Question

One positive integer is 77 less than twice another. the sum of their squares is 530530. find the integers.

Answer 1

a integer is 7 less than twice another
x is 7 less than 2y
x=-7+2y
x=2y-7


the sum of their squares is 530
x²+y²=530

subsitute 2y-7 for x(2y-7)²+y²=530
expand
4y²-28y+49+y²=530
5y²-28y+49=530
minus 530 both sides
5y²-28y-481=0
now comes the tricky part
we need to factor
or use quadratic formula
for ax²+bx+c=0
$x=\frac{-b+/-\sqrt{b^2-4ac}}{2a}$

so
for $5y^2-28y-481=0$
$y=\frac{-(-28)+/-\sqrt{(-28)^2-4(5)(-481)}{2(5)}$
$y=\frac{28+/-\sqrt{784+9620}}{10}$
$y=\frac{28+/-\sqrt{10404}}{10}$
$y=\frac{28+/-102}{10}$

so
$y=\frac{28+102}{10}$ or $y=\frac{28-102}{10}$
$y=\frac{130}{10}$ or $y=\frac{-74}{10}$
y=13 or y=-7.4 (nope because it is negative and it is not an integer)

therefor y=13

sub back
x=2y-7
x=2(13)-7
x=26-7
x=19


the integers are 13 and 19