binom : a² + 2ab + b² = ( a +b)²
sau a² - 2ab + b² = ( a - b)²
adun / scad in ex. 1
x⁴ + 2x² + 1 - 1 - 3 = 0 ; stim x⁴ = ( x²)²
formam binom : [ ( x² )² + 2x² + 1 ] - 4 = 0
binom
( x² + 1)² - 4 = 0 folosim formula : a² - b² = ( a -b) · ( a +b)
( x² + 1)² - 2² =0
( x² + 1 - 2 ) · ( x² + 1 + 2 ) = 0
( x² - 1 ) · ( x² + 3 ) = 0
suma de numere pozitive x² + 3 ≠ 0
nu exista radacini reale
dar x² - 1 = 0
( x -1) · ( x +1) = 0
x - 1 = 0 x₁ =1 radacina reala
x + 1 =0 x₂ = - 1 radacina reala
SAU : rezolvam ecuatia ca ecuatie bipatrata , cu substitutie
x⁴ = ( x²)²
notam x² = y
transformam in ecuatie de gradul II :
y² + 2y - 3= 0 ; Δ = 2² - 4 ·1· ( -3) = 4 + 12 =16 ; √Δ=√16 = 4
y = ( -2 - 4) / 2 = - 6 / 2 = - 3
y = ( -2 + 4) / 2 = 2 /2 = 1
atunci x² = - 3 nu are solutie in multimea ne. reale
∀ x∈ R , x² ≥ 0
x² = 1
x² - 1 = 0
( x -1) · ( x +1) = 0
solutii ; x =1 ; x = -1
S = { - 1 ; 1 }
