FINDING TWO NUMBERS WITH THE GIVEN DIFFERENCE AND PRODUCT

Example 1 :

Two positive numbers differ by 4 and their product is 192. Find the numbers.

Solution :

Let x and y be the two numbers.

Given : Two positive numbers differ by 4.

 x - y = 4

x = y + 4 ----(1)

Given : The product of two numbers is 192.

xy = 192

Substitute x = y + 4 into the above equation.

(y + 4)y = 192

y² + 4y = 192

y² + 4y - 192 = 0

Solve by factoring.

y² - 12y + 16y - 192 = 0

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

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

y - 12 = 0  or  y + 16 = 0

y = 12  or  y = -16

Since the numbers are positive, y can not be -16.

So, y = 12.

Substitute y = 12 into (1).

x = 12 + 4

x = 16

Therefore, the two numbers are 16 and 12.

Verification :

Two positive numbers differ by 4.

16 - 12 = 4

4 = 4

The product of the two numbers is 192.

16 ⋅ 12 = 192

192 = 192

The answer is justied.

Example 2 :

Two positive numbers differ by ⁸⁄₃ and their product is 1. Find the numbers.

Solution :

Let x and y be the two numbers.

Given : Two positive numbers differ by ⁸⁄₃.

 x - y = ⁸⁄₃

 x = y + ⁸⁄₃----(1)

Given : The product of two numbers is 1.

xy = 1

Substitute x = y + ⁸⁄₃ into the above equation.

(y + ⁸⁄₃)y = 1

y² + (⁸⁄₃)y = 1

Multiply both sides by 3.

3[y² + (⁸⁄₃)y] = 3(1)

3y² + 8y = 3

3y² + 8y - 3 = 0

Solve by factoring.

3y² - y + 9y - 3 = 0

y(3y - 1) + 3(3y - 1) = 0

(3y - 1)(y + 3) = 0

3y - 1 = 0  or  y + 3 = 0

y = ⅓  or  y = -3

Since the numbers are positive, y can not be -3.

So, y = ⅓.

Substitute y = ⅓ into (1).

x = ⅓ + ⁸⁄₃

x = ⅓ + ⁸⁄₃

x = ⁽¹ ⁺ ⁸⁾⁄₃

x = ⁹⁄₃

x = 3

Therefore, the two numbers are 3 and ⅓.

Verification :

Two positive numbers differ by ⁸⁄₃.

3 - ⅓ = ⁸⁄₃

⁽⁹ ⁻ ¹⁾⁄₃ = ⁸⁄₃

⁸⁄₃ = ⁸⁄₃

The product of the two numbers is 192.

3 ⋅ ⅓ = 1

1 = 1

The answer is justied.

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