Question 28610: The sum of the squares of three consecutive, positive integers is equal to the sum of the squares of the next two integers. Find the five integers.
Answer by narayaba(40)   (Show Source):
You can put this solution on YOUR website! Let the 5 consecutive integers be represented as
x, x+1, x+2, x+3, x+4
According to the problem,
x^2 + (x+1)^2 + (x+2)^2 = (x+3)^2 + (x+4)^2
expanding each term on both the right and left side of the equation
using the formula (a + b)^2 = a^2 + 2*a*b + b^2
x^2 + x^2 + 2*x + 1 + x^2 + 4*x + 4 = x^2 + 6*x + 9 + x^2 + 8x + 16
grouping all the x^2 terms, x terms and the constants on both the right and left side of the equation we get
3*x^2 + 6*x + 5 = 2x^2 + 14*x + 25
rewriting this equation as
3x^2 - 2x^2 + 6*x - 14*x + 5-25 = 0
x^2 - 8*x -20 = 0
we can use the quadratic formula
x = (-b +/- sqrt(b^2 - 4*a*c)) /(2*a)
or factorise the equation x^2 - 8*x -20 = 0 as (x-10)(x+2) = 0
x = 10 or x =-2
The problems defines that consecutive positive integers
which means x = -2 is not a valid answer
Thus x =10
and the 5 consecutive positive integres are 10,11,12,13,14
|
|
|
