The sum of the squares of two consecutive odd positive integers is 202. find the integers
Question
Answer:
----------------------------------------------------------Define x :
----------------------------------------------------------
Let one of the number be x. The other number is x + 2.
----------------------------------------------------------
Construction Equation :
----------------------------------------------------------
The sum of the square of the two number is 202.
[tex]x^2 + (x+2)^2 = 202[/tex]
----------------------------------------------------------
Solve for x :
----------------------------------------------------------
[tex]x^2 + (x+2)^2 = 202[/tex]
[tex]x^2 + x^2 + 4x + 4 = 202[/tex]
[tex]2x^2 + 4x - 198 = 0[/tex]
[tex]x^2 + 2x - 99 = 0[/tex]
[tex] (x+11)(x-9) =0[/tex]
[tex] (x+11)= 0 \ \ or \ \ (x-9) =0[/tex]
[tex]x = - 11 \ or \ x = 9 [/tex]
x = -11 is rejected, since it only want positive integer.
----------------------------------------------------------
Find the two numbers :
----------------------------------------------------------
x = 9.
x + 3 = 11Β
----------------------------------------------------------
Answer: The numbers are 9 and 11.
----------------------------------------------------------
solved
general
11 months ago
1092