Question

Rozwiąż x²+3x-18+4√(x²+3x-6)=0

w nawiasie jest całe pod pierwiastkiem

Answer 1

 

Answer 2

x² + 3x- 18 + 4√(x²+3x-6) = 0

D: x²+3x-6 ≥ 0

Δ = 9 - 4 * 1 * (-6) = 9 + 24 = 33

√Δ = √33

x1  = (- 3 - √33)/2 ~ -4,4

x2 = (-3 + √33)/2 ~ 1,4

D: x ∈ (-oo, x1) u (x2, +oo)

 

x² + 3x- 6 - 12 + 4√(x²+3x-6) = 0

x²+3x-6 = t   i t ≥ 0

t - 12 + 4√t = 0  

4√t = 12 - t       D: 12 - t ≥ 0    czyli t ≤ 12  i t ≥ 0   to    D:  t ∈ <0, 12>

 4√t    = 12 - t   /²

(12 - t )² = (4√t)²

144 - 24t + t²= 16t

t² - 24t - 16t + 144 = 0

t² - 40t + 144 = 0

Δ = 40² - 4 * 1 * 144 = 1600 - 576 = 1024

√Δ = 32

t1 = (40 - 32)/2 = 8/2 = 4       ∈ D

t2 = (40 + 32)/2 = 72/2 = 36      ∉ D

 

x²+3x-6 = t  

x²+3x-6 = 4                       

x²+3x-10 = 0                     

Δ = 9 + 40 = 49                                     

√Δ = 7                                                   

x1 = (-3 - 7)/2 = -10/2 = -5 ∈ D              

x2 = (-3 + 7)/2 = 4/2 = 2  ∈ D                

 

odp. x = -5,  x = 2