Question 115744

the sum of the squares of two consecutive even integers is 452. find the numbers.


Let two consecutive even integers be:

{{{x}}} and {{{y = x+2}}}

the sum of the squares:

{{{x^2 + y^2 = 452}}}

{{{x^2 + (x+2)^2 = 452}}}

{{{x^2 + (x^2 + 4x + 4) = 452}}}

{{{2x^2 + 4x + 4 = 452}}}……..move {{{452}}} to the left to get a standard form of quadratic equation

{{{2x^2 + 4x + 4 – 452 = 0}}}……..

{{{2x^2 + 4x - 448 = 0}}}…….. divide by {{{2}}}

{{{x^2 + 2x - 224 = 0}}}……..use quadratic formula to solve for {{{x}}}


{{{x[1,2]=(-b +- sqrt (b^2 -4*a*c )) / (2*a)}}}

{{{x[1,2]=(-2 +- sqrt (2^2 -4*1*(-224) )) / (2*1)}}}

{{{x[1,2]=(-2 +- sqrt (4 + 896)) / 2}}}

{{{x[1,2]=(-2 +- sqrt (900)) / 2}}}

{{{x[1,2]=(-2 +- 30) / 2}}}


{{{x[1]=(-2 + 30) / 2}}}

{{{x[1]= 28 / 2}}}

{{{x[1]= 14}}}


{{{x[2]=(-2 - 30) / 2}}}

{{{x[2]= - 32 / 2}}}
{{{x[2]= - 16}}}



So, if {{{x[1] = 14}}} => {{{y[1] = x+2= 14 + 2= 16}}}……….1.

and if {{{x[2] = -16}}} => {{{y[2] = x+2= -16 + 2= -14}}}………2.

check solutions:


1. {{{x[1]^2 + y[1]^2 = 452}}}

{{{14^2 + 16^2 = 452}}}

{{{196 + 256 = 452}}}

{{{452 = 452}}}

2. {{{x[2]^2 + y[2]^2 = 452}}}

{{{(-16)^2 + (-14)^2 = 452}}}

{{{256 + 196 = 452}}}

{{{452 = 452}}}