Question-187835

Question Number 187835 by pascal889 last updated on 22/Feb/23

Commented by Frix last updated on 22/Feb/23

(√(x−a))+(√(x−b))=(√c) x=(((a−b)^2 )/(4c))+((a+b)/2)+(c/4)

$$\sqrt{{x}−{a}}+\sqrt{{x}−{b}}=\sqrt{{c}} \\ $$$${x}=\frac{\left({a}−{b}\right)^{\mathrm{2}} }{\mathrm{4}{c}}+\frac{{a}+{b}}{\mathrm{2}}+\frac{{c}}{\mathrm{4}} \\ $$

Answered by HeferH last updated on 22/Feb/23

$$\:{x}=\mathrm{2} \\ $$

Answered by Rasheed.Sindhi last updated on 22/Feb/23

(√(x−1)) +(√(x−2)) =1 ((√(x−1)) +(√(x−2)) )((√(x−1)) −(√(x−2)) ) =1((√(x−1)) −(√(x−2)) ) (√(x−1)) −(√(x−2)) =((√(x−1)) )^2 −((√(x−2)) )^2 =x−1−x+2=1 { (((√(x−1)) +(√(x−2)) =1)),(((√(x−1)) −(√(x−2)) =1)) :} ⇒2(√(x−1)) =2⇒x−1=1⇒x=2

$$\sqrt{{x}−\mathrm{1}}\:+\sqrt{{x}−\mathrm{2}}\:=\mathrm{1} \\ $$$$\left(\sqrt{{x}−\mathrm{1}}\:+\sqrt{{x}−\mathrm{2}}\:\right)\left(\sqrt{{x}−\mathrm{1}}\:−\sqrt{{x}−\mathrm{2}}\:\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{1}\left(\sqrt{{x}−\mathrm{1}}\:−\sqrt{{x}−\mathrm{2}}\:\right) \\ $$$$\sqrt{{x}−\mathrm{1}}\:−\sqrt{{x}−\mathrm{2}}\:=\left(\sqrt{{x}−\mathrm{1}}\:\right)^{\mathrm{2}} −\left(\sqrt{{x}−\mathrm{2}}\:\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={x}−\mathrm{1}−{x}+\mathrm{2}=\mathrm{1} \\ $$$$\begin{cases}{\sqrt{{x}−\mathrm{1}}\:+\sqrt{{x}−\mathrm{2}}\:=\mathrm{1}}\\{\sqrt{{x}−\mathrm{1}}\:−\sqrt{{x}−\mathrm{2}}\:=\mathrm{1}}\end{cases} \\ $$$$\Rightarrow\mathrm{2}\sqrt{{x}−\mathrm{1}}\:=\mathrm{2}\Rightarrow{x}−\mathrm{1}=\mathrm{1}\Rightarrow{x}=\mathrm{2} \\ $$

Answered by Rasheed.Sindhi last updated on 22/Feb/23

(√(x−1)) =a , (√(x+2)) =b a+b=1.....A a^2 −b^2 =((√(x−1)) )^2 −((√(x−2)) )^2 =x−1−x+2=1 ((a^2 −b^2 )/(a+b))=(1/1) a−b=1.....B A+B: 2a=2⇒a=1 (√(x−1)) =1⇒x−1=1⇒x=2

$$\sqrt{{x}−\mathrm{1}}\:={a}\:,\:\:\sqrt{{x}+\mathrm{2}}\:={b} \\ $$$${a}+{b}=\mathrm{1}…..{A} \\ $$$${a}^{\mathrm{2}} −{b}^{\mathrm{2}} =\left(\sqrt{{x}−\mathrm{1}}\:\right)^{\mathrm{2}} −\left(\sqrt{{x}−\mathrm{2}}\:\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:={x}−\mathrm{1}−{x}+\mathrm{2}=\mathrm{1} \\ $$$$\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{{a}+{b}}=\frac{\mathrm{1}}{\mathrm{1}} \\ $$$${a}−{b}=\mathrm{1}…..{B} \\ $$$${A}+{B}:\:\mathrm{2}{a}=\mathrm{2}\Rightarrow{a}=\mathrm{1} \\ $$$$\sqrt{{x}−\mathrm{1}}\:=\mathrm{1}\Rightarrow{x}−\mathrm{1}=\mathrm{1}\Rightarrow{x}=\mathrm{2} \\ $$

Answered by manxsol last updated on 22/Feb/23

(√(x−a))+(√(x−b))=(√c) I (^ (√(x−a))+(√(x−b)))((√(x−a))−(√(x−b)))=(√c)((√(x−a))−(√(x−b))) (√(x−a))−(√(x−b))=((b−a)/( (√c))) II +I+II 2(√(x−a))=(√c)+((b−a)/( (√c))) 2(√c)(√(x−a))=(c+b−a) 4c(x−a)=(c+b−a)^2 condition x≥a x≥b c+b≥a exercise a=1 b=2 c=1 4c(x−a)=(c+b−a)^2 4(1)(x−1)=(2+1−1)^2 4(x−1)=4 x−1=1 x=2

$$\sqrt{{x}−{a}}+\sqrt{{x}−{b}}=\sqrt{{c}}\:\:\:\:\:\:{I} \\ $$$$\overset{} {\left(}\sqrt{{x}−{a}}+\sqrt{{x}−{b}}\right)\left(\sqrt{{x}−{a}}−\sqrt{{x}−{b}}\right)=\sqrt{{c}}\left(\sqrt{{x}−{a}}−\sqrt{{x}−{b}}\right) \\ $$$$\sqrt{{x}−{a}}−\sqrt{{x}−{b}}=\frac{{b}−{a}}{\:\sqrt{{c}}}\:\:\:{II} \\ $$$$+{I}+{II} \\ $$$$\mathrm{2}\sqrt{{x}−{a}}=\sqrt{{c}}+\frac{{b}−{a}}{\:\sqrt{{c}}} \\ $$$$\mathrm{2}\sqrt{{c}}\sqrt{{x}−{a}}=\left({c}+{b}−{a}\right) \\ $$$$\mathrm{4}\boldsymbol{{c}}\left(\boldsymbol{{x}}−\boldsymbol{{a}}\right)=\left(\boldsymbol{{c}}+\boldsymbol{{b}}−\boldsymbol{{a}}\right)^{\mathrm{2}} \\ $$$${condition} \\ $$$${x}\geqslant{a} \\ $$$${x}\geqslant{b} \\ $$$${c}+{b}\geqslant{a} \\ $$$${exercise} \\ $$$${a}=\mathrm{1} \\ $$$${b}=\mathrm{2} \\ $$$${c}=\mathrm{1} \\ $$$$\mathrm{4}\boldsymbol{{c}}\left(\boldsymbol{{x}}−\boldsymbol{{a}}\right)=\left(\boldsymbol{{c}}+\boldsymbol{{b}}−\boldsymbol{{a}}\right)^{\mathrm{2}} \\ $$$$\mathrm{4}\left(\mathrm{1}\right)\left({x}−\mathrm{1}\right)=\left(\mathrm{2}+\mathrm{1}−\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\mathrm{4}\left({x}−\mathrm{1}\right)=\mathrm{4} \\ $$$${x}−\mathrm{1}=\mathrm{1} \\ $$$${x}=\mathrm{2} \\ $$$$ \\ $$

Answered by SEKRET last updated on 23/Feb/23

(√(x−1 )) =a (√(x−2)) =b { ((a+b=1)),((a−b=1)) :} a=1 b=0 x=2

$$\:\:\:\sqrt{\boldsymbol{\mathrm{x}}−\mathrm{1}\:}\:=\boldsymbol{\mathrm{a}}\:\:\:\:\:\:\:\sqrt{\boldsymbol{\mathrm{x}}−\mathrm{2}}\:=\boldsymbol{\mathrm{b}} \\ $$$$\:\:\:\begin{cases}{\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}=\mathrm{1}}\\{\boldsymbol{\mathrm{a}}−\boldsymbol{\mathrm{b}}=\mathrm{1}}\end{cases}\:\:\:\:\:\boldsymbol{\mathrm{a}}=\mathrm{1}\:\:\:\boldsymbol{\mathrm{b}}=\mathrm{0}\:\:\:\:\boldsymbol{\mathrm{x}}=\mathrm{2} \\ $$$$ \\ $$

Commented by manxsol last updated on 23/Feb/23

$${excellent} \\ $$

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