We will learn how to solve Word Problems on quadratic equations by factoring.

**1.** The product of two numbers is 12. If their sum added to the sum of their squares is 32, find the numbers.

**Solution:**

Let the numbers be x and y.

As their product is 12, we get xy = 12 ..................... (i)

According to the question, x + y + x\(^{2}\) + y\(^{2}\) = 32 ..................... (ii)

From (i), y = \(\frac{12}{x}\)

Putting y = \(\frac{12}{x}\) in (ii), we get

x + \(\frac{12}{x}\) + x\(^{2}\) + (\(\frac{12}{x}\))\(^{2}\) = 32

⟹ (x + \(\frac{12}{x}\)) + (x + \(\frac{12}{x}\))\(^{2}\) - 2 x
**∙** \(\frac{12}{x}\) = 32

⟹ (x + \(\frac{12}{x}\))\(^{2}\) + (x + \(\frac{12}{x}\)) - 56 = 0

Putting x + \(\frac{12}{x}\) = t,

t\(^{2}\) + t - 56 = 0

⟹ t\(^{2}\) + 8t – 7t – 56 = 0

⟹ t(t + 8) - 7(t + 8) = 0

⟹ (t + 8)(t - 7) = 0

⟹ t + 8 = 0 or, t – 7 = 0

⟹ t = -8 or, t = 7

When t = -8,

x + \(\frac{12}{x}\) = t = -8

⟹ x\(^{2}\) + 8x + 12 = 0

⟹ x\(^{2}\) + 6x + 2x + 12 = 0

⟹ x(x + 6) + 2(x + 6) = 0

⟹ (x + 6)(x + 2) = 0

⟹ x + 6 = 0 or, x + 2 = 0

⟹ x = -6 or, x = -2

When t = 7

x + \(\frac{12}{x}\) = t = 7

⟹ x\(^{2}\) - 7x + 12 = 0

⟹ x\(^{2}\) - 4x - 3x + 12 = 0

⟹ x(x – 4) - 3(x – 4) = 0

⟹ (x - 4)(x - 3) = 0

⟹ x - 4 = 0 or, x - 3 = 0

⟹ x = 4 or 3

Thus, x = -6, -2, 4, 3

Then, the other number y = \(\frac{12}{x}\) = \(\frac{12}{-6}\), \(\frac{12}{-2}\), \(\frac{12}{4}\), \(\frac{12}{3}\) = -2, -6, 3, 4.

Thus, the two numbers x, y are -6, -2, or -2, -6, or 4, 3 or 3, 4.

Therefore, the required two numbers are -6, -2 or 4, 3.

**2.** An association has
a fund of $195. In addition that, each member of the association contributes
the number of dollars equal to the number of members. The total money is divided
equally among the members. If each of the members gets $ 28, find the number of
members in the association.

**Solution: **

Let the number of members be x.

Total contributions from them = $ x\(^{2}\) and the association has a fund of $ 195.

According to the problem,

x\(^{2}\) + 195 = 28x

⟹ x\(^{2}\) - 28x + 195 = 0

⟹ x\(^{2}\) - 15x - 13x + 195 = 0

⟹ x(x - 15) - 13(x - 15) = 0

⟹ (x - 15)(x - 13) = 0

Therefore, x = 15 or 13

There are 15 or 13 members in the association.

**Note:** Two answers are acceptable in this case.

**Quadratic Equation**

**Introduction to Quadratic Equation**

**Formation of Quadratic Equation in One Variable**

**General Properties of Quadratic Equation**

**Methods of Solving Quadratic Equations**

**Examine the Roots of a Quadratic Equation**

**Problems on Quadratic Equations**

**Quadratic Equations by Factoring**

**Word Problems Using Quadratic Formula**

**Examples on Quadratic Equations **

**Word Problems on Quadratic Equations by Factoring**

**Worksheet on Formation of Quadratic Equation in One Variable**

**Worksheet on Quadratic Formula**

**Worksheet on Nature of the Roots of a Quadratic Equation**

**Worksheet on Word Problems on Quadratic Equations by Factoring**

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