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.
