Class 10 - Mathematics
Quadratic Equations -Exercise 4.3
Top Block 1
Exercise 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
Answer :
(i) 2x2 – 7x + 3 = 0
Divide by 2 on both sides, we get
2x2/2 – 7x/2 + 3/2 = 0/2
⇒ x2 – 7x/2 + 3/2 = 0
⇒ x2 – 7x/2 + 3/2 + (7/4)2 – (7/4)2 = 0
⇒ x2 – 7x/2 + (7/4)2 + 3/2 – (7/4)2 = 0
⇒ (x – 7/4)2 + 3/2 – 49/16 = 0
⇒ (x – 7/4)2 + (3 * 8 – 49 * 1)/16 = 0
⇒ (x – 7/4)2 + (24 – 49)/16 = 0
⇒ (x – 7/4)2 – 25/16 = 0
⇒ (x – 7/4)2 = 25/16
⇒ x – 7/4 = ±√(25/16)
⇒ x – 7/4 = ±5/4
Case 1: When 5/4 is positive
x – 7/4 = 5/4
⇒ x = 5/4 + 7/4
⇒ x = (5 + 7)/4
⇒ x = 12/4
⇒ x = 3
Case 2: When 5/4 is negative
x – 7/4 = -5/4
⇒ x = -5/4 + 7/4
⇒ x = (-5 + 7)/4
⇒ x = 2/4
⇒ x = 1/2
Hence, the required roots are: 3 and 1/2
(ii) 2x2 + x – 4 = 0
Divide by 2 on both sides, we get
2x2/2 + x/2 – 4/2 = 0/2
⇒ x2 + x/2 – 2 = 0
⇒ x2 + x/2 – 2 + (1/4)2 – (1/4)2 = 0
⇒ x2 + x/2 + (1/4)2 – 2 – (1/4)2 = 0
⇒ (x + 1/4)2 – 2 – 1/16 = 0
⇒ (x + 1/4)2 – 33/16 = 0
⇒ (x + 1/4)2 = 33/16
⇒ x + 1/4 = ±√(33/16)
⇒ x + 1/4 = ±√33/4
Case 1: When √33/4 is positive
⇒ x + 1/4 = √33/4
⇒ x = √33/4 – 1/4
⇒ x = (√33 – 1)/4
Case 2: When √33/4 is negative
⇒ x + 1/4 = -√33/4
⇒ x = -√33/4 – 1/4
⇒ x = -(√33 + 1)/4
Hence, the required roots are: (√33 – 1)/4 and -(√33 + 1)/4
(iii) 4x2 + 4√3x + 3 = 0
Divide by 4 on both sides, we get
4x2/4 + 4√3x/4 + 3/4 = 0/4
⇒ x2 + √3x + 3/4 = 0
⇒ x2 + √3x + 3/4 + (√3/2)2 – (√3/2)2 = 0
⇒ (x + √3/2)2 + 3/4 – 3/4 = 0
⇒ (x + √3/2)2 = 0
⇒ (x + √3/2) (x + √3/2) = 0
⇒ x = -√3/2, -√3/2
Hence, the required roots are: √3/2 and -√3/2
(iv) 2x2 + x + 4 = 0
Divide by 2 on both sides, we get
2x2/2 + x/2 + 4/2 = 0/2
⇒ x2 + x/2 + 2 = 0
⇒ x2 + x/2 + 2 + (1/4)2 – (1/4)2 = 0
⇒ (x + 1/4)2 + 2 – 1/16 = 0
⇒ (x + 1/4)2 + 31/16 = 0
⇒ (x + 1/4)2 = -31/16
Since square of a number cannot be negative,
So, (x + 1/4)2 cannot be a real value.
Hence, there is no real value of x satisfying the given equation.
Question : 2:Find the roots of the quadratic equations given in Q.1 above by applying the quadratic formula.
Answer :
(i) 2x2 – 7x + 3 = 0
Comparing the given quadratic equation with ax2 + bx + c = 0, we have:
a = 2, b= – 7, c = 3
Now discriminant = b2 – 4ac
= (-7) 2 – 4 * 2 * 3
= 49 – 24
= 25 > 0
Since b2 – 4ac is positive.
Hence, the given quadratic equation has real roots. The roots are given as:
x = {-b ± √(b2 – 4 * a * c)}/2a
⇒ x = [-(-7) ± √{(-7)2 – 4 * 2 * 3)}]/(2 * 2)
⇒ x = {7 ± √(49 – 24)}/4
⇒ x = (7 ± √25)/4
⇒ x = (7 ± 5)/4
Taking positive sign, we get
⇒ x = (7 + 5)/4
⇒ x = 12/4
⇒ x = 3
Taking negative sign, we get
⇒ x = (7 – 5)/4
⇒ x = 2/4
⇒ x = 1/2
Thus, the roots are: 3 and 1/2
(ii) 2x2 + x – 4 = 0
Comparing the given quadratic equation with ax2 + bx + c = 0, we have:
a = 2, b= 1, c = -4
Now discriminant = b2 – 4ac
= 1 2 – 4 * 2 * (-4)
= 1 + 32
= 33 > 0
Since b2 – 4ac is positive.
Hence, the given quadratic equation has real roots. The roots are given as:
x = {-b ± √(b2 – 4 * a * c)}/2a
⇒ x = [-1 ± √{12 – 4 * 2 * (-4)}]/(2 * 2)
⇒ x = {-1 ± √(1 + 32)}/4
⇒ x = (-1 ± √33)/4
Taking positive sign, we get
⇒ x = (-1 + √33)/4
Taking negative sign, we get
⇒ x = (-1 – √33)/4
Thus, the roots are: (-1 + √33)/4 and (-1 – √33)/4
(iii) 4x2 + 4√3x + 3 = 0
Comparing the given quadratic equation with ax2 + bx + c = 0, we have:
a = 4, b= 4√3, c = 3
Now discriminant = b2 – 4ac
= (4√3) 2 – 4 * 4 * 3
= 48 – 48
= 0
Since b2 – 4ac is zero.
Hence, the given quadratic equation has equal roots. The roots are given as:
x = {-b ± √(b2 – 4 * a * c)}/2a
⇒ x = [-4√3 ± √{(4√3)2 – 4 * 4 * 3)}]/(2 * 4)
⇒ x = {-4√3 ± √(48 – 48)}/8
⇒ x = (-4√3 ± 0)/8
⇒ x = -4√3/8
⇒ x = -√3/2
Thus, the roots are: -√3/2 and -√3/2
(iv) 2x2 + x + 4 = 0
Comparing the given quadratic equation with ax2 + bx + c = 0, we have:
a = 2, b= 1, c = 4
Now discriminant = b2 – 4ac
= 1 2 – 4 * 2 * 4
= 1 – 32
= -31 < 0
Since b2 – 4ac is negative.
Hence, the given quadratic equation has no real roots.
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
Answer :
(i) Given, x – 1/x = 3
⇒ (x2 – 1)/x = 3
⇒ x2 – 1 = 3x
⇒ x2 – 3x – 1 = 0
Comparing the given quadratic equation with ax2 + bx + c = 0, we have:
a = 1, b= -3, c = -1
Now discriminant = b2 – 4ac
= (-3) 2 – 4 * 1 * (-1)
= 9 + 4
= 13 > 0
Since b2 – 4ac is positive.
Hence, the given quadratic equation has real roots. The roots are given as:
x = {-b ± √(b2 – 4 * a * c)}/2a
⇒ x = [-(-3) ± √{(-3)2 – 4 * 1 * (-1)}]/(2 * 1)
⇒ x = {3 ± √(9 + 4)}/2
⇒ x = (3 ± √13)/2
Taking positive sign, we get
⇒ x = (3 + √13)/2
Taking negative sign, we get
⇒ x = (3 – √13)/2
Thus, the roots are: (3 + √13)/2 and (3 – √13)/2
(ii) Given, 1/(x + 4) – 1/(x – 7) = 11/30
⇒ {(x – 7) – (x + 4)}/{(x + 4)(x – 3)} = 11/30
⇒ (x – 7 – x – 4)/(x2 – 3x – 28) = 11/30
⇒ -11/(x2 – 3x – 28) = 11/30
⇒ -1/(x2 – 3x – 28) = 1/30
⇒ x2 – 3x – 28 = -30
⇒ x2 – 3x – 28 + 30 = 0
⇒ x2 – 3x + 2 = 0
⇒ x2 – x – 2x + 2 = 0
⇒ x(x – 1) – 2(x – 1) = 0
⇒ (x – 1)(x – 2) = 0
⇒ x = 1, 2
Thus, the roots of the equation are: 1, 2
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.
Answer :
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,
1/(x – 3) + 1/(x + 5) = 1/3
⇒ {(x + 5) + (x – 3)}/ {(x + 5)(x – 3)} = 1/3
⇒ 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
⇒ x2 – 7x + 3x – 21 = 0
⇒ x(x – 7) + 3(x – 7) = 0
⇒ (x – 7)(x + 3) = 0
⇒ x = 7, -3
Since age cannot be negative,
So, x = 7
Hence, the present age of Rehman = 7 years
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.
Answer :
Let, Shefali’s marks in Mathematics = x
So, the marks in English = (30 – x) [Since 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 + 27x – 2x + 54 = 210
⇒ -x2 + 25x + 54 = 210
⇒ -x2 + 25x + 54 – 210 = 0
⇒ -x2 + 25x – 156 = 0
⇒ x2 – 25x + 156 = 0
⇒ x2 – 13x – 12x + 156 = 0
⇒ x(x – 13) – 12(x – 13) = 0
⇒ (x – 13)(x – 12) = 0
⇒ x = 12, 13
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
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.
Answer :
Let the shorter side (i.e., breadth) = x metres.
So, the longer side (length) = (x + 30) metres.
In a rectangle,
Diagonal = √{(breadth)2 + (length)2}
⇒ x + 60 = √{x2 + (x + 30)2}
Squaring on both sides, we get
(x + 60)2 = x2 + (x + 30)2
⇒ (x + 60)2 = x2 + x2 + 60x + 900
⇒ (x + 60)2 = 2x2 + 60x + 900
⇒ x2 + 120x + 3600 = 2x2 + 60x + 900
⇒ 2x2 – x2 + 60x – 120x + 900 – 3600 = 0
⇒ x2 – 60x – 2700 = 0
⇒ x2 – 90x + 30x – 2700 = 0
⇒ x(x – 90) + 30(x – 90) = 0
⇒ (x – 90)(x + 30) = 0
⇒ x = 90, -30
Since breadth cannot be negative
So, x = 90
Now, x + 30 = 90 + 30=120
Thus, the shorter side = 90 m
The longer side = 120 m.
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.
Answer :
Let the larger number be x.
Since (smaller number) 2 = 8 (larger number)
⇒ (smaller number)2 = 8x
⇒ smaller number = √(8x)
According to question,
x2 – √(8x)2 = 180
⇒ 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
Since x ≠ -10 as x is the larger of two numbers.
So, x = 18
Hence, the larger number = x = 18
Smaller number = √(8 * 18) = √144 = 12
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.
Answer :
Let the speed of the train is x km/hr
Given, the train travelling at a uniform speed for 360 km would have taken 48 minutes less to
travel the same distance
if its speed work 5 km per hour more.
⇒ 360/x – 360/(x + 5) = 48/60 {since 48 minutes = 48/60 hours}
⇒ 360{1/x – 1/(x + 5)} = 4/5
⇒ 360[{(x + 5) – x}/{x(x + 5)}] = 4/5
⇒ 360[{x + 5 – x}/{x(x + 5)}] = 4/5
⇒ 360[5/{x2 + 5x}] = 4/5
⇒ 1800/{x2 + 5x}] = 4/5
⇒ 1800*5 = 4(x2 + 5x)
⇒ (1800*5)/4 = x2 + 5x
⇒ x2 + 5x = 450 * 5
⇒ x2 + 5x = 2250
⇒ x2 + 5x – 2250 = 0
⇒ x2 – 45x + 50x – 2250 = 0
⇒ x(x – 45) + 50(x – 45) = 0
⇒ (x – 45)*(x + 50) = 0
⇒ x = 45, -50
Since speed cannot be negative
So, x = 45
Hence, the speed of the train is 45 km/hr
Mddle block 1
Question : 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.
Answer :
Let the larger tap fills the tank in x hours.
So, the smaller tap fills the tank (x + 10) hours.
Time taken to fill tank by larger tap in one hour = 1/x hours
Time taken to fill tank by Smaller tap in one hour = 1/(x + 10) hours
Now, the tank filled by the two taps together in one hour = 1/x + 1/(x + 10)
= (x + 10 + x)/(x2 + 10x)
= (2x + 10)/(x2 + 10x)
Now, according to question,
(2x + 10)/(x2 + 10x) = 1/9 3/8
⇒ (2x + 10)/(x2 + 10x) = 1/(75/8)
⇒ (2x + 10)/(x2 + 10x) = 8/75
⇒ 75(2x + 10) = 8(x2 + 10x)
⇒ 150x + 750 = 8x2 + 80x
⇒ 8x2 + 80x – 150x – 750 = 0
⇒ 8x2 – 70x – 750 = 0
⇒ 4x2 – 35x – 375 = 0 [divide by 4]
⇒ 4x2 – 60x + 25x – 375 = 0
⇒ 4x(x – 15) + 25(x – 15) = 0
⇒ (x – 15)(4x + 25) = 0
⇒ x = 15, -25/4
Since x is the time taken to fill the tank cannot be negative
So, x = 15
Hence, the larger tap fills the tank in 15 hours.
and the smaller tap fills the tank in (x + 10) = (15 + 10) = 25 hours.
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 speed of the express train is 11km/h
more than that of the passenger train, find the average speed of the two trains.
Answer :
Let the speed of the passenger train = x km/h
Therefore, the speed of express train = (x + 11) km/h
Distance travelled = 132 km
Time taken by passenger train = 132/x hours
Time taken by express train = 132/(x + 11) hours
According to question,
132/x = 132/(x + 11) + 1
⇒ 132/x – 132/(x + 11) = 1
⇒ {132(x + 11) – 132x}/{x(x + 11)} = 1
⇒ 132x + 1452 – 132x = x(x + 11)
⇒ 1452 = x(x + 11)
⇒ x2 + 11x – 1452 = 0
⇒ x2 + 44x – 33x – 1452 = 0
⇒ x(x + 44) – 33(x + 44) = 0
⇒ (x – 33)(x + 44) = 0
⇒ x = 33, -44
Since, the speed of the train cannot be negative
So, x = 33
Hence, the speed of the passenger train = 33 km/h
Therefore, the speed of express train = (33 + 11) = 44 km/h
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.
Answer :
Let the side of larger square = x m
And side of the smaller square = y m
According to question,
x2 + y2 = 468 …………1
Difference between perimeters is
4x – 4y = 24
⇒ 4(x – y) = 24
⇒ x – y = 24/4
⇒ x – y = 6
⇒ x = y + 6 ………2
Put value of x in equation 1, we get
(y + 6)2 + y2 = 468
⇒ y2 + 36 + 12y + y2 = 468
⇒ 2y2 + 12y + 36 – 486 = 0
⇒ 2y2 + 12y – 432 = 0
⇒ y2 + 6y – 216 = 0 [Divide by 2]
⇒ y2 + 18y – 12y – 216 = 0
⇒ y(y + 18) – 12(y + 18) = 0
⇒ (y + 18)(y – 12) = 0
⇒ y = 12, -18
Since length of square cannot be negative
So, y = 12
So, the side of the smaller square = y = 12 m
and the side of larger square = x = y + 6 = 12 + 6 = 18 m
