Question

$2x^{2} + \sqrt{2 x^{2} -x} =2+x$

Answer 1

$\rm 2x^2+ \sqrt{2x^2-x}=2+x \\ \\ 2x^2-x+ \sqrt{2x^2-x}-2=0 \\ \\ Notez ~ \sqrt{2x^2-x}=a;~a \geq 0. \\ \\ Ecuatia~devine: \\ \\ a^2+a-2=0 \\ \\ \Delta=1-4 \cdot 1 \cdot (-2) =9. \\ \\ a_{1,2}= \frac{-1 \pm \sqrt{\Delta}}{2 \cdot 1}= \frac{-1 \pm 3}{2} = \left \{ {{-2} \atop {1}} \right. .$

$\rm Insa ~a \geq 0,~de~unde~rezulta~a=1. \\ \\ a=1 \Leftrightarrow \sqrt{2x^2-x}=1. \\ \\ Deci~2x^2-x=1 \Leftrightarrow 2x^2-x-1=0 \\ \\ \Delta_{x}= (-1)^2-4 \cdot 2 \cdot (-1)=9. \\ \\ x_{1,2}= \frac{1 \pm \sqrt{\Delta}}{2 \cdot 2}= \frac{1 \pm 3}{4} = \left \{ {{ -\frac{1}{2} } \atop {1}} \right. . \\ \\ Ecuatia~admite~solutiile~- \frac{1}{2} ~si~1.$