Numeros: ``x´´ e ``y´´
x - y = 2 ------> x = y + 2
x^2 + y^2 = 580 Sustituimos la ``x´´
(y + 2)^2 + y^2 = 580
y^2 + 2*2*y + 4 + y^2 = 580
y^2 + 4y + 4 + y^2 = 580
2y^2 + 4y - 576 = 0
[tex] y = \frac{-4 \frac{+}{} \sqrt{4^2-4*2*(-576)} }{2*2} [/tex]
[tex]\frac{-4 \frac{+}{} \sqrt{16+4608} }{4} [/tex]
[tex]\frac{-4 \frac{+}{} \sqrt{4624} }{4} = \frac{-4 \frac{+}{} 68 }{4} [/tex]
y = 16
y = 18
x - 16 = 2
x = 18
x - 18 = 2
x = 20
Soluciones:
x = 18 ; y = 16
x = 20 ; y = 18
