(x-√3)²=(2√3-4)²=
(x-√3)²=( 4-2√3)² pt ca a²=(-a)²
(x-√3)²=(3-2√3+1)²
(x-√3)²=((√3-1)²)²
(x-√3)²=(√3-1)^4
(x-√3)²-(√3-1)^4=0
(x-√3+(√3-1)²) (x-√3-(√3-1)²)=0
a)
(x-√3+(√3-1)²=0
x=√3-(√3-1)²
x=√3-(4-2√3)
x=-4+3√3 care verifica
intr-adevar
(-4+3√3-√3)²=(-4+2√3)²=(4-2√3)² cum am zis a²=(-a)²
b)
x-√3-(√3-1)²=0
x=√3+(√3-1)²
x=√3+4-2√3
x=4-2√3
care verifica
intr-adevar
(4-√3-√3)²=(4-2√3)²=(2√3-4)²
ecuatiede gard 2, 2 radacini aflate si verificate, bine rezolvat
ALTFEL, MAI SIMPLU
(x-√3)²=(2√3-4)²
ecuatia
x²=b² are doua solutii x=b si x=-b
atunci
x-√3=2√3-4
x=3√3-4
si
x-√3=-2√3+4=
x=-√3+4
