Answer:
8 and 10
Step-by-step explanation:
Given
[tex]\frac{1}{x}[/tex] + [tex]\frac{1}{x+2}[/tex] = [tex]\frac{9}{40}[/tex] ← combine the fractions on the left side
[tex]\frac{x+2+x}{x(x+2)}[/tex] = [tex]\frac{9}{40}[/tex]
[tex]\frac{2x+2}{x^2+2x}[/tex] = [tex]\frac{9}{40}[/tex] ( cross- multiply )
9(x² + 2x) = 40(2x + 2) ← distribute both sides
9x² + 18x = 80x + 80 ( subtract 80x + 80 from both sides )
9x² - 62x - 80 = 0
Consider the factors of the product of the coefficient of the x² term and the constant term which sum to give the coefficient of the x- term.
product = 9 × - 80 = - 720 and sum = - 62
The factors are - 72 and + 10
Use these factors to split the x- term
9x² - 72x + 10x - 80 = 0 ( factor the first/second and third/fourth terms )
9x(x - 8) + 10(x - 8) = 0 ← factor out (x - 8) from each term
(x - 8)(9x + 10) = 0
Equate each factor to zero and solve for x
x - 8 = 0 ⇒ x = 8
9x + 10 = 0 ⇒ 9x = - 10 ⇒ x = - [tex]\frac{10}{9}[/tex]
x is an integer , then x = 8 and x + 2 = 8 + 2 = 10
The integers are 8 and 10
