Daca ecuatia nu admite solutii reale <=> delta<0
x^2-2ax+1+5a^2=0
a=1
b=-2a+1
c=5a^2
delta=b^2-4ac
delta=(-2a+1)^2-4•1•5a^2
delta = 4a^2-4a+1-20a^2
delta=-16a^2-4a+1=0
Dar delta<0 <=>
16a^2-4a+1<0
16a^2-4a+1=0
a=16
b=-2
c=1
delta'=b^2-4ac
delta'=16-4•16•1
delta'=16-64
delta=-48 < 0 Deci nu admite solutii reale.
Ceea ce trebuia demonstrat!
Bafta la scoala!
