Question

Rezolvati in multimea nr. reale ecuatia:
x-1= radical(5+x)

Answer 1

$\sqrt{5+x}$ este definit pentru x≥-5.
$\sqrt{5-x} \geq 0=\ \textgreater \ x-1 \geq 0=\ \textgreater \ \boxed{x \geq 1}.$

Ridicam ecuatia la patrat:

=> $x^{2} -2x+1=5+x\ \textless \ =\ \textgreater \ \\ \ \textless \ =\ \textgreater \ x^{2} -3x-4=0\ \textless \ =\ \textgreater \ \\ \ \textless \ =\ \textgreater \ x^{2} -3x-3-1=0\ \textless \ =\ \textgreater \ \\ \ \textless \ =\ \textgreater \ ( x^{2} -1)-3(x+1)=0\ \textless \ =\ \textgreater \ \\ \ \textless \ =\ \textgreater \ (x+1)(x-1)-3(x+1)=0\ \textless \ =\ \textgreater \ \\ \ \textless \ =\ \textgreater \ (x+1)(x-4)=0=\ \textgreater \ x=-1~sau~x=4.$

x=-1 nu convine pentru ca -1<1.

Deci unica solutie este: $\boxed{x=4}.$