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Rezolvati in R ecuatia:
[tex]|x^2-4| - |x-2|=0[/tex]
Cu demonstratii de clasa a 8. Multumesc. Pana la ora 14:00.


Răspuns :

   
[tex]\displaystyle\\ \text{Explicitam modulele:}\\ x^2-4\text{ are solutiile: } x_1=-2~si~x_2=2\\ \Rightarrow~x^2-4~\text{este negativa intre radacini si pozitiva in rest}\\ |x^2-4|=-x^2+4\text{ pentu }x\in (-2;~2)\\ |x^2-4|=x^2-4\text{ pentu }x\in (-\infty;~ -2]\bigcup[2;~+\infty)\\ x - 2 \text{ are solutia: }x=2\\ \Rightarrow~x-2~\text{este negativa pentru: } x\ \textless \ 2~\text{si pozitiva pentru: } x\ \textgreater \ 2\\ |x-2|=x-2~pt~x\ \textgreater \ 2\\ |x-2|=-x+2~pt~x\ \textless \ 2 [/tex]


[tex]\displaystyle\\ \text{Rezulta ca ecuatia: } |x^2-4| - |x-2|=0~\text{ se descompune in ecuatiile:}\\\\ (E1)\\ \text{ Pentru } x \leq -2\\ x^2-4 -(-x+2)=0\\ x^2 - 4 +x-2= 0\\ x^2+x-6=0\\\\ x_{12}= \frac{-b\pm \sqrt{b^2-4ac}}{2a}= \frac{-1\pm \sqrt{1+24}}{2}=\frac{-1\pm \sqrt{25}}{2}=\frac{-1\pm 5}{2}\\\\ x_1 = \frac{-1+ 5}{2}=\frac{4}{2} = \boxed{\bf 2 \ \textgreater \ -2 \text{ solutie eliminata}} \\\\ x_2 = \frac{-1- 5}{2}=\frac{-6}{2}=\boxed{\bf -3}[/tex]


[tex]\displaystyle\\ (E2)\\ \text{ Pentru } x\in[-2;~2)\\ -x^2+4 -(-x+2)=0\\ -x^2+4+x-2=0\\ -x^2 + x+2=0\\\\ x_{34}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{-1\pm \sqrt{1+8}}{-2}=\frac{-1\pm \sqrt{9}}{-2}=\frac{-1\pm 3}{-2}\\\\ x_3 = \frac{-1+ 3}{-2}=\frac{2}{-2}=\boxed{\bf-1}\\\\ x_4 = \frac{-1-3}{-2}=\frac{-4}{-2}=\boxed{\bf-2}[/tex]


[tex]\displaystyle\\ (E3)\\ \text{ Pentru } x \geq 2\\ x^2-4 -(x-2)=0\\ x^2-4 -x+2=0\\ x^2-x-2=0\\\\ x_{56}= \frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{1\pm\sqrt{1+8}}{2}=\frac{-1\pm \sqrt{9}}{2}=\frac{-1\pm 3}{2}\\\\ x_5 = \frac{-1+ 3}{2} = \frac{2}{2} =\boxed{\bf 1\ \textless \ 2~\text{solutie eliminata}}\\\\ x_6 = \frac{-1- 3}{2} = \frac{-4}{2} =\boxed{\bf -2\ \textless \ 2~\text{solutie eliminata}}\\\\ [/tex]