Question

rozwiąż równanie
a) x² + 4|x|+1=0
b) |x² - 3|= 1
​

Answer 1

a)

• [tex]x\in(-\infty,0\rangle[/tex]
  
  [tex]x^2+4\cdot(-x)+1=0\\x^2-4x+1=0\\x^2-4x+4-3=0\\(x-2)^2=3\\x-2=-\sqrt3 \vee x-2=\sqrt3\\x=2-\sqrt3 \vee x=2+\sqrt3\\\\2-\sqrt3,2+\sqrt3\not \in(-\infty,0\rangle[/tex]

• [tex]x\in(0,\infty)[/tex]
  
  [tex]x^2+4x+1=0\\x^2+4x+4-3=0\\(x+2)^2=3\\x+2=-\sqrt3 \vee x+2=\sqrt3\\x=-2-\sqrt3 \vee x=-2+\sqrt3\\\\-2-\sqrt3,-2+\sqrt3\not \in (0,\infty)[/tex]
  

Zatem [tex]x\in\emptyset[/tex].

b)

[tex]|x^2-3|=1\\x^2-3=-1 \vee x^2-3=1\\x^2=2 \vee x^2=4\\x=-\sqrt 2 \vee x=\sqrt2 \vee x=-2 \vee x=2[/tex]