We can follow the steps given below to solve word problems on quadratic equations.

**Step 1 :**

Convert the information given into to a quadratic equation.

**Step 2 :**

Solve the quadratic equation obtained using one of the methods given below.

1. Factoring

2. Quadratic formula

3. Completing square method

**Step 3 :**

Relate the solution obtained from one of the methods to the statement asked in the question.

**Problem 1 :**

If the difference between a number and its reciprocal is 24/5, find the number.

**Solution :**

**Let "x" be the required number "1/x" be its reciprocal.**

**x - (1/x) = 24/5**

**(x ^{2} - 1)/x = 24/5**

**5****(x ^{2} - 1) = 24x**

**5****x ^{2} - 5 = 24x**

**5****x ^{2} - 24x - 5 = 0**

**5****x ^{2} - 25x + 1x - 5 = 0**

**5x(x - 5) + 1(x - 5) = 0**

**(5x - 1) (x - 5) = 0**

**5x - 1 = 0 or x - 5 = 0**

**x = 1/5 or x = 5**

So, the required numbers are 5 and 1/5.

**Problem 2 :**

A garden measuring 12m by 16m is to have a pedestrian pathway that is ‘w’ meters wide installed all the way around so that it increases the total area to 285 m^{2}. What is the width of the pathway?

**Solution :**

From the picture given above, length of the garden including pathway is 12 + 2w and width is 16 + 2w.

length ⋅ width = 285 m^{2}

(12 + 2w) ⋅ (16 + 2w) = 285

192 + 24w + 32w + 4w^{2} = 285

4w^{2 }+ 56w + 192 - 285 = 0

4w^{2 }+ 56w - 93 = 0

a = 4, b = 56 , c = -93

x = [-b ± √b^{2} - 4ac]/2a

x = [-56 ± √(56)^{2} - 4(4)(-93)]/2(4)

x = [-56 ± √(3136 + 1488)]/8

x = [-56 ± 68]/8

x = (-56 + 68) /8 and x = (-56 - 68)/8

x = 12/8

x = 1.5 m

So, the required width is 1.5 m.

**Problem 3 :**

A bus covers a distance of 90 km at a uniform speed. Had the speed been 15 km/hour more it would have taken 30 minutes less for the journey. Find the original speed of the bus.

**Solution :**

Distance covered = 90 km

Let "x" be the original speed of the bus

Increased speed = x + 15

Time = Distance / Speed

Time taken by the bus when it travels in original speed = 90/x

Time taken by the bus when it travels in increased speed = 90/(x + 15)

[90/x] - [90/(x + 15)] = 1/2

90[(x + 15 - x)/x(x + 15)] = 1/2

15/x^{2} + 15x = 1/180

2700 = x^{2} + 15x

x^{2} + 15x - 2700 = 0

x^{2} + 60x - 45x - 2700 = 0

x(x + 60) - 45(x + 60) = 0

(x - 45) (x + 60) = 0

x = 45

So, the original speed of the bus is 45 km per hour.

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