Question

Se x+y=12 e x•y=4.entao o valor de x²-y² é?

Answer 1

$x+y=12\\ (x+y)^2=12^2\\ x^2+2xy+y^2=144\\\\ xy=4 \implies -4xy=-16\\\\ Somando-se \ x^2+2xy+y^2=144 \ com \ -4xy=-16 \temos:\\\\ x^2-2xy+y^2=128\\ (x-y)^2=128\\ (x-y)=\pm \sqrt{128} \\ (x-y)=\pm 8\sqrt{2}\\\\ Agora \ temos:\\\\ x^2-y^2=(x+y)(x-y)=(12).(8\sqrt{2})=96\sqrt{2}\\ ou\\ x^2-y^2=(x+y)(x-y)=(12).(-8\sqrt{2})=-96\sqrt{2}$