!!! este o ecuatie de gradul 4
teoretic poate avea 4 (patru ) solutii reale
Fie x²-16x=u
Ecuatia devine
u²-2u-63=0
u²+7u-9u-63=0
u(u+7)-9(u+7)=0
(u+7)(u-9)=0
u1=-7
u2=9
x²-16x=-7
x²-16x+7=0
x=[16+-√(256-49)]/2
x1,2= (16+-√207)/2 , deci 2 solutii reale, Δ>0
x²-16x=9
x²-16x+9=0
x3,4=[16+-√(256-36)]/2
x3,4=(16+-√220)/2= (16+-2√55)/2= 8+-√55, alte 2 solutii reale, Δ>0
deci S= {(8-√55);(8+√55); (16-√207)/2; (16+√207)/2}
