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Question 414573: The sum of the squares of two positive numbers is 808. One number is four more than the other. Find the two numbers.
Answer by mananth(16946) (Show Source):
You can put this solution on YOUR website! x^2+y^2=808
x=y+4
substitute the value of x in the equation
(y+4)^2+y^2=808
y^2+8y+16+y^2=808
2y^2+8y-792=0
/2
y^2+4y-396=0
y^2+22y-18y-396=0
y(y+22)-18(y+22)=0
(y+22)(y-18)=0
y=-22 OR 18
x=-18 OR 22
(-18,-22), (18,22)
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