We shall now discuss problems that can be solved with quadratic equations.

Consider these examples.

**Example 1**

The product of two consecutive numbers is 72. Find the numbers.

**Solution :**

Let the smaller number = x

Let the larger number = x+1

Their product = x(x+1)

We are given that

x(x + 1) = 72

⇒ x2 + x = 72

⇒ x2 + x - 72 = 0

Factorizing

x2 + 9x - 8x - 72 = 0

x(x + 9) - 8(x + 9) = 0

(x - 8) (x + 9) = 0

⇒ x - 8 = 0 or x + 9 = 0

⇒ x= 8 or x = - 9

⇒ x + 1 = 8 + 1 or x + 1 = -9 + 1

= 9 = -8

Therefore, the two consecutive numbers are 8, 9 or -9, -8

**Example 2**

A man walks a distance of 48 kilometers in a certain amount of time. If he increased his speed by 2 km/hr,

he would have reached his destination four hours earlier. Find his usual speed.

** Solution :**

Let the man's usual speed = x km/hr

Time taken to walk 48 km = (48/x) hrs

Time = Distance/Speed

Increased speed = (x + 2) km/hr

Time taken to walk 48 km =48/x+2

At a speed of (x + 2) km/hr, the man reaches his destination 4 hours earlier.

⇒ 96 = 4(x^{2} + 2x)

96 = 4x^{2}+ 8x

Transposing

4x^{2} + 8x - 96 = 0

⇒ 4(x^{2} + 2x - 96) = 0

⇒ x^{2 }+ 2x - 24 = 0/4

⇒ x^{2}+ 2x - 24 = 0

Factorizing

⇒ x^{2 }+ 6x - 4x - 24 = 0

x(x + 6) - 4 (x+6) = 0

(x - 4) (x + 6) = 0

⇒ x - 4 = 0 or x + 6 = 0

⇒ x = 4 or x = -6.

Since the man cannot walk at a negative pace, we reject x = -6

Speed = x =4km/hr

**Example 3**

Ann can row her boat at a speed of 5 km/hr in still water. If it takes her one hour more to row her boat 5.25 km

upstream than to return downstream, find the speed of the stream.

Speed of boat = 5 km/hr

Let the speed of the stream current = x km/hr.

Speed of the boat upstream (against the current)

= (5 - x) km/hr

Speed of the boat downstream (with the current)

= (5 + x) km/hr.

Let t_{1} = time taken to travel 5.25 km upstream

then t_{1} = 5.25/5-x since time = Distance/Speed

Let t_{2} = time taken to travel 5.25 km downstream then

t_{2} = 5.25/5+x

Now, t_{1} t_{2}

Since it takes 1 hour more to travel upstream than downstream

t_{1} = t_{2} + 1

5.25 [2x] = 1(25-x^{2})

10.50 x = 25 - x^{2}

Transposing

x^{2} + 10.5x - 25 = 0

⇒ 2x^{2} + 21x - 50 = 0 * 2

⇒ 2 x^{2} + 21x - 50 = 0

⇒ 2x^{2} + 25x - 4x - 50 = 0

⇒ x(2x + 25) - 2(2x + 25) = 0

⇒ (x-2) (2x + 25) = 0

⇒ x-2 = 0 or 2x + 25 = 0

⇒ x = 2 or 2x = - 25

or x=-25/2

= -12.5 km/hr.

Since the speed of the stream cannot be negative we reject

x = -12.5km/hr.

∴Speed of the stream = 2 km/hr.

**I. To solve by completing the square**

- The product of 2 numbers exceeds their sum by 14. If their sum is 10 find the numbers.
- A plan of a rectangular hall has a ground area of 180 sq.m. The plan is altered so that the length

diminishes by 3m and the breadth increases by 2m without altering the area. Find the original length

and breadth and the new length and breadth of the hall. - A number consists of two digits, the digit in the units place is the square of the digit in the tens place.

The number formed by reversing the digits exceeds twice the number by 15. Find the number. - Two taps when opened together can fill a cistern in 30 minutes. The larger tap takes 25 minutes less to

fill the cistern, than the smaller one. Find the time taken by each tap, to fill the cistern, separately. - If a body is projected upward from the surface of the earth with a velocity, v, the altitude, h, reached at

any time t, is given by h=vt - 16t2 . If v = 40, h = 9 find the corresponding value of t. - The bending moment for a certain uniform beam is given by the formula

M = 25x -(wx2/2) where M is the bending moment at a

distance of x meters from one end of the beam, and w is the weight parameter of the beam. Find how

far from the farther end is the bending moment 30 when the weight parameter is 10kg. - A plane left 40 minutes late due to bad weather and in order to reach its destination 1600 km away in

time, it had to increase its speed by 400 km/hr from its usual speed. Find its usual speed, the time

normally taken, and the new speed. - A piece of wire 3m long is to be bent in the form of a right angled triangle with a hypotenuse of length

1.3m. Find the lengths of the other two sides. - A head wind reduced a cyclist's speed by 3km/hr, as a result he took 30 minutes more to travel a

distance of 30 km. What would have been his speed had there been no head wind. - Two trains leave a railway station at the same time. The first train travels due west and the second train

due north. The first train travels 5 km/h faster than the second train. If after two hours they are 50 km

apart, find the speed of each train.

1. Let the numbers be x and y

Their sum is 10

∴ x + y = 10

or y = 10 – x

The product exceeds their sum by 14

So

x (10 – x) = 10+14

⇒ 10x – x^{2} = 24

Transposing

x^{2} – 10x + 24 = 0

Factorizing

x^{2} – 6x – 4x + 24 = 0

x(x – 6) – 4 (x – 6) = 0

(x – 4) (x – 6) = 0

⇒x – 4 = 0 or x – 6 = 0

⇒ x = 4 or x = 6

⇒ y = 10 – x or y = 10 – x

⇒ y = 10 – 4 or y = 10 – 6

⇒ y = 6 or y = 4

∴ The number are 4, 6

2. Let the original length = l meters

Let the original breadth = b meters

Area = length * breadth

⇒ 180 = l * b

⇒ l = 180/b

In the altered plans

New length = (l – 3)meters

New breadth = (b + 2) meters

Area = (l – 3) (b + 2)

or 180 = (l – 3) (b + 2) ----------(1)

cross multiplying

180 b = 1(180 – 3b) (b + 2)

180 b = (180 – 3b)b + (180-3b)^{2}

180 b = 180b – 3B^{2} + 360 – 6b

Transposing

180b – 180b + 3B^{2} – 360 + 6b = 0

3B^{2} + 6b – 360 = 0

3(B^{2} + 2b – 120) = 0

B^{2} + 2b – 120 = 0/3

B^{2} + 2b – 120 = 0

Factorizing

B^{2}+ 12b – 10b – 120 = 0

b(b + 12) – 10 (b + 12) = 0

(b – 10) (b + 12) = 0

⇒ b – 10 = 0 or b + 12 = 0

⇒ b = 10 or b = - 12

The breadth cannot be negative

∴ b = 10 m.

l = 180/b

= 180/10

= 18 m.

original length = 18m

original breadth = 10m

new length = l-3 = 18-3 = 15m

new length = b+2 = 10 + 2 = 12m.

3. If we could write the 2 digit number in the form original number reversed number

Let x be the digit in the tens place then x^{2} is the digit in the units place.

Original number = 10 * x + x^{2}

= 10x + x^{2}

Twice the original number = 2(10x + x^{2})

Reversed number = 10 * x^{2} + x

= 10x^{2} + x

The reversed number exceeds twice the original number by 15

10x^{2} + x = 2(10x + x^{2}) + 15

10x^{2} + x = 20x + 2x^{2} + 15

Transposing

10 x^{2} + x – 20x – 2x^{2} – 15 = 0

8 x^{2} – 19x – 15 = 0

Factorizing

8x^{2} –24x + 5x – 15 = 0

8x (x-3) + 5(x-3) = 0

(8x+5) (x-3) = 0

⇒ 8x + 5 = 0 or x – 3 = 0

⇒ 8x = - 5 or x = 3

⇒ x =- 5/8 or x = 3

Since a digit cannot be a rational number

x = 3

∴ x^{2} = (3)^{2} = 9

The original number = 10x + x^{2}

= 30 + 9

= 39

4. Let the time taken by the smaller tap to fill the cistern = x minutes

Time taken by larger tap to fill the listern =(x-25 ) minutes.

In 1 minute the smaller tap fills1/x of the cistern

In 1 minute the larger tap fills 1/x-25 of the cistern.

Since it takes 30 minutes to fill the cistern. The work done by both the taps in 1 minute is

(2x – 25) * 30 = 1(x^{2} – 25x)

60x – 750 = x^{2} – 25x

Transposing

x^{2} –25x – 60x + 750 = 0

x^{2} –85x + 750 = 0

x^{2} –75x – 10x + 750 = 0

x(x–75) – 10 (x–75) = 0

(x-10) (x-75) = 0

⇒ x – 10 = 0 or x – 75 = 0

⇒ x = 10 or x = 75

If x = 10 mins

Then x –25 = 10-25 = -15 minutes.

Since time is not negative we reject x = 10

∴ x = 75 mins

Time taken by the smaller tap = 75 minutes

Time taken by the larger tap = 75 – 25 = 50 minutes

5. Given h= vt – 16t_{2} (1)

and v = 40m/s h = 9m.

Substituting in ---------------------- (1)

9 = 40t – 16t_{2}

Transposing

9 - 40t + 16t_{2} = 0

16t_{2} – 40t + 9 = 0

Factorizing

16t_{2} – 36t –4t + 9 = 0

4t(4t-9) –1 (4t-9) = 0

(4t – 1) (4t – 9) = 0

⇒ 4t – 1 = 0 or 4t – 9 = 0

⇒ 4t = 1 or 4t = 9

⇒ t =1/4 or t = 9/4

⇒ t = 0.25 sec or t = 2.25 secs.

6.

30 = 25x – 5x^{2}

Dividing throughout by 5

Transposing

x^{2} – 5x + 6 = 0

x^{2} – 3x – 2x + 6 = 0

x (x-3) – 2(x-3) = 0

(x-2) (x-3) = 0

⇒ x – 2 = 0 or x – 3 = 0

⇒ x = 2 m or x = 3m.

Since the further end is to be obtained

x = 3m.

7. Let the planes usual speed = x km/hr.

Distance = 1600 km.

Usual time = 1600/x

Time = Distance/Speed

New speed = (x+400) km/hr.

New time = 1600/(x+400)

Since the plane is late by 40 minutes, it has to gain those 40 minutes to reach its destination on time.

∴ 1600/x =1600/(x+400) +40/60 (Minutes and seconds are generally converted into hours

Cross Multiplying

3(640000) = 2 (x^{2} + 400x)

⇒ 1920000 = 2x^{2} + 800x

Dividing both sides by 2 and transposing

x^{2} + 400x = 960000

x^{2} + 400x – 960000 = 0

x^{2} + 1200x – 800x – 960000 = 0

x(x+1200) – 800 (x+1200) = 0

(x-800) (x+1200) = 0

x – 800 = 0 or x + 1200 = 0

x = 800 or x = -1200

Since the speed cannot be negative

x = 800 km/hr

Time usually taken = 1600/800 = 2 hours.

New speed = x + 400

= 800 + 400

= 1200 km/hr.

8. Total length of the wire = 3m

Let ABC be the right triangle

Let AB = x

and BC = y

CA = 1.3 m

Perimeter of Δ ABC = length of wire

∴ AB + BC + CA = 3

x + y + 1.3 = 3

x + y = 3 – 1.3

x + y = 1.7

y = 1.7 - x

By Pythagoras theorem

AB2 + BC2 = CA^{2}

x^{2} + y2 = (1.3)^{2}

Substituting y = 1.7 – x

x 2 + (1.7–x)^{2}= (1.3)^{2}

⇒ x^{2} + 2.89 – 3.4x+ x^{2} = 1.69

Transposing

2x^{2} – 3.4x + 2.89 – 1.69 = 0

2 x^{2} – 3.4x + 1.20 = 0

2 x^{2} – 3.4x + 1.2 = 0

9. Let the cyclist's speed = x km/hr

Time taken to travel 30 km = Distance/Speed

= 30/x

Because of the head wind, his speed is reduced by 3 km/hr.

So his speed against the head wind = (x-3) km/hr

Time taken to travel at reduced speed = 30/x-3

Time taken is 30 minutes more than it would have been if there were no head wind.

Cross Multiplying

90*2 = 1*(x^{2>} – 3x)

180 = x^{2} – 3x

Transposing

x^{2} –3x – 180 = 0

Factorizing

x^{2}–15x+12x–180 = 0

x(x-15)+12(x-15) = 0

(x+12) (x-15) = 0

⇒ x+12 = 0 or x-15 = 0

⇒ x= -12 or x = 15

Since the cyclists speed cannot be negative we reject x = -12

Speed = 15 km/hr.

⇒ 20x^{2} – 34x + 12 = 0

⇒ 20x^{2} – 24x-10x + 12 = 0

4x (5x-6) –2(5x-6) = 0

⇒ (4x-2) (5x-6) = 0

⇒ 4x-2 = 0 or 5x-6 = 0

⇒ 4x = 2 or 5x = 6

⇒ x = 2/4 or x = 6/5

⇒ x = 0.5 or x = 1.2

If x = 0.5m y = 1.7-x = 1.7 – 0.5 = 1.2 m

If x = 1.2m y = 1.7-x = 1.7 – 1.2 = 0.5 m

The sides are 0.5m and 1.2m.

10. Let O be the station from which the trains start their journey. In 2 hours time the first train is at A

and the second train is at B.

Distance AB = 50 km.

Let the speed of the 2nd train = x km/hr

The speed of the 1st train = (x+5) km/hr

Distance traveled by 1st train = speed x time

In 2 hours = 2*(x+5)

= OA = (2x + 10) km

Distance travelled by the 2nd train

In 2 hours = OB = 2*x

= (2x) km.

By Pythagoras theorem in a right angled triangle

OA^{2} + OB^{2} = AB^{2}

By Pythagoras theorem in a right angled triangle

OA^{2} + OB^{2} = AB^{2}

(2x+10)^{2} + (2x)^{2} = (50)^{2}

4x^{2} + 40x + 100 + 4x^{2} = 2500

Transposing.

8x^{2} + 40x + 100 – 2500 = 0

8x^{2} + 40x – 2400 = 0

Dividing through out by 8

x^{2} + 5x – 300 = 0

Factorizing

x^{2} + 20x – 15x – 300 = 0

x(x+20) – 15(x+20) = 0

⇒ x-15 = 0 or x+20 = 0

⇒ x = 15 or x = -20

Since the speed cannot be negative x =-20 is rejected

X = 15 km/hr.

First trains speed = x+5 = 15+5 = 20 km/hr.

2nd train speed = x = 15 km

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