Question

Решите 4x^3+3x^2=0 x^2-4x+4=0 (x-7)^2+2(x-7)=0

Answer 1

$\\1) x^2(4x+3)=0\\ \left[\begin{gathered} x=0\\ x=-3/4 \end{gathered}\right.\\ 2) x^2-4x+4=0\\ (x-2)^2=0\\ x=2\\\\3)(x-7)^2+2(x-7)=0\\ (x-7)(x-7+2)=0\\ (x-7)(x-5)=0\\ \left [ \begin{gathered} x=7\\ x=5 \end{gathered}\right.\\\\$

Answer 2

1)
4x^3+3^2=0
x^2*(4x+3)=0
x^2=0
x=0
4x+3=0
x=-0,75
2)
x^2-4x+4=0
(x-2)^2=0
x-2=0
x=2
3)
(x-7)^2+2(x-7)
(x-7)*(x-7+2)=0
(x-7)*(x-5)=0
x-7=0
x=7
x-5=0
x=5
x1=5; x2=7