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

To solve such problems, we first formulate the quadratic equation from the information we have, and then we

solve it.

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(x2 + 2x)

96 = 4x2+ 8x

** Transposing **

4x2 + 8x - 96 = 0

⇒ 4(x2 + 2x - 96) = 0

⇒ x2 + 2x - 24 = 0/4

⇒ x 2+ 2x - 24 = 0

**Factorizing **

x2 + 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

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.

** Answer:**

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 t1 = time taken to travel 5.25 km upstream

then t1 = 5.25/5-x since time = Distance/Speed

Let t2 = time taken to travel 5.25 km downstream then

t2 = 5.25/5+x

Now, t1> t2

Since it takes 1 hour more to travel upstream than downstream

t1 = t2 + 1

5.25 [2x] = 1(25-x2)

10.50 x = 25 - x2

**Transposing**

x2 + 10.5x - 25 = 0

⇒ 2x2 + 21x - 50 = 0 * 2

⇒ 2 x2 + 21x - 50 = 0

⇒ 2x2 + 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.

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