Question Number 125538 by Mammadli last updated on 11/Dec/20
$$\sqrt{−\boldsymbol{{z}}}\:=\:\boldsymbol{{z}}\:+\:\mathrm{6}\:;\:\left(\boldsymbol{{z}}=?\right) \\ $$
Answered by mr W last updated on 12/Dec/20
$${let}\:{x}=\sqrt{−{z}}>\mathrm{0} \\ $$$$\Rightarrow{z}=−{x}^{\mathrm{2}} \\ $$$$\Rightarrow{x}=−{x}^{\mathrm{2}} +\mathrm{6} \\ $$$$\Rightarrow{x}^{\mathrm{2}} +{x}−\mathrm{6}=\mathrm{0} \\ $$$$\Rightarrow\left({x}+\mathrm{3}\right)\left({x}−\mathrm{2}\right)=\mathrm{0} \\ $$$$\Rightarrow{x}=\:\mathrm{2},\:−\mathrm{3}<\mathrm{0}\:\left({rejected}\right) \\ $$$$\Rightarrow{z}=−\mathrm{2}^{\mathrm{2}} =−\mathrm{4} \\ $$
Commented by Mammadli last updated on 12/Dec/20
Dear sir, how to answer only -4 or not...
Commented by mr W last updated on 12/Dec/20
$${you}\:{are}\:{right}.\:{x}\geqslant\mathrm{0}! \\ $$
Commented by Mammadli last updated on 12/Dec/20
$$\boldsymbol{{dear}}\:\boldsymbol{{sir}},\:\boldsymbol{{x}}\geqslant−\mathrm{6}\:\boldsymbol{{must}}\:\boldsymbol{{pay}}\:\boldsymbol{{the}}\:\boldsymbol{{condition}}.. \\ $$
Commented by Mammadli last updated on 12/Dec/20
$$\boldsymbol{{thks}}\:\boldsymbol{{dear}}\:\boldsymbol{{sir}} \\ $$