Question

Utilizand o necunoscuta auxiliara, rezolvati ecuatiile:
a) $x^{2} + \sqrt{ x^{2} +2x+8} =12-2x$
b) $x-5 \sqrt{x} +6=0$

Answer 1

$\displaystyle a) ~ x^2+2x+8=x^2+2x+1+7=(x+1)^2+7\ \textgreater \ 0,~deci~expresia \\ \\ are~sens~pentru~orice~numar~real~x. \\ \\ x^2+ \sqrt{x^2+2x+8}=12-2x \\ \\ x^2+2x-12+ \sqrt{x^2+2x+8}=0 \\ \\ x^2+2x+8+ \sqrt{x^2+2x+8} -20=0 \\ \\ Notam~u=\sqrt{x^2+2x+8}\ \textgreater \ 0,~si~obtinem: \\ \\ u^2+u-20=0. \\ \\ \Delta=81~;~u_{1,2}= \frac{-1 \pm 9}{2}= \left \{ {{-5,~nu~convine,~caci~u \ \textgreater \ 0} \atop {4 ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~}} \right.$

$\displaystyle u=4 \Rightarrow \sqrt{x^2+2x+8}=4 \Rightarrow x^2+2x+8=16 \Leftrightarrow \\ \\ \Leftrightarrow x^2+2x-8=0. \\ \\ \Delta'=36 \\ \\ x_{1,2}= \frac{-2 \pm 6}{2}= \left \{ {{-4} \atop {2}} \right. . \\ \\ Solutie:~ \{-4,2 \}.$

$\displaystyle b)~Conditii~de~existenta:~x \ge 0. \\ \\ Notam~ v=\sqrt{x} \ge 0. ~Deci~x=v^2.\\ \\ Ecuatia~devine~v^2-5v+6=0. \\ \\ \Delta=1~;~v_{1,2}= \frac{5 \pm 1}{2}= \left \{ {{3} \atop {2}} \right. . \\ \\ v=3 \Rightarrow x=9. \\ \\ v=2 \Rightarrow x=4. \\ \\ Solutie:~ \{4;9 \}.$