Table of Contents
QUADRATIC EQUATIONS Key Points
- Standard form of quadratic equation in variable π₯ is πππ + ππ + π = π, where π, π, π are
real numbers and π β 0. - y = ππ₯2+ ππ₯ + π is called a quadratic function.
- Uses of Quadratic functions.
(i) When the rocket is fired upward, then the path of the rocket is defined by a
βquadratic function.β
(ii) Shapes of the satellite dish, reflecting mirror in a telescope, lens of the eye glasses
and orbits of the celestial objects are defined by the quadratic equations.
(iii) The path of a projectile is defined by quadratic function.
(iv) When the breaks are applied to a vehicle, the stopping distance is calculated by using
quadratic equation - Roots of quadratic equation: A real number πΌ is called a root of the quadratic equation
ππ₯2 + ππ₯ + π = 0, if ππΌ2 + ππΌ + π = 0 - Quadratic formula: The roots of a quadratic equation ππ₯2 + ππ₯ + π = 0 are given by

Nature of roots:
The nature of roots of a quadratic equation ππ₯2 + ππ₯ + π = 0 depends on π2 β 4ππ is
called the discriminant
(i) πΌπ ππ β πππ > π π‘βππ π‘βπ ππππ‘π πππ ππππ πππ πππ π‘ππππ‘
(ii) πΌπ ππ β πππ = π π‘βππ π‘βπ ππππ‘π πππ ππππ πππ πππ’ππ
(iii) πΌπ ππ β πππ < π π‘βππ π‘βπ ππππ‘π πππ πππ‘ ππππ
Try This
Check whether the following equations are quadratic or not ?
(i) π₯ 2– 6π₯ – 4 = 0
Sol: Quadratic equation
(ii) π₯ 3– 6π₯ 2 + 2π₯ -1 = 0
Sol: Not a quadratic equation(Cubic equation)
(iii) 7x = 2x 2
Sol: 2π₯2 β 7π₯ = 0
Quadratic equation


π£) (2π₯ + 1) (3π₯ + 1) = π(π₯ -1) (π₯ -2)
πππ: (2π₯ + 1) (3π₯ + 1) = π(π₯ -1) (π₯ -2)
2π₯(3π₯ + 1) + 1(3π₯ + 1) = π[π₯(π₯ β 2) β 1(π₯ β 2)]
6π₯2 + 2π₯ + 3π₯ + 1 = π[π₯2 β 2π₯ β π₯ + 2]
6π₯2 + 5π₯ + 1 = π[π₯2 β 3π₯ + 2]
6π₯2 + 5π₯ + 1 = ππ₯2 β 3ππ₯ + 2π
6π₯2 β ππ₯2 + 5π₯ + 3ππ₯ + 1 β 2π = 0
(6 β π)π₯2 + (5 + 3π)π₯ + 1 β 2π = 0
This is a quadratic equation.
(vi) 3y2 = 192
Sol: 3y2 β 192=0
This is a quadratic equation
Example-1.
i. Raju and Rajendar together have 45 marbles. Both of them lost 5 marbles each, and the
product of the number of marbles now they have is 124. We would like to find out how
many marbles they had previously. Represent the situation mathematically.
Sol: Total marbles=45
Let the number of marbles at Raju = π₯
Then the number of marbles at Rajendar = 45 β π₯
If both of them lost 5 marbles each then
The number of marbles at Raju = π₯ β 5
The number of marbles at Rajendar = 45 β π₯ β 5 = 40 β π₯
The product of remaining marbles=124
β (π₯ β 5)(40 β π₯) = 124
β 40π₯ β π₯2 β 200 + 5π₯ = 124
β βπ₯2 + 45π₯ β 200 β 124 = 0
β βπ₯2 + 45π₯ β 324 = 0
β π₯2 β 45π₯ + 324 = 0 (multiply with β1)
ii. The hypotenuse of a right triangle is 25 cm. We know that the difference in lengths of the
other two sides is 5 cm. We would like to find out the length of the two sides?
Sol: Let the length of smaller side = π₯ ππ
The length of larger side = (π₯ + 5)ππ
Length of hypotenuse = 25 ππ
In a right angle triangle we know that
(π πππ)2 + (π πππ)2 = (βπ¦πππ‘πππ’π π)2
(π₯)2 + (π₯ + 5)2 = (25)2
π₯2 + π₯2 + 2 Γ π₯ Γ 5 + 52 = 625
2π₯2 + 10π₯ + 25 β 625 = 0
2π₯2 + 10π₯ + β600 = 0
π₯2 + 5π₯ β 300 = 0
Required quadratic equation: π₯2 + 5π₯ β 300
Example-2. Check whether the following are quadratic equations:
i. (π₯ β 2)2 + 1 = 2π₯ β 3
Sol: (π₯ β 2)2 + 1 = 2π₯ β 3
β π₯2 β 4π₯ + 4 + 1 β 2π₯ + 3 = 0
β π₯2 β 6π₯ + 8 = 0
The given equation is a quadratic equation.
ii. π₯(π₯ + 1) + 8 = (π₯ + 2)(π₯ β 2)
Sol: π₯(π₯ + 1) + 8 = (π₯ + 2)(π₯ β 2)
β π₯2 + π₯ + 8 = π₯2 β 22
β π₯2 + π₯ + 8 β π₯2 + 4 = 0
β π₯ + 12 = 0
The given equation is not a quadratic equation.
iii. π₯(2π₯ + 3) = π₯2 + 1
Sol: π₯(2π₯ + 3) = π₯2 + 1
β 2π₯2 + 3π₯ β π₯2 β 1 = 0
β π₯2 + 3π₯ β 1 = 0
The given equation is a quadratic equation
iv. (π₯ + 2)3 = π₯3 β 4
Sol: (π₯ + 2)3 = π₯3 β 4 (π + π)3 = π3 + π3 + 3ππ(π + π)
β π₯3 + 23 + 3 Γ π₯ Γ 2(π₯ + 2) = π₯3 β 4
β π₯3 + 8 + 6π₯(π₯ + 2) = π₯3 β 4 β π₯3 + 8 + 6π₯2 + 12π₯ β π₯3 + 4 = 0
β 6π₯2 + 12x + 12 = 0
The given equation is a quadratic equation
EXERCISE-5.1
- Check whether the following are quadratic equations:
i. (π₯ + 1)2 = 2(π₯ β 3)
Sol: (π₯ + 1)2 = 2(π₯ β 3)
β π₯2 + 2π₯ + 1 = 2π₯ β 6
β π₯2 + 2π₯ + 1 β 2π₯ + 6 = 0
β π₯2 + 7 = 0
The given equation is a quadratic equation.
ii. π₯2 β 2π₯ = (β2)(3 β π₯)
Sol: π₯2 β 2π₯ = (β2)(3 β π₯)
β π₯2 β 2π₯ = β6 + 2π₯
β π₯2 β 2π₯ + 6 β 2π₯ = 0
β π₯2 β 4π₯ + 6 = 0
The given equation is a quadratic equation.
iii. (π₯ β 2)(π₯ + 1) = (π₯ β 1)(π₯ + 3)
Sol: (π₯ β 2)(π₯ + 1) = (π₯ β 1)(π₯ + 3)
β π₯2 + π₯ β 2π₯ β 2 = π₯2 + 3π₯ β π₯ β 3
β π₯2 β π₯ β 2 = π₯2 + 2π₯ β 3
β π₯2 β π₯ β 2 β π₯2β 2π₯ + 3 = 0
β β3π₯ + 1 = 0
The given equation is not a quadratic equation.
iv. (π₯ β 3)(2π₯ + 1) = π₯(π₯ + 5)
Sol: (π₯ β 3)(2π₯ + 1) = π₯(π₯ + 5)
β 2π₯2 + π₯ β 6π₯ β 3 = π₯2 + 5π₯
β 2π₯2 β 5π₯ β 3 β π₯2 β 5π₯ = 0
β π₯2 β 10π₯ β 3 = 0
The given equation is a quadratic equation.
v. (2π₯ β 1)(π₯ β 3) = (π₯ + 5)(π₯ β 1)
Sol: (2π₯ β 1)(π₯ β 3) = (π₯ + 5)(π₯ β 1)
β 2π₯2 β 6π₯ β π₯ + 3 = π₯2 β π₯ + 5π₯ β 5
β 2π₯2 β 7π₯ + 3 = π₯2 + 4π₯ β 5
β 2π₯2 β 7π₯ + 3 β π₯2 β 4π₯ + 5 = 0 β π₯2 β 11π₯ + 8 = 0 The given equation is a quadratic equation.
vi. π₯2 + 3π₯ + 1 = (π₯ β 2)2
Sol: π₯2 + 3π₯ + 1 = (π₯ β 2)2
β π₯2 + 3π₯ + 1 = π₯2 β 4π₯ + 4
β π₯2+ 3π₯ + 1 β π₯2 + 4π₯ β 4 = 0
β 7π₯ β 3 = 0
The given equation is not a quadratic equation.
vii. (π₯ + 2)3 = 2π₯(π₯2 β 1)
Sol: (π₯ + 2)3 = 2π₯(π₯2 β 1)
β π₯3 + 23 + 3 Γ π₯ Γ 2(π₯ + 2) = 2π₯3 β 2π₯
β π₯3 + 8 + 6π₯(π₯ + 2) = 2π₯3 β 2π₯
β π₯3 + 8 + 6π₯2 + 12π₯ β 2π₯3 + 2π₯ = 0
β βπ₯3 + 6π₯2 + 14π₯ + 8 = 0
The given equation is not a quadratic equation.
viii. π₯3 β 4π₯2 β π₯ + 1 = (π₯ β 2)3
Sol: π₯3 β 4π₯2 β π₯ + 1 = (π₯ β 2)3
β π₯3 β 4π₯2 β π₯ + 1 = π₯3 β 23 β 3 Γ π₯ Γ 2(π₯ β 2)
β π₯3 β 4π₯2 β π₯ + 1 = π₯3 β 8 β 6π₯2 + 12π₯
β π₯3 β 4π₯2 β π₯ + 1 β π₯3 + 8 + 6π₯2 β 12π₯ = 0
β 2π₯2 β 13π₯ + 9 = 0
The given equation is a quadratic equation.
- i. The area of a rectangular plot is 528 m2 . The length of the plot (in metres) is one more
than twice its breadth. We need to find the length and breadth of the plot.
Sol: Let breadth of rectangular plot (π) = π₯ π
Length of rectangular plot (π) = (2π₯ + 1)π
Given area of the rectangular plot = 528 π2
π Γ π = 528
(2π₯ + 1) Γ π₯ = 528
2π₯2 + π₯ β 528 = 0
This is the required quadratic equation.
ii. The product of two consecutive positive integers is 306. We need to find the integers.
Sol: Let the two consecutive positive integers be π₯ , π₯ + 1
Given the product of two consecutive positive integers = 306
π₯ Γ (π₯ + 1) = 306 βΉ π₯2 + π₯ β 306 = 0
This is the required quadratic equation
iii. Rohanβs mother is 26 years older than him. The product of their ages after 3 years will
be 360 years. We need to find Rohanβs present age.
Sol: Let Rohanβs age=π₯ years
Rohanβs mother age=(π₯ + 26)π¦ππππ
After 3 years
Rohanβs age=π₯ + 3 years
Rohanβs mother age=(π₯ + 26 + 3) = (π₯ + 29)π¦ππππ
Given the product of their ages after 3 years=360 years
β (π₯ + 3)(π₯ + 29) = 360
β π₯2 + 29π₯ + 3π₯ + 87 β 360 = 0
β π₯2 + 32π₯ β 273 = 0
This is the required quadratic equation.
iv. A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h
less, then it would have taken 3 hours more to cover the same distance. We need to
find the speed of the train.
Sol: Let the speed of the train= π₯ ππ/β
Distance= 480 km
ππππ(π1) =π·ππ π‘ππππ/πππππ=480-π₯/β
If the speed had been 8 km/h less then speed=(π₯ β 8) ππ/β
ππππ(π2) =π·ππ π‘ππππ/πππππ=480/π₯ β 8*β
Difference of times(π2 β π1) =3h

SOLUTIONS OF A QUADRATIC EQUATIONS BY FACTORISATION
A real number ο‘ is called a root of the quadratic equation ππ₯2 + ππ₯ + π = 0, if ππΌ2 + π π + π = 0
We also say that x = πΌ is a solution of the quadratic equation.
Example-3. Find the roots of the equation 2π₯2 β 5π₯ + 3 = 0, by factorisation.
Sol: 2π₯2 β 5π₯ + 3 = 0
2π₯2 β 2π₯ β 3π₯ + 3 = 0
2π₯(π₯ β 1) β 3(π₯ β 1) = 0
(π₯ β 1)(2π₯ β 3) = 0
π₯ β 1 = 0 ππ 2π₯ β 3 = 0
π₯ = 1 ππ π₯ =3/2
β΄ 1 πππ 3/2 πππ π‘βπ ππππ‘π ππ π‘βπ πππ’ππ‘πππ 2π₯2 β 5π₯ + 3 = 0
EXERCISE-5.2
- Find the roots of the following quadratic equations by factorisation.
(i) π₯2 β 3π₯ β 10 = 0
Sol: π₯2 β 3π₯ β 10 = 0
π₯2 β 2π₯ + 5π₯ β 10 = 0
π₯(π₯ β 2) + 5(π₯ β 2) = 0
(π₯ β 2)(π₯ + 5) = 0
π₯ β 2 = 0 ππ π₯ + 5 = 0
π₯ = 2 ππ π₯ = β5
The roots of π₯2 β 3π₯ β 10 = 0 are 2 πππ β 5
(ii) 2π₯2 + π₯ β 6 = 0
Sol: 2π₯2 + π₯ β 6 = 0
2π₯2 β 3π₯ + 4π₯ β 6 = 0
π₯(2π₯ β 3) + 2(2π₯ β 3) = 0
(2π₯ β 3)(π₯ + 2) = 0
2π₯ β 3 = 0 ππ π₯ + 2 = 0
π₯ =3/2 ππ π₯ = β2
The roots of 2π₯2 + π₯ β 6 = 0 πππ 3/2 πππ β 2
(iii) β2π₯2 + 7π₯ + 5β2 = 0
Sol: β2π₯2 + 7π₯ + 5β2 = 0
β2π₯2 + 2π₯ + 5π₯ + 5β2 = 0
β2π₯(π₯ + β2) + 5(π₯ + β2) = 0
(π₯ + β2)(β2π₯ + 5) = 0
π₯ + β2 = 0 ππ β2π₯ + 5 = 0
π₯ = ββ2 ππ π₯ =β5/β2
πβπ ππππ‘π ππ β2π₯2 + 7π₯ + 5β2 = 0 πππ β β2 πππ β5/β2
(iv) 2π₯2 β π₯ + 1/8= 0
Sol: 2π₯2 β π₯ + 1/8= 0
Multiply with β8β
8 Γ 2π₯2 β 8 Γ π₯ + 8 Γ1/8= 8 Γ 0
16π₯2 β 8π₯ + 1 = 0
16π₯2 β 4π₯ β 4π₯ + 1 = 0
4π₯(4π₯ β 1) β 1(4π₯ β 1) = 0
(4π₯ β 1)(4π₯ β 1) = 0
4π₯ β 1 = 0 ππ 4π₯ β 1 = 0
π₯ =1/4 ππ π₯ =1/4
πβπ ππππ‘π ππ 2π₯2 β π₯ + 1/8= 0 πππ 1/4 πππ 1/4
.
(v) 100π₯2 β 20π₯ + 1 = 0
Sol: 100π₯2 β 20π₯ + 1 = 0
100π₯2 β 10π₯ β 10π₯ + 1 = 0
10π₯(10π₯ β 1) β 1(10π₯ β 1) = 0
(10π₯ β 1)(10π₯ β 1) = 0
10π₯ β 1 = 0 ππ 10π₯ β 1 = 0
π₯ =1/10 ππ π₯ =1/10
πβπ ππππ‘π ππ 100π₯2 β 20π₯ + 1 = 0 πππ 1/10 πππ 1/10
(vi) π₯(π₯ + 4) = 12
Sol: π₯(π₯ + 4) = 12
π₯2 + 4π₯ β 12 = 0
π₯2 β 2π₯ + 6π₯ β 12 = 0
π₯(π₯ β 2) + 6(π₯ β 2) = 0
(π₯ β 2)(π₯ + 6) = 0
π₯ β 2 = 0 ππ π₯ + 6 = 0
π₯ = 2 ππ π₯ = β6
πβπ ππππ‘π ππ π₯(π₯ + 4) = 12 πππ 2 πππ β 6.
(vii) 3π₯2 β 5π₯ + 2 = 0
Sol: 3π₯2 β 5π₯ + 2 = 0
3π₯2 β 3π₯ β 2π₯ + 2 = 0
3π₯(π₯ β 1) β 2(π₯ β 1) = 0
(π₯ β 1)(3π₯ β 2) = 0
π₯ β 1 = 0 ππ 3π₯ β 2 = 0
π₯ = 1 ππ π₯ =2/3
πβπ ππππ‘π ππ 3π₯2β 5π₯ + 2 = 0 πππ 1 πππ 2/3
.

(ix) 3(π₯ β 4)2 β 5(π₯ β 4) = 12
Sol: 3(π₯ β 4)2 β 5(π₯ β 4) = 12
3(π₯2 β 8π₯ + 16) β 5π₯ + 20 β 12 = 0
3π₯2 β 24π₯ + 48 β 5π₯ + 8 = 0
3π₯2 β 29π₯ + 56 = 0
3π₯2 β 21π₯ β 8π₯ + 56 = 0
3π₯(π₯ β 7) β 8(π₯ β 7) = 0
(π₯ β 7)(3π₯ β 8) = 0
π₯ β 7 = 0 ππ 3π₯ β 8 = 0
π₯ = 7 ππ π₯ =8/3
πβπ ππππ‘π ππ 3(π₯ β 4)2 β 5(π₯ β 4) = 12 πππ 7 πππ 8/3
2.Find two numbers whose sum is 27 and product is 182.
Sol: Let one number = π₯, The second number= 27 β π₯
Product of numbers=182
π₯(27 β π₯) = 182
27π₯ β π₯2 = 182
βπ₯2 + 27π₯ β 182 = 0
π₯2 β 27π₯ + 182 = 0
π₯2 β 13π₯ β 14π₯ + 182 = 0
π₯(π₯ β 13) β 14(π₯ β 13) = 0
(π₯ β 13)(π₯ β 14) = 0
π₯ β 13 = 0 ππ π₯ β 14 = 0
π₯ = 13 ππ π₯ = 14
If π₯ = 13 the required numbers are 13 and 14.
If π₯ = 14 the required numbers are 14 and 13.
3. Find two consecutive positive integers, sum of whose squares is 613.
Sol: Let the two consecutive positive integers be π₯, π₯ + 1.
Sum of whose squares = 613
π₯2 + (π₯ + 1)2 = 613
π₯2 + π₯2 + 2π₯ + 1 β 613 = 0
2π₯2 + 2π₯ β 612 = 0
π₯2 + π₯ β 306 = 0
π₯2 β 17π₯ + 18π₯ β 306 = 0
π₯(π₯ β 17) + 18(π₯ β 17) = 0
(π₯ β 17)(π₯ + 18) = 0
π₯ = 17 ππ π₯ = β18
β΄ π₯ = 17 ( π ππππ π₯ ππ π πππ ππ‘ππ£π πππ‘ππππ π π π₯ β β18)
The required two consecutive positive integers are 17 , 18.

π₯2 + π₯2 β 14π₯ + 49 β 169 = 0
2π₯2 β 14π₯ β 120 = 0
π₯2 β 7π₯ β 60 = 0
π₯2 β 12π₯ + 5π₯ β 60 = 0
π₯(π₯ β 12) + 5(π₯ β 12) = 0
(π₯ β 12)(π₯ + 5) = 0
π₯ β 12 = 0 ππ π₯ + 5 = 0
π₯ = 12 ππ π₯ = β5
β΄ π₯ = 12 (π ππππ π πππ ππ π π‘πππππππ ππ πππ ππ‘ππ£π ππ‘ππππ π π π₯ β β5)
5.A cottage industry produces a certain number of pottery articles in a day. It was observed
on a particular day that the cost of production of each article (in rupees) was 3 more than
twice the number of articles produced on that day. If the total cost of production on that
day was Rs 90, find the number of articles produced and the cost of each article.
Sol: Let the number of articles produced=π₯
The cost of each article=π
π (2π₯ + 3)
Given the total cost of production on that day=Rs 90
π₯(2π₯ + 3) = 90
2π₯2 + 3π₯ β 90 = 0
2π₯2 β 12π₯ + 15π₯ β 90 = 0
2π₯(π₯ β 6) + 15(π₯ β 6) = 0
(π₯ β 6)(2π₯ + 15) = 0
π₯ β 6 = 0 ππ 2π₯ + 15 = 0
π₯ = 6 ππ π₯ =β15/2
β΄ π₯ = 6 ( ππ’ππππ ππ πππ‘πππππ ππ πππ€ππ¦π πππβ²π‘ππ πππππ‘ππ£π)
The number of articles produced= π₯ =6
The cost of each article=π
π (2π₯ + 3)=π
π (2 Γ 6 + 3) = π
π 15.
- Find the dimensions of a rectangle whose perimeter is 28 meters and whose area is 40
square meters.
Sol: Let the length of the rectangle=π , breadth =π
Perimeter of the rectangle=28 m
2(π + π) = 28 β π + π =28/2
β π + π = 14 β π = 14 β π
Area of the square=40 square meters.
β π Γ π = 40
β π(14 β π) = 40
β 14π β π2 β 40 = 0
β βπ2 + 14π β 40 = 0
β π2 β 14π + 40 = 0
β π2 β 10π β 4π + 40 = 0
β π(π β 10) β 4(π β 10) = 0
β (π β 10)(π β 4) = 0
β π β 10 = 0 ππ π β 4 = 0
β π = 10 ππ π = 4
πΌπ π = 10 π π‘βππ π = 14 β 10 = 4 π
πΌπ π = 4 π π‘βππ π = 14 β 4 = 10 π
The dimensions of the rectangle are 10 m and 4 m.
- The base of a triangle is 4cm longer than its altitude. If the area of the triangle is 48 sq.cm
then find its base and altitude.
Sol: Let altitude (h)=π₯
The base of a triangle(b)=(π₯ + 4)
The area of the triangle = 48 sq.cm
1/2Γ π Γ β = 48
(π₯ + 4) Γ π₯ = 48 Γ 2
π₯2 + 4π₯ β 96 = 0
π₯2 β 8π₯ + 12π₯ β 96 = 0
π₯(π₯ β 8) + 12(π₯ β 8) = 0
(π₯ β 8)(π₯ + 12) = 0
π₯ β 8 = 0 ππ π₯ + 12 = 0
π₯ = 8 ππ π₯ = β12
β΄ π₯ = 8 ( π΄ππ‘ππ‘π’ππ πππβ²π‘ππ πππππ‘ππ£π)
Altitude of the triangle=π₯ = 8 ππ
Base of the triangle=π₯ + 4 = 8 + 4 = 12 ππ.
8.Two trains leave a railway station at the same time. The first train travels towards west
and the second train towards north. The first train travels 5 km/hr faster than the second
train. If after two hours they are 50 km. apart find the average speed of each train.
Sol: Let the speed of second train=π₯ ππ/βπ
The speed of first train=(π₯ + 5)ππ/βπ
Time (t)=2 hr
Distance travelled by second train=π Γ π‘ = 2 Γ π₯ = 2π₯ ππ
Distance travelled by first train=π Γ π‘ = 2 Γ (π₯ + 5) = (2π₯ + 10)ππ

- In a class of 60 students, each boy contributed rupees equal to the number of girls and
each girl contributed rupees equal to the number of boys. If the total money then collected
was βΉ1600. How many boys are there in the class?
Sol: Total number of students=60
Let the number of boys=π₯
The number of girls=60 β π₯
Money contributed by the boys=π₯ Γ (60 β π₯) = 60π₯ β π₯2
Money contributed by the boys=(60 β π₯) Γ π₯ = 60π₯ β π₯2
Total money collected=βΉ1600
60π₯ β π₯2 + 60π₯ β π₯2 = 1600
β2π₯2 + 120π₯ β 1600 = 0
π₯2 β 60π₯ + 800 = 0(πππ£πππππ ππ¦ β² β 2β²)
π₯2 β 40π₯ β 20π₯ + 800 = 0
π₯(π₯ β 40) β 20(π₯ β 40) = 0
(π₯ β 40)(π₯ β 20) = 0
π₯ β 40 = 0 ππ π₯ β 20 = 0
π₯ = 40 ππ π₯ = 20
β΄ The number of boys in the class=40 or 20.
- A motor boat heads upstream a distance of 24km on a river whose current is running at 3
km per hour. The trip up and back takes 6 hours. Assuming that the motor boat
maintained a constant speed, what was its speed?
Sol: Let the speed of the boat in water=π₯ ππ/βπ
The speed of the river=3 ππ/βπ
The distance of river=24 km
The speed of the boat in upstream= (π₯ β 3) ππ/βπ

π₯2 β 8π₯ β 9 = 0
π₯2 β 9π₯ + 1π₯ β 9 = 0
π₯(π₯ β 9) + 1(π₯ β 9) = 0
(π₯ β 9)(π₯ + 1) = 0
π₯ β 9 = 0 ππ π₯ + 1 = 0
π₯ = 9 ππ π₯ = β1
β΄ π₯ = 9 ( π ππππ ππ π‘βπ ππππ‘ ππ πππβ²π‘ πππππ‘ππ£π)
The speed of the boat in water= 9 km/hr.


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