Question

Sa se arate ca ecuatia x²-2ax+1+5a²=0nu admite solutii reale oricare ar fi a€R .

Answer 1

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!

Answer 2

Δ=(2a)²-4*1*(1+5a²)=4a²- 4-20a²=-4(4a²+1)<0, deci ecuatia data nu are radacini reala ∀a∈R.