Question

AJUTORRR! Cine e bun la mate? Am nevoie,va rog,macar rezultatele la ele daca nu :c

Answer 1

Răspuns:

Explicație pas cu pas:

a) x⁴-3x²+2 = 0

x² = t => t²-3t+2 = 0 <=> t²-2t-t+2 = 0 => t(t-2) -(t-2) = 0 <=>

(t-2)(t-1) = 0 => t₁ = 1 ; t₂ = 2 =>

x² = 1 => x₁,₂ = ± 1 ; x₃,₄ = ± √2

b) x⁴ - 4x² +3 = 0

x² = t => t²-4t+3 = 0 <=> t²-3t-t+3 = 0 <=> t(t-3)-(t-3) = 0 <=>

(t-3)(t-1) = 0 => t₁ = 1 ; t₂ = 3

x² = 1 => x₁,₂ =  ± 1 ; x² = 3 => x₃,₄ =  ± √3

c) x⁴-5x²+4 = 0 ; x² = t

t² -5t+4 = 0 <=> t²-4t-t+4 = 0 <=> t(t-4)-(t-4) =0 <=>

(t-4)(t-1) = 0 => t₁ = 1 ; t₂ = 4

x² = 1 =>  x₁,₂ =  ± 1 ; x² = 4 => x₃,₄ =  ± √4 =  ± 2

d) x⁴-7x²+12 = 0 ; x² = t => t²-7t+12 = 0

a = 1 ; b = -7 , c = 12 ; Δ = b²-4ac = 49-48 = 1 ; √Δ = 1

t₁,₂ = (-b±√Δ)/2a = (7±1)/2 => t₁ = 3 ; t₂ = 4

x² = 3 => x₁,₂ =  ± √3 ; x² = 4 => x₃,₄ =  ± 2

e) x⁴+3x²+2 = 0 ; x² = t ; t²+3t+2 = 0 ; a = 1 ; b = 3 ; c = 2

Δ = 9-8 = 1 ; √Δ = 1 ; t₁,₂ = (-3±1)/2 => t₁ = -2 ; t₂ = -1

x² = -1 => x₁,₂ = ± i ∈ C ; x² = -2 => x₃,₄ = ± i√2 ∈ C

f) x⁴+4x²+3 = 0 ; x² = t ; x²+4t+3 = 0 ; a = 1 ; b = 4 ; c = 3

Δ = 16-12 = 4 ; √Δ = √4 = 2

t₁,₂ = (-4± 2)/2 => t₁ = -3 ; t₂ = -1

x² = -3 => x₁,₂ = ± i√3 ∈ C ; x² = -1 => x₃,₄ = ± i ∈ C

g) x⁴+7x²+12 = 0 ; x² = t ; t²+7t+12 = 0 ; a = 1 ; b = 7 ; c = 12

Δ = 49-48 = 1 ; √Δ = 1 ; t₁,₂ = (-7±1)/2 => t₁ = -4 ; t₂ = -3

x² = -4 => x₁,₂ = ±2i ∈ C ; x₃,₄ = ±i√3 ∈ C

h) x⁴ +11x²+30 = 0 ; x² = t ; t²+11t+30 = 0 ; a = 1 ; b = 11; c = 30

Δ = 121-120 = 1 ; √Δ = 1 ; t₁,₂ = (-11±1)/2 => t₁ = -6 ; t₂ = -5

x² = -6 => x₁,₂ = ± i√6 ∈ C ; x² = -5 => x₃,₄ = ± i√5 ∈ C

a) x⁴-6x²+25 = 0 ; x² = t ; t²-6t+25 = 0 , a = 1 ; b = -6 ; c = 25

Δ = 36-100 = -64 => √Δ = 8i ∈ C

t₁,₂ = (6±8i) /2 = 3±4i

x² = 3-4i => x₁,₂  = ± √(3-4i) ∈ C

x² = 3+4i => x₃,₄ = ±√(3+4i) ∈ C

b) x⁴ -16x²+100 = 0 ; x² = t => t²-16t+100 = 0

t₁,₂ = [16±√(256-400)]/2

t₁,₂ = 8±6i ∈ C

x₁,₂,₃,₄ = ±√(8±6i) ∈ C

c) x⁴-30x²+289 = 0 ; x² = t => t²-30t+289 = 0

t₁,₂ = [30±√(900-1156)]/2

t₁,₂ = 15±8i ∈ C

x₁,₂,₃,₄ = ±√(15±8i) ∈ C

d) x⁴-48x²+576 = 0 ; x² = t => t²-48t+576 = 0

t₁,₂ = [48±√(2304-2304)] / 2

t₁,₂ = 24

x² = 24 => x₁,₂ = x₃,₄ = ± √24 = ±2√6