**Word Problems on Quadratic Equations :**

In this section, we are going to learn, how to solve quadratic equation word problems step by step.

**Problem 1 :**

**Difference between a number and its positive square root is 12. Find the number. **

**Solution :**

Let "x" be the required number.

Its positive square root is √x

**Given :** Difference between x and √x = 12

x - √x = 12

x - 12 = √x

(x - 12)² = x

x² - 24x + 144 = x

x² - 25x + 144 = 0

(x - 9)(x - 16) = 0

x = 9 or x = 16

x = 9 does not satisfy the condition given in the question.

Hence the required number is 16.

**Problem 2 :**

A piece of iron rod costs $60. If the rod was 2 meter shorter and each meter costs $1 more, the cost would remain unchanged. What is the length of the rod?

**Solution :**

Let "x" be the length of the given rod.

Then the length of the rod 2 meter shorter is (x - 2) and the total cost of both the rods is $60 (Because cost would remain unchanged).

Cost of one meter of the given rod is

= 60 / x

Cost of one meter of the rod which is 2 meter shorter is

= 60 / (x - 2)

**Given : **If the rod was 2 meter shorter and each meter costs $1 more.

That is, 60/(x-2) is $1 more than 60/x.

[60 / (x - 2)] - [60 / x] = 1

Simplify.

[60x - 60(x - 2)] / [x(x - 2)] = 1

[60x - 60x + 120] / [x² - 2x] = 1

120 / (x² - 2x) = 1

120 = x² - 2x

0 = x² + 2x - 120

x² + 2x - 120 = 0

(x + 10)(x - 12) = 0

x = - 10 or x = 12

Because length can not be a negative number, we can ignore "- 10".

Hence, the length of the given rod is 12 m.

**Problem 3 :**

Divide 25 in two parts so that sum of their reciprocals is 1/6.

**Solution :**

Let "x" be one of the parts of 25. Then the other part is (25 - x). **Given : **Sum of the reciprocals of the parts is 1/6.

Then, we have

1/x + 1/(25 - x) = 1/6

Simplify.

(25 - x + x) / x(25 - x) = 1/6

25 / (25x - x²) = 1/6

6(25) = 25x - x²

150 = 25x - x²

x² - 25x + 150 = 0

(x - 15)(x - 10) = 0

x = 15 or x = 10

When x = 15,

25 - x = 25 - 15

25 - x = 10

When x = 10,

25 - x = 25 - 10

25 - x = 15

Hence, the two parts of the 25 are 10 and 15.

**Problem 4 :**

The hypotenuse of a right angled triangle is 20 cm. The difference between its other two sides is 4 cm. Find the length of the sides.

**Solution :**

Let "x" and "x + 4" be the lengths of other two sides.

Using Pythagorean theorem, we have

(x + 4)² + x² = 20²

Simplify.

x² + 8x + 16 + x² = 400

2x² + 8x + 16 = 400

Subtract 400 from both sides.

2x² + 8x - 384 = 0

Divide both sides by 2.

x² + 4x - 192 = 0

(x + 16)(x - 12) = 0

x = -16 or x = 12

x = -16 can not be accepted. Because length can not be negative.

If x = 12,

x + 4 = 12 + 4 = 16

Hence, the other two sides of the triangle are 12 cm and 16 cm.

**Problem 5 :**

The sides of an equilateral triangle are shortened by 12 units, 13 units and 14 units respectively and a right angle triangle is formed. Find the length of each side of the equilateral triangle.

**Solution :**

Let "x" be the length of each side of the equilateral triangle.

Then, the sides of the right angle triangle are

(x - 12), (x - 13) and (x - 14)

In the above three sides, the side represented by (x - 12) is hypotenuse (Because that is the longest side).

Using Pythagorean theorem, we have

(x - 12)² = (x - 13)² + (x - 14)²

x² - 24x + 144 = x² - 26x + 169 + x² - 28x + 196

x² - 30x + 221 = 0

(x - 13)(x - 17) = 0

x = 13 or x = 17.

x = 13 can not be accepted.

Because, if x = 13, the side represented by (x - 14) will be negative.

Hence, the side of the equilateral triangle is 17 units.

Apart from the problems given above, if you need more word problems on quadratic equations

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