two integers differ by 12 and the sum of their squares is 74. find the integers ???? :(

Question

anonymous · Answer

x = integer 1
y = integer 2

y - x = 12
y^2 + x^2 = 74

now you have a system of equations

solve the 1st equation for x, in terms of y
substitute that into the 2nd equation and solve for y

campbell_st · Answer

well let 1 integer = x

the other is x + 12 

then 

$$x^2 + (x + 12)^2 = 74$$

so 

simplifying 

$$2x^2 + 24x + 144 = 74$$

or 

$$x^2 + 12x + 35 = 0$$

fyou can solve the quadratic by factoring....

anonymous · Answer

thank you

anonymous · Answer

wait but isn't the integer ment to "differ" so it's -
instead of +

anonymous · Answer

x differrs from x+ 12 by 12 units

this is the same as (x+12) - 12 = x

anonymous · Answer

x + 12 = y

y - 12 = x

anonymous · Answer

sorry I still don't get it because....this is what I got (x^2-12)=74

anonymous · Answer

mistake   (x-12)^2=74

anonymous · Answer

makes no difference

anonymous · Answer

then I simplified it and got x^2 - 24x + 144=74
= x^2 - 24x + 70=0

anonymous · Answer

it is the sum  of the squares
either 
$$x^2+(x-12)^2=74$$ or 
$$x^2+(x+12)^2=74$$

anonymous · Answer

thank you :)