Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev

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Class 10 : Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev

The document Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev is a part of the Class 10 Course Mathematics (Maths) Class 10.
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Ques 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√3 x+ 3 = 0
(iv) 2x2 + x + 4 = 0

Sol: (i) 2x− 7x + 3 = 0

Dividing throughout by the co-efficient of x2, we get
Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev
Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev
Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev
Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev
Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev
Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev
Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev
Case I:
Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev
Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev

Case II:
Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev
Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev
Thus, required roots are
x = 3 and x = 1/2

(ii) 2x2 + x−  4 = 0

We have:
2x2 + x − 4 = 0
Dividing throughout by 2,

Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev
Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev

Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev
Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev
Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev
Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev
Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev
Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev
Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev
Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev
(iv) 2x2 + x + 4 = 0
Dividing throughout by 2, we have:

Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev
Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev

But the square of a number cannot be negative.

Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev cannot be a real value.

So, no real roots exist.

⇒ There is no real value of x satisfying the given equation.

Ques 2: Find the roots of the following quadratic equations, using the quadratic formula:

(i) 2x2 − 7x + 3 = 0
(ii) 2x2 + x − 4 = 0
(iii) 4x2 + 4√3 x + 3 = 0
(iv) 2x2 + x + 4 = 0

Sol: (i) 2x− 7x + 3 = 0

Comparing the given equation with ax2 + bx + c = 0, we have:

a = 2
b = − 7
c = 3

∴ b2 − 4ac =(− 7)2 − 4 (2) (3)
= 49 − 24 = 25 ≥ 0

Since b2 − 4ac > 0

∴ The given equation has real roots. The roots are given by

Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev
Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev
Thus, the roots of the given equation are

Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev

(ii) 2x2 + x−4 = 0

Comparing the given equation with ax2 + bx + c = 0, we have:

a = 2
b = 1 c = − 4
∴ b2 − 4ac = (1)2 − 4 (2) (− 4) = 1 + 32
= 33 > 0

Since b2 − 4ac > 0
∴ The given equation has equal roots. The roots are given by

Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev

Thus, the required roots are:

 Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev

Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev

Comparing the given equation with ax+ bx + c = 0, we have:

a = 4
b = 43
c = 3
Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev
= [16 × 3] − 48

= 48 − 48 = 0
Since b2 − 4ac = 0
∴ The given equation has real roots, which are given by:
Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev
Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev
Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev
Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev

(iv) 2x2 + x + 4 = 0

Comparing the given equation with ax2 + bx + c = 0, we have:

a = 2
b = 1
c = 4

∴ b2 − 4ac = (1)2 − 4 (2) (4)
= 1 − 32 = − 31 < 0

Since b2 − 4ac is less than 0, therefore the given equation does not have real roots.

Ques 3: Find the roots of the following equations:

Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev
Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev
Sol: Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev

Here, we have:
Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev

⇒ x2 − 1= 3 x

⇒ x2 − 3x − 1 = 0       ...(1)

Comparing (1) with ax2 + bx + c = 0, we have:

a = 1
b = − 3
c = − 1

∴ b2 − 4ac =(− 3)2 − 4 (1) (− 1)
= 9 + 4 = 13 > 0
Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev
Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev
Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev

Now, taking +ve sign,

Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev

Taking −ve sign,

Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev

Thus, the required roots of the given equation are:

Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev
Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev

We have:
Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev
Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev

⇒ − 11 × 30 = 11 (x2 − 3x − 28)
⇒ − 30 = x2 − 3x − 28
⇒ x2 − 3x − 28 + 30 = 0
⇒ x2 − 3x + 2 = 0

Comparing (1) with ax2 + bx + c = 0, we have:

a =1
b = − 3
c = 2

∴ b2 − 4ac =(− 3)2 − 4 (1) (2)
= 9 − 8 = 1 > 0

∴ The quadratic equation (1) has real roots, which are given by:
Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev
Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev
Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev
Taking +ve sign, we have:

Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev
Taking −ve sign, we have:

Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev

Thus, the roots of the given equation are:

x = 2 and x = 1.

Ques 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.
Sol: Let the present age of Rehman = x
∴ 3 years ago Rehman’s age = (x − 3) years  
5  years later Rehman’s age = (x + 5) years

Now, according to the condition,

Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev
Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev

⇒ 3 [x + 5 + x − 3] = (x − 3) (x + 5)
⇒ 3 [2x + 2] = x2 + 2x − 15
⇒ 6x + 6 = x2 + 2x − 15
⇒ x2 + 2x − 6x − 15 − 6=0
⇒ x2 − 4x − 21 = 0                 ...(1)

Now, comparing (1) with ax2 + bx + c = 0, we have:

a =1
b = − 4
c = − 21

∴ b2 − 4ac =(− 4)2 − 4 (1) (− 21)
= 16 + 84
= 100
Since, Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev
Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev
Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev
Taking positive sign, we have:

Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev
Taking negative sign, we have:

Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev

Since age cannot be negative,

∴ x ≠− 3 ⇒ x = 7

So, the present age of Rehman = 7 years.

Ques 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.

Sol: Let, Shefali’s marks in Mathematics = x
∴ Marks in English = (30 − x) [∵ Sum of their marks in Eng. and Maths = 30]

Now, according to the condition,

(x + 2) × [(30 − x) − 3] = 210
⇒ (x + 2) × (30 − x − 3) = 210
⇒ (x + 2) (−x + 27) = 210
⇒ −x2 + 25x + 54 = 210
⇒ −x2 + 25x + 54 − 210 = 0
⇒ −x2 + 25x − 156 = 0
⇒ x2 − 25x + 156 = 0                  ...(1)

Now, comparing (1) with ax2 + bx + c = 0

a = 1
b = −25
c = 156

∴ b2 − 4ac =(−25)2 − 4(1) (156)
= 625 − 624 = 1
Since, Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev
Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev
Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev

Taking +ve sign, we have

Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev

Taking − ve sign, we get

Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev

When x = 13, then 30 − 13 = 17
When x = 12, then 30 − 12 = 18
Thus, marks in Maths = 13, marks in English = 17
marks in Maths = 12, marks in English = 18

Ques 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.

Sol: Let the shorter side of rectangle be x m, then longer side be (x + 30) m and diagonal be (x + 60) m
In right-angled ΔABC,
(x + 60)2 = (x + 30)2 + (x)2 [Pythagoras Theorem]
⇒ x2 + 120x + 3600 = x2 + 60x + 900 + x2 
⇒ x2 - 60x - 2700 = 0
  x2 - 90x + 30x - 2700 = 0
x (x - 90) + 30 (x - 90)= 0
 (x + 30) (x - 90) = 0
x + 30 = 0 or x - 90 = 0
x = -30 (rejected) or x = 90
Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev
Hence, shorter side = 90 m
and longer side = 90 + 30 = 120 m

Ques 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.
Sol: Let the larger number be x.
Since, (smaller number)2 = 8 (larger number)
⇒ (smaller number)2 = 8x
⇒ smaller number =  Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev
Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev
⇒ x2 − 8x = 180
⇒ x2 − 8x − 180 = 0

Comparing (1) with ax+ bx + c = 0, we have:

a = 1
b = − 8
c = − 180

∴ b2 − 4ac =(− 8)2 − 4 (1) (− 180)
= 64 + 720 = 784

Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev
Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRevEx 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev
Taking +ve sign, we get

Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev

Taking −ve sign, we get

Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev

But x = − 10 is not admissible,
∴ The smaller number = 18
Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev

Thus, the larger number = 12 or − 12
Thus, the two numbers are:

18  and  12  or  18  and  −12

Ques 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.

Sol: Let the uniform speed of the train be x km/hr.

Since, time taken by the train Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev

Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev

When speed is 5 km/hr more then time is 1 hour less.
Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev
Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev
Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev
Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev
Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev
Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev
⇒ –5 × 360 = − 1 (x2 + 5x)
⇒ –5 × 360 = − x2 − 5x    ∵ 360 × 5 = 1800
⇒ x2 + 5x − 1800 = 0   ...(1) 

Comparing (1) with
ax2 + bx + c = 0
⇒ a = 1
b = 5
c = −1800
∴ b2 − 4ac = (5)2 − 4 (1) (− 1800)
= 25 + 7200
= 7225
Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev
Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev
Taking +ve sign, we get:

Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev

Taking −ve sign, we get

Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev

Since, the speed of a vehicle cannot be negative,

∴ x ≠− 45
Thus, x = 40 ⇒ speed of the train = 40 km/hr.

Ques 9: Two water taps together can fill a tank in  Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev 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.

Sol: Let the smaller tap fills the tank in x hours.
∴ The larger tap fills the tank (x − 10) hours.

⇒ Time to fill the tank by smaller tap = 1/x hours.

Time to fill the tank by larger tap = 1/x-10 hours.

Since, the tank filled by the two taps together in 1 hour
Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev
Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev

Now, according to the condition,

Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRevEx 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev
Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev
⇒ 8x− 80x = 150x − 750
⇒ 8x2 − 80x − 150x= − 750
⇒ 8x2 − 230x + 750 = 0                 ..(1)

Comparing (1) with ax2 + bx + c = 0, we get

a =8
b = −230
c = 750

∴ b2 − 4ac =(−230)2 − 4 (8) (750)
= 52900 − 24000
= 28900
Since Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev
Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev
Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRevEx 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev
Taking, the +ve sign, we get

Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev
Taking the −ve sign, we get
Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev
For Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev
which is not possible,       [∴ Time cannot be negative]

∴ x =25
⇒ x − 10 = 25 − 10 = 15
Thus, time to fill the tank by the smaller tap alone = 25 hours and larger tap alone = 15 hours.

Ques 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.

Sol: Let the average speed of the passenger train be x km/hr
∴ Average speed of the express train = (x + 11) km/hr Total distance covered = 132 km

Also, Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev

Time taken by the passenger train = 132/x hour

Time taken by the express train = 132/x + 11 hour

According to the condition, we get

Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev
Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev

⇒ − 11 (132) = − 1 (x2 + 11x)
⇒ − 1452 = − 1 (x2 + 11x)
⇒ x2 + 11x − 1452 = 0     ...(1)

Comparing (1) with ax2 + bx + c = 0,

a = 1
b = 11
c = −1452

∴ b2 − 4ac = (11)2 − 4 (1) (−1452)
⇒ b2 − 4ac = 121 + 5808 = 5929

Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev
Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev    Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev

∴ Taking +ve sign, we get

Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev

Taking −ve sign, we get

Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev

But average speed cannot be negative ∴
x ≠−44
∴ x = 33
⇒ Average speed of the passenger train = 33 km/hr
∴ Average speed of the express train = (x + 11) km/hr
= (33 + 11) km/hr
= 44 km/hr

Ques 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.

Sol: Let the side of the smaller square be x m.
⇒ Perimeter of the smaller square = 4x m.
∴ Perimeter of the larger square = (4x + 24) m

⇒ Side of the larger square = Perimeter / 4

Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev

∴ Area of the smaller square = (side)2 = (x)2 m2     
Area of the larger square = (x + 6)m2   
According to the condition,

x2 + (x + 6)2 = 468
⇒ x2 + x2 + 12x + 36 = 468
⇒ 2x2 + 12x − 432 = 0
⇒ x2 + 6x − 216 = 0  ...(1)

Comparing (1) with ax+ bx + c = 0
∴ a = 1
b = 6
c = − 216

∴ b2 − 4ac = (6)2 − 4 (1) (− 216)
= 36 + 864 = 900
Since, Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev
Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev
Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev

Taking +ve sign, we have:

Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev

Taking −ve sign, we get

Ex 4.3 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev

But the length of a square cannot be negative

∴ x = 12

⇒ Length of the smaller square = 12 m

Thus, the length of the larger square = x + 6

= 12 + 6
= 18 m

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