SOLUTION: Use algebra to find two consecutive even integers such that the sum of their squares is 20. x= first even integer (x+2)= second even integer x^2 + (x+2)^2 = 20

Algebra ->  Problems-with-consecutive-odd-even-integers -> SOLUTION: Use algebra to find two consecutive even integers such that the sum of their squares is 20. x= first even integer (x+2)= second even integer x^2 + (x+2)^2 = 20      Log On


   

Question 812967: Use algebra to find two consecutive even integers such that the sum of their squares is 20.
x= first even integer (x+2)= second even integer
x^2 + (x+2)^2 = 20
Found 2 solutions by ewatrrr, richwmiller:
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
 

Hi,

x= first even integer (x+2)= second even integer

x^2 + (x+2)^2 = 20   |Nice work on the Set Up!

x^2 + x^2 + 4x + 4 = 20

2x^2 + 4x - 16 = 0

 x^2 + 2x - 8 = 0

(x+4)(x-2) = 0

(x+4) = 0 ,  x = -4

(x-2) = 0 ,  x = 2

Integers -4,-2   and 2,4

CHECKING our Answer***

 sum of squares = 20

Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
good so far
continue and solve the equation
x^2 + (x+2)^2 = 20
x^2 + (x+2)*(x+2) = 20
(2x^2+4x+4) = 20
x^2+2x+2=10
x^2+2x-8=0
(x+4)*(x-2)=0
x=-4
x+2=-2
and
x=2
x+2=4
check
(-4)^2+(-2)^2=20
16+4=20
ok
2^2+4^2=20
4+16=20
ok