Question

sa se rezolve in multimea numerelor reale $\sqrt[3]{ x^{2} -x-3} =-1$

Answer 1

(∛x²-x-3)³=(-1)³
x²-x-3=-1
x²-x-2=0
Δ=b²-4ac=1+8=9
x1=(1-3)/2=-1
x2=(1+3)/2=2

Answer 2

$\sqrt[3]{ x^{2} -x-3}=-1|^3 \\ (\sqrt[3]{ x^{2} -x-3})^3 =(-1)^3 \\ x^{2} -x-3=-1 \\ x^{2} -x-3+1=0 \\ x^{2} -x-2=0 \\ \Delta=1-4*(-2)*1=1+8=9 \\ x_1= \frac{1-3}{2}= \frac{-2}{2}=-1 \\ x_2= \frac{1+3}{2}= \frac{4}{2}=2$
S={-1, 2}