Question

Solutia reala a ecuatiei √x2-6x+9 +2|(3-x)(2x+1)|=0 .

Answer 1

$\bold{Solutia~1.} \\ \sqrt{ x^{2} -6x+9}+2|(3-x)(2x+1)|=0 \\ \sqrt{(x-3)^2} +2|(3-x)(2x+1)| =0 \\ |x-3|+2|3-x|*|2x+1|=0 \\ |x-3|+2|x-3|*|2x+1|=0 \\ |x-3|(1+2*|2x+1|)=0 \Rightarrow |x-3|=0~sau~1+2*|2x+1|=0. \\ \\ |x-3|=0 \Rightarrow x-3=0 \Rightarrow x=3. \\ 1+2*|2x+1|=0 \Rightarrow |2x+1|= -\frac{1}{2},~imposibil. \\ Solutie: \boxed{x=3} .$

$\bold{Solutia ~2.} \\ \sqrt{ x^{2} -6x+9} \geq 0~si~2|(3-x)(2x+1)| \geq 0,~dar \\ \sqrt{ x^{2} -6x+9}+2|(3-x)(2x+1)|=0 \Rightarrow \sqrt{ x^{2} -6x+9}=0~si~ \\ 2|(3-x)(2x+1)|=0.~va~rezulta~solutia~x=3.$

La prima metoda am folosit urmatoarele proprietati ale modulului:
$|a*b|=|a|*|b| \\ |a|=|-a|$

Answer 2

$\sqrt{x^2-6x+9} +2|(3-x)(2x+1)|=0 \\\\ \sqrt{(x-3)^2}+2|(3-x)(2x+1)|=0 \\\\ |(x-3)|+2|(3-x)(2x+1)|=0 \\\\\\ |x-3|=0 \\\\ x-3=0 \\\\ \boxed{x=3} \\\\\\ 2|(3-x)(2x+1)|=0 \\\\(3-x)(2x+1)=0 \\\\\\ 3-x=0 \\\\ -x=-3 \\\\ \boxed{x=3} \\\\\\ 2x+1=0 \\\\ 2x=-1 \\\\ x=-\frac{1}{2} \not \in N\\\\\\\\ \boxed{\boxed{S \in \{3\}}}$