Question Number 30148 by .none. last updated on 17/Feb/18
$${how}\:{to}\:{solve}\:\sqrt{\left({c}−\sqrt{−{c}}\right)^{\mathrm{2}} }+\sqrt{\left({c}+\mathrm{6}−\sqrt{−{c}}\right)^{\mathrm{2}} } \\ $$$$\mathrm{2}<{a}<\mathrm{3},\:\:−\mathrm{4}<{b}<−\mathrm{3},\:\:{c}=\frac{{a}+{b}}{\mathrm{2}} \\ $$
Answered by mrW2 last updated on 17/Feb/18
$$−\mathrm{2}<{a}+{b}<\mathrm{0} \\ $$$$−\mathrm{1}<{c}=\frac{{a}+{b}}{\mathrm{2}}<\mathrm{0} \\ $$$${let}\:{d}=−{c} \\ $$$$\mathrm{0}<{d}<\mathrm{1} \\ $$$$\:\sqrt{\left({c}−\sqrt{−{c}}\right)^{\mathrm{2}} }+\sqrt{\left({c}+\mathrm{6}−\sqrt{−{c}}\right)^{\mathrm{2}} } \\ $$$$=\sqrt{\left(−{d}−\sqrt{{d}}\right)^{\mathrm{2}} }+\sqrt{\left(−{d}+\mathrm{6}−\sqrt{{d}}\right)^{\mathrm{2}} } \\ $$$$=\sqrt{\left({d}+\sqrt{{d}}\right)^{\mathrm{2}} }+\sqrt{\left(\mathrm{6}−{d}−\sqrt{{d}}\right)^{\mathrm{2}} } \\ $$$$={d}+\sqrt{{d}}+\mathrm{6}−{d}−\sqrt{{d}} \\ $$$$=\mathrm{6} \\ $$
Commented by MJS last updated on 17/Feb/18
$$\left.\mathrm{you}\:\mathrm{had}\:\mathrm{been}\:\mathrm{slightly}\:\mathrm{faster}\:\mathrm{than}\:\mathrm{me}\:;−\right) \\ $$
Answered by MJS last updated on 17/Feb/18
$$\Rightarrow−\mathrm{1}\leqq{c}<\mathrm{0} \\ $$$$\Rightarrow\sqrt{\left({c}−\sqrt{−{c}}\right)^{\mathrm{2}} }=\mid{c}−\sqrt{−{c}}\mid=\sqrt{−\mathrm{c}}−\mathrm{c} \\ $$$$\sqrt{\left({c}+\mathrm{6}−\sqrt{−{c}}\right)^{\mathrm{2}} }=\mid\mathrm{c}+\mathrm{6}−\sqrt{−{c}}\mid=\mathrm{6}−\left(\sqrt{−\mathrm{c}}−\mathrm{c}\right) \\ $$$$\Rightarrow\mathrm{the}\:\mathrm{solution}\:\mathrm{is}\:\mathrm{6} \\ $$