Question

The sum of the squares of two consecutive even integers is 1252. Find the integers. Please show work!!

Answer 1

» let's the First integer = a

» and second integer = a+2

Now

» according to question :-

=> a² + (a+2)² = 1252

=> a² + a² + 4a + 4 = 1252

=> 2a² + 4a - 1248 = 0

=> a² + 2a - 624 = 0

=> a² + 26a - 24a - 621 = 0

=> a(a+26) -24(a+26) = 0

=> (a+26)(a-24) = 0

then

=> a = -26 OR. a = 24
______

hence,

» if 1st integer = -26
so 2nd integer = -26+2 -24
___________________-[AMSWER]

» if 1st integer = 24
so 2nd integer = 24+2 = 26
__________________[ANSWER]

=================================
_-_-_-_✌☆$BE \: \: \: \: BRAINLY$☆✌_-_-_-_

Answer 2

Johey mate your answer is here

first integer =i
second integer=i+2
according to questions

=>i^2+(i+2)^2=1252
=>i^2+i^2+4i+4=1252
=>2i^2+4i-1248=0
=>i^2+2i-624=0
first integer
i=-26 or i=24

second integer
i+2=
-26+2=-24
or
i+2=
24+2=26

hope its help you.....