WORD PROBLEMS INVOLVING QUADRATIC EQUATIONS BY FACTORING

Problem 1 :

Two natural numbers differ by 2 and their product is 360. Find the numbers.

Solution :

Let x and y be the two natural numbers.

x - y = 2 ----(1)

xy = 360 ----> y = ³⁶⁰⁄ₓ ----(2)

Substitute y = ³⁶⁰⁄ₓ into (1).

x - ³⁶⁰⁄ₓ = 2

Multiply both sides by x.

x(x - ³⁶⁰⁄ₓ) = 2x

x² - 360 = 2x

x² - 2x - 360 = 0

Solve by factoring.

x² - 20x + 18x - 360 = 0

x(x - 20) + 18(x - 20) = 0

(x + 18)(x - 20) = 0

x + 18 = 0  or  x - 20 = 0

x = -18  or  x = 20

Since the numbers are natural numbers, x can not be negative. So, x = 20.

Substitute x = 20 into (2).

y = ³⁶⁰⁄₂₀

y = 180

Therefore, the two natural numbers are 20 and 180.

Problem 2 :

There are three consecutive positive integers such that the sum of the square of first and the product of the other two is 154. Find the integers.

Solution :

Let x, (x + 1) and (x + 2) be the first three consecutive integers

The sum of the squares of first and the product of the other two is 154

x² + (x + 1)(x + 2) = 154

x² + x² + 2x + 1x + 2 = 154

2x² + 3x + 2 = 154

2x² + 3x + 2 - 154 = 0

2x² + 3x - 152 = 0

(2x + 19)(x - 8) = 0

2x + 19 = 0  or  x - 8 = 0

x = -9.5  or  x = 8

Since the given integers are positive, x can not be -9.5.

So,

x = 8

x + 1 = 9

x + 2 = 10

Therefore three consecutive integers are 8, 9 and 10.

Problem 3 :

The product of two positive successive multiples of 3 is 810. Find the multiples

Solution :

Let x and (x + 3) be the two positive successive multiples of 3.

The product of two positive successive multiples = 810

x(x + 3) = 810

x² + 3x = 810

Solve by factoring.

x² + 3x - 810 = 0

x² - 27x + 30x - 810 = 0

x(x - 27) + 30(x - 27) = 0

(x - 27)(x + 30) = 0

x - 27 = 0  or  x + 30 = 0

x = 27  or  x = -30

Since the multiples are positive, x can not be negative.

So,

x = 27

x + 3 = 30

Therefore, two positive successful multiples of 3 are 27 and 30.

Problem 4 :

The sum of the squares of two natural numbers is 34 and sum of 5 times the smaller and 3 times the larger is 30. Find the numbers.

Solution :

Let x be the smaller natural number and y be the larger natural number

The sum of their squares = 34

x² + y² = 34 ----(1)

Sum of 5 times the smaller and 3 times the larger is 30.

5x + 3y = 30

5x = 30 - 3y

x = ⁽³⁰ ⁻ ³ʸ⁾⁄₅ ----(2)

Substitute x = ⁽³⁰ ⁻ ³ʸ⁾⁄₅ into (1).