Question Number 212550 by mnjuly1970 last updated on 17/Oct/24
$$ \\ $$$$\:\:\:\:{the}\:{following}\:{equation}\:{has} \\ $$$$\:\:\:\:\:{no}\:{root}\:.\:{find}\:{the}\:{relationship} \\ $$$$\:\:\:{between}\:{a}\:,\:{b}\:,\:{c}\:: \\ $$$$\:\:\:\:\mathrm{1}:\:\:{c}\leqslant\mathrm{2} \\ $$$$\:\:\:\:\mathrm{2}:\:{c}\:>\mathrm{2} \\ $$$$\:\:\:\:\mathrm{3}:\:{c}\:>{ab} \\ $$$$\:\:\:\:\mathrm{4}:\:{c}\leqslant\:{ab} \\ $$$$\:\:\:\:{eq}^{{n}} \::\:\:\sqrt{\:{x}+\mathrm{1}+{b}\:+\mathrm{2}\sqrt{{x}+{b}}\:}\:+\:\sqrt{{x}+\mathrm{1}+{a}\:+\mathrm{2}\sqrt{{x}+{a}}}\:=\:{c} \\ $$$$ \\ $$
Answered by Ghisom last updated on 17/Oct/24
$${x}+\mathrm{1}+\alpha+\mathrm{2}\sqrt{{x}+\alpha}=\left(\mathrm{1}+\sqrt{{x}+\alpha}\right)^{\mathrm{2}} \\ $$$$\mathrm{solving}\:\mathrm{in}\:\mathbb{R}\:\Rightarrow\:\sqrt{{x}+\alpha}\geqslant\mathrm{0} \\ $$$$\sqrt{{x}+{a}}+\sqrt{{x}+{b}}={c}−\mathrm{2} \\ $$$${c}−\mathrm{2}>\mathrm{0}\:\Rightarrow\:{c}>\mathrm{2} \\ $$
Commented by mnjuly1970 last updated on 17/Oct/24
$$\: \\ $$