Solutions Class 10 Maths
Chapter – 4 (Quadratic Equations)

The NCERT Solutions in English Language for Class 10 Mathematics Chapter – 4 Quadratic Equations Exercise 4.3 has been provided here to help the students in solving the questions from this exercise.

Chapter : 4 Quadratic Equations

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

Solutions –

(i) 2x² – 7x + 3 = 0
⇒ 2x² – 7x = – 3
Dividing by 2 on both the sides
⇒ x² – 7x/2 = -3/2
Since 7/2 ÷ 2 = (7/4), Therefore, (7/4)² should be added to both sides of the quadratic equation
⇒ x² – 7x/2 + (7/4)² = – 3/2 + (7/4)²⇒ (x – 7/4)² = – 3/2 + (49/16) [Since, a² – 2ab + b² = (a – b)²]
⇒ (x – 7/4)²= (-24 + 49) / 16
⇒ (x – 7/4)² = 25/16
⇒ (x – 7/4)² = (5/4)²⇒ x – 7/4 = 5/4 and x – 7/4 = – 5/4
⇒ x = 5/4 + 7/4 and x = – 5/4 + 7/4
⇒ x = 12/4 and x = 2/4
⇒ x = 3 and x = 1/2
Roots are : 3, 1/2

(ii) 2x² + x – 4 = 0
⇒ 2x² + x = 4
Dividing by 2 on both the sides)
⇒ x² + x/2 = 2
Since 1/2 ÷ 2 = 1/2 × 1/2 = 1/4,
(1/4)² should be added to both sides
⇒ x² + x/2 + (1/4)² = 2 + (1/4)²⇒ [x + (1/4)]² = 2 + 1/16 [Since, a² + 2ab + b² = (a + b)²]
⇒ [x + (1/4)]² = (32 + 1) / 16
⇒ [x + (1/4)]² = 33/16
⇒ x + 1/4 = √33/4 and x + 1/4 = – √33/4
⇒ x = √33/4 – 1/4 and x = – √33/4 – 1/4
⇒ x = (√33 – 1)/4 and x = (- √33 – 1)/4
Thus, the roots are (√33 – 1)/4 and (- √33 – 1)/4

(iii) 4x² + 4√3x + 3 = 0
⇒ 4x² + 4√3x = -3
Dividing by 4 on both the sides
⇒ x² + √3x = – 3/4
⇒ x² + √3x + (√3/2)² = – 3/4 + (√3/2)² [(√3/2)² is added on both sides]
⇒ [x + (√3/2)]² = – 3/4 + 3/4 [Since, a² + 2ab + b² = (a + b)²]
⇒ [x + (√3/2)]² = 0
⇒ x = – √3/2 and x = – √3/2
Roots are – √3/2 and – √3/2

(iv) 2x² + x + 4 = 0
⇒ 2x² + x = -4
Dividing both the sides by 2
⇒ x² + x/2 = 2
Adding (1/4)²on both sides of the equation,
⇒ x² + x/2 + (1/4)² = – 2 + (1/4)²⇒ [x + (1/4)]² = – 2 + 1/16 [Since, a² + 2ab + b² = (a + b)²]
⇒ [x + (1/4)]² = (- 32 + 1)/16
⇒ (x + (1/4))² = – 31/16 < 0
Square of any real number can’t be negative.
Hence, real roots don’t exist.

2. Find the roots of the quadratic equations given in Q.1 above by applying the quadratic formula.
(i) 2x² – 7x + 3 = 0
On comparing the given equation with ax² + bx + c = 0, we get,
a = 2, b = -7 and c = 3
By using the quadratic formula, we get,
NCERT Class 10 Maths Solution
⇒ 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
The roots are : 3, /2

(ii) 2x² + x – 4 = 0
On comparing the given equation with ax² + bx + c = 0, we get,
a = 2, b = 1 and c = -4
By using the quadratic formula, we get,
NCERT Class 10 Maths Solution
⇒ x = (-1 ± √1 + 32)/4
⇒ x = (-1 ± √33)/4
∴ x = (-1 + √33)/4 or x = (-1 – √33)/4
The roots are : (-1 + √33)/4, (-1 – √33)/4

(iii) 4x² + 4√3x + 3 = 0
On comparing the given equation with ax² + bx + c = 0, we get,
a = 4, b = 4√3 and c = 3
By using the quadratic formula, we get,
NCERT Class 10 Maths Solution
⇒ x = (-4√3 ± √48 – 48)/8
⇒ x = (-4√3 ± 0)/8
∴ x = -√3/2 or x = -√3/2
The roots are : -√3/2, -√3/2

(iv) 2x² + x + 4 = 0
On comparing the given equation with ax² + bx + c = 0, we get,
a = 2, b = 1 and c = 4
By using the quadratic formula, we get
NCERT Class 10 Maths Solution
⇒ x = (-1 ± √1 – 32)/4
⇒ x = (-1 ± √-31)/4
As we know, the square of a number can never be negative. Therefore, there is no real solution for the given equation.

3. Find the roots of the following equations:
(i) x – $\mathbf{\frac{1}{x}}$ = 3, x ≠ 0

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

Solution:

(i) x – $\mathbf{\frac{1}{x}}$ = 3
Multiplying x on both the sides
⇒ x² – 1 = 3x
⇒ x² – 3x – 1 = 0
On comparing the given equation with ax² + bx + c = 0, we get
a = 1, b = -3 and c = -1
By using the quadratic formula, we get
NCERT Class 10 Maths Solution
⇒ x = (3 ± √9 + 4)/2
⇒ x = (3 ± √13)/2
∴ x = (3 + √13)/2 or x = (3 – √13)/2
The roots are : (3 + √13) / 2, (3 – √13) / 2

(ii) 1/x+4 – 1/x-7 = 11/30
⇒ [(x – 7) – (x + 4)] / (x + 4)(x – 7) = 11/ 30
⇒ [x – 7 – x – 4] / x² + 4x – 7x – 28 = 11/ 30
⇒ (- 11) / (x² – 3x – 28) = 11/30
⇒ – 11 × 30 = 11(x² – 3x – 28)
⇒ – 30 = (x² – 3x – 28) [Cancelling 11 from both sides of the equation]
⇒ x² – 3x – 28 + 30 = 0
⇒ x² – 3x + 2 = 0
Comparing x² – 3x + 2 = 0 against the standard form ax² – bx + c = 0 ,
We find that a = 1, b = – 3 and c = 2
b² – 4ac = (-3)² – 4(1)(2)
= 9 – 8
= 1 > 0
Therefore, real roots exist for this quadratic equation.
Therefore,
NCERT Class 10 Maths Solution
x = (3 ± √1) / 2
x = (3 + 1)/2 and x = (3 – 1)/2
x = 4 / 2 and x = 2 / 2
x = 2 and x = 1
Therefore, roots are 2, 1.

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 years.
Three years ago, his age was (x – 3) years.
Five years hence, his age will be (x + 5) years.
It is given that the sum of the reciprocals of Rehman’s ages 3 years ago and 5 years from now is 1/3.
∴ $\frac{1}{x-3}$ + $\frac{1}{x-5}$ = $\frac{1}{3}$

⇒ [(x + 5) + (x – 3)] / (x – 3)(x + 5) = 1/3
⇒ (2x + 2) / x² + 2x – 15 = 1/3
⇒ (2x + 2)(3) = x² + 2x – 15
⇒ 6x + 6 = x² + 2x – 15
⇒ x² + 2x – 15 = 6x + 6
⇒ x² – 4x – 21 = 0
Let us now find the roots by factorization method:
⇒ x² – 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
But, age can’t be a negative value.
Therefore, Rehman’s present age is 7 years.

5. In a class test, the sum of Shefali’s marks in Mathematics and English is 30. If she had 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 in Maths be x.
Then, the marks in English will be 30 – x.
According to the question,
(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
⇒ x = 12, 13
If the marks in Maths are 12, then marks in English will be 30 – 12 = 18
If the marks in Maths are 13, then marks in English will be 30 – 13 = 17

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 rectangle be x m.
Then, larger side of the rectangle = (x + 30) m
By applying Pythagoras theorem:
Hypotenuse² = Side 1² + Side 2²(60 + x)² = x² + (30 + x)²60² + 2(60)x + x² = x² + 30² + 2(30)x + x²3600 + 120x + x² = x² + 900 + 60x + x²3600 + 120x + x² – x² – 900 – 60x – x² = 0
2700 + 60x – x² = 0
Multiplying both sides by -1:
x² – 60x – 2700 = 0
Comparing x² – 60x – 2700 = 0 with ax² + bx + c = 0 we get,
a = 1, b = – 60, c = – 2700
b² – 4ac = (-60)² – 4(1)(-2700)
b² – 4ac = 3600 + 10800
b² – 4ac = 14400 > 0
∴ Roots exist.
NCERT Class 10 Maths Solution
= [-(- 60) ± √(14400)] / 2
= [(60) ± 120] / 2
x = (60 + 120) / 2 and x = (60 – 120) / 2
x = 180 / 2 and x = – 60 / 2
x = 90 and a = – 30
Length can’t be a negative value.
Hence, x = 90
Length of shorter side is x = 90 m
Length of longer side = 30 + x = 30 + 90 = 120 m

7. The difference between the 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 and smaller number be x and y respectively.
According to the question,
x²– y² = 180 and y² = 8x
⇒ x²– 8x = 180
⇒ x²– 8x – 180 = 0
⇒ x²– 18x + 10x – 180 = 0
⇒ x(x – 18) +10(x – 18) = 0
⇒ (x – 18)(x + 10) = 0
⇒ x = 18, -10

Case 1: If the larger number is 18, then square of smaller number = 8 × 18
Therefore, smaller number = ± √8 × 18
= ± √ (2 × 2 × 2 × 3 × 3 × 2)
= ± (2 × 2 × 3)
= ± 12

Case 2: If larger number is – 10, then square of smaller number = 8 × (- 10) = – 80
The square of any number cannot be negative.
∴ x = -10 is not applicable.
Thus, the numbers are 18, 12 (or) 18, – 12.

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/hr.
Distance = Speed × Time
360 = x × t
⇒ t = 360 / x
Increased speed of the train can be written as x + 5
New time to cover the same distance = t – 1
(x + 5) × (t – 1) = 360 ———— (i)
xt – x + 5t – 5 = 360
360 – x + 5(360/x) – 5 = 360 [Since, xt = 360 and t = 360 / x]
– x + 1800/x – 5 = 0
– x² + 1800 – 5x = 0
x² + 5x – 1800 = 0
Comparing x² + 5x – 1800 = 0 with ax² + bx + c = 0 we get,
a = 1, b = 5, c = – 1800
b² – 4ac = (5)² – 4(1)(- 1800)
b² – 4ac = 25 + 7200
b² – 4ac = 7225 > 0
∴ Roots exist.
NCERT Class 10 Maths Solution
x = (- 5 ± √7225) / 2
x = (- 5 ± 85) / 2
x = (- 5 + 85) / 2 and x = (- 5 – 85) / 2
x = 80 / 2 and x = – 90 / 2
x = 40 and x = – 45
Speed of the train cannot be a negative value.
Therefore, speed of the train is 40 km /hr.

9. Two water taps together can fill a tank in $\mathbf{9\frac{3}{8}}$ hours. The tap of the larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time at which each tap can separately fill the tank.

Solution – Let the time taken by the smaller pipe to fill the tank be x hr.
Time taken by the larger pipe = (x – 10) hr
Part of tank filled by smaller pipe in 1 hour = 1/x
Part of the tank filled by larger pipe in 1 hour = 1/(x – 10)
As given, the tank can be filled in $9\frac{3}{8}$ = 75/8 hours by both the pipes together.
Therefore,
1/x + 1/(x – 10) = 8/75

$\frac{(x-10)+x}{x(x-10)}$ = 8/75

⇒ $\frac{2x - 10}{x(x-10)}$ = 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, 30/8
Time taken by the smaller pipe cannot be 30/8 = 3.75 hours, as the time taken by the larger pipe will become negative, which is logically not possible.
Therefore, the time taken individually by the smaller pipe and the larger pipe will be 25 and 25 – 10 =15 hours, respectively.

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 average speed of passenger train be x km/h.
Average speed of express train = (x + 11) km/h
It is given that the time taken by the express train to cover 132 km is 1 hour less than the passenger train to cover the same distance.
(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, 33
As we know, speed cannot be negative.
Therefore, the speed of the passenger train will be 33 km/h and thus, the speed of the express train will be 33 + 11 = 44 km/h.

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 sides of the two squares be x m and y m.
Therefore, their perimeter will be 4x and 4y, respectively
And the area of the squares will be x² and y^2,respectively.
Given,
4x – 4y = 24
x – y = 6
x = y + 6
Also, x²+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(y +18) -12(y + 18) = 0
⇒ (y + 18)(y – 12) = 0
⇒ y = -18, 12
As we know, the side of a square cannot be negative.
Hence, the sides of the squares are 12 m and (12 + 6) m = 18 m.

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NCERT Solutions Class 10 Maths Chapter 4 Quadratic Equations Ex 4.3