Question Number 223228 by fantastic last updated on 18/Jul/25
Commented by fantastic last updated on 18/Jul/25
$${Right}! \\ $$
Commented by mr W last updated on 18/Jul/25
$$\mathrm{10}^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} \geqslant\mathrm{2}{ab}=\mathrm{2}×\mathrm{10}×\mathrm{6}=\mathrm{120} \\ $$$$\mathrm{100}\geqslant\mathrm{120}\:\Rightarrow{impossible}\:{figure} \\ $$
Answered by mehdee7396 last updated on 18/Jul/25
$${xy}=\mathrm{36}\:\&\:\:{x}+{y}=\mathrm{10} \\ $$$$\Rightarrow{X}^{\mathrm{2}} −\mathrm{10}{X}+\mathrm{36}=\mathrm{0} \\ $$$$\Rightarrow{X}=\mathrm{5}\pm\sqrt{\mathrm{25}−\mathrm{36}}\notin\:\Re \\ $$