Problema Solution
The sum of the squares of two positive numbers is 808. One number is four more than the other. Find the two numbers. The sum of a number and its reciprocal is 106/45. Find the number.
Answer provided by our tutors
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)