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Question Number 30148 by .none. last updated on 17/Feb/18

how to solve (√((c−(√(−c)))^2 ))+(√((c+6−(√(−c)))^2 ))  2<a<3,  −4<b<−3,  c=((a+b)/2)

$${how}\:{to}\:{solve}\:\sqrt{\left({c}−\sqrt{−{c}}\right)^{\mathrm{2}} }+\sqrt{\left({c}+\mathrm{6}−\sqrt{−{c}}\right)^{\mathrm{2}} } \\ $$ $$\mathrm{2}<{a}<\mathrm{3},\:\:−\mathrm{4}<{b}<−\mathrm{3},\:\:{c}=\frac{{a}+{b}}{\mathrm{2}} \\ $$

Answered by mrW2 last updated on 17/Feb/18

−2<a+b<0  −1<c=((a+b)/2)<0  let d=−c  0<d<1   (√((c−(√(−c)))^2 ))+(√((c+6−(√(−c)))^2 ))  =(√((−d−(√d))^2 ))+(√((−d+6−(√d))^2 ))  =(√((d+(√d))^2 ))+(√((6−d−(√d))^2 ))  =d+(√d)+6−d−(√d)  =6

$$−\mathrm{2}<{a}+{b}<\mathrm{0} \\ $$ $$−\mathrm{1}<{c}=\frac{{a}+{b}}{\mathrm{2}}<\mathrm{0} \\ $$ $${let}\:{d}=−{c} \\ $$ $$\mathrm{0}<{d}<\mathrm{1} \\ $$ $$\:\sqrt{\left({c}−\sqrt{−{c}}\right)^{\mathrm{2}} }+\sqrt{\left({c}+\mathrm{6}−\sqrt{−{c}}\right)^{\mathrm{2}} } \\ $$ $$=\sqrt{\left(−{d}−\sqrt{{d}}\right)^{\mathrm{2}} }+\sqrt{\left(−{d}+\mathrm{6}−\sqrt{{d}}\right)^{\mathrm{2}} } \\ $$ $$=\sqrt{\left({d}+\sqrt{{d}}\right)^{\mathrm{2}} }+\sqrt{\left(\mathrm{6}−{d}−\sqrt{{d}}\right)^{\mathrm{2}} } \\ $$ $$={d}+\sqrt{{d}}+\mathrm{6}−{d}−\sqrt{{d}} \\ $$ $$=\mathrm{6} \\ $$

Commented byMJS last updated on 17/Feb/18

you had been slightly faster than me ;−)

$$\left.\mathrm{you}\:\mathrm{had}\:\mathrm{been}\:\mathrm{slightly}\:\mathrm{faster}\:\mathrm{than}\:\mathrm{me}\:;−\right) \\ $$

Answered by MJS last updated on 17/Feb/18

⇒−1≦c<0  ⇒(√((c−(√(−c)))^2 ))=∣c−(√(−c))∣=(√(−c))−c  (√((c+6−(√(−c)))^2 ))=∣c+6−(√(−c))∣=6−((√(−c))−c)  ⇒the solution is 6

$$\Rightarrow−\mathrm{1}\leqq{c}<\mathrm{0} \\ $$ $$\Rightarrow\sqrt{\left({c}−\sqrt{−{c}}\right)^{\mathrm{2}} }=\mid{c}−\sqrt{−{c}}\mid=\sqrt{−\mathrm{c}}−\mathrm{c} \\ $$ $$\sqrt{\left({c}+\mathrm{6}−\sqrt{−{c}}\right)^{\mathrm{2}} }=\mid\mathrm{c}+\mathrm{6}−\sqrt{−{c}}\mid=\mathrm{6}−\left(\sqrt{−\mathrm{c}}−\mathrm{c}\right) \\ $$ $$\Rightarrow\mathrm{the}\:\mathrm{solution}\:\mathrm{is}\:\mathrm{6} \\ $$

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