Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 201991 by MATHEMATICSAM last updated on 18/Dec/23

Solve  ((1/x) − (1/x^3 ))^(1/2)  + ((1/x^2 ) − (1/x^3 ))^(1/2)  = 1

$$\boldsymbol{\mathrm{Solve}} \\ $$$$\left(\frac{\mathrm{1}}{{x}}\:−\:\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \:+\:\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:−\:\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \:=\:\mathrm{1} \\ $$

Answered by mr W last updated on 18/Dec/23

((√(1/x))/1)=((√(1/x^3 ))/( (√(1/x^2 )))) =(√(1/x))   ⇒((√(1/x)))^2 +((√(1/x^2 )))^2 =1  ⇒x^2 −x−1=0  ⇒x=((1+(√5))/2) ✓

$$\frac{\sqrt{\frac{\mathrm{1}}{{x}}}}{\mathrm{1}}=\frac{\sqrt{\frac{\mathrm{1}}{{x}^{\mathrm{3}} }}}{\:\sqrt{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}}\:=\sqrt{\frac{\mathrm{1}}{{x}}}\: \\ $$$$\Rightarrow\left(\sqrt{\frac{\mathrm{1}}{{x}}}\right)^{\mathrm{2}} +\left(\sqrt{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}\right)^{\mathrm{2}} =\mathrm{1} \\ $$$$\Rightarrow{x}^{\mathrm{2}} −{x}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{x}=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\:\checkmark \\ $$

Commented by mr W last updated on 19/Dec/23

Answered by Rasheed.Sindhi last updated on 19/Dec/23

((1/x) − (1/x^3 ))^(1/2)  + ((1/x^2 ) − (1/x^3 ))^(1/2)  = 1  ((x/x^2 ) − (1/x^3 ))^(1/2)  + ((1/x^2 ) − (1/x^3 ))^(1/2)  = 1  (1/x){(x−(1/x))^(1/2) +(1−(1/x))^(1/2) }=1  (x−(1/x))^(1/2) +(1−(1/x))^(1/2) =x       Let (√(x−(1/x))) =a , (√(1−(1/x))) =b        a+b=x ∧ a^2 −b^2 =x−1  ⇒a−b=((a^2 −b^2 )/(a+b))=((x−1)/x)   { (( a+b=x)),((a−b=((x−1)/x)=1−(1/x))) :}     ⇒ { ((2a=x+((x−1)/x)=x−(1/x)+1=a^2 +1)),((2b=x−((x−1)/x)=x+(1/x)−1=x−(1−(1/x))=x−b^2 )) :}  ⇒ { ((a^2 −2a+1=0⇒(a−1)^2 =0⇒a=1⇒b^2 +b=a=1)),((b^2 +2b=x)) :}  ⇒(√(x−(1/x))) =1⇒x−(1/x)=1⇒x^2 −x−1=0  ⇒x_(≥0) =((1+(√(1+4)))/2)=((1+(√5) )/2)

$$\left(\frac{\mathrm{1}}{{x}}\:−\:\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \:+\:\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:−\:\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \:=\:\mathrm{1} \\ $$$$\left(\frac{{x}}{{x}^{\mathrm{2}} }\:−\:\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \:+\:\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:−\:\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \:=\:\mathrm{1} \\ $$$$\frac{\mathrm{1}}{{x}}\left\{\left({x}−\frac{\mathrm{1}}{{x}}\right)^{\mathrm{1}/\mathrm{2}} +\left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right)^{\mathrm{1}/\mathrm{2}} \right\}=\mathrm{1} \\ $$$$\left({x}−\frac{\mathrm{1}}{{x}}\right)^{\mathrm{1}/\mathrm{2}} +\left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right)^{\mathrm{1}/\mathrm{2}} ={x} \\ $$$$ \\ $$$$\:\:\:{Let}\:\sqrt{{x}−\frac{\mathrm{1}}{{x}}}\:={a}\:,\:\sqrt{\mathrm{1}−\frac{\mathrm{1}}{{x}}}\:={b} \\ $$$$\:\:\:\:\:\:{a}+{b}={x}\:\wedge\:{a}^{\mathrm{2}} −{b}^{\mathrm{2}} ={x}−\mathrm{1} \\ $$$$\Rightarrow{a}−{b}=\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{{a}+{b}}=\frac{{x}−\mathrm{1}}{{x}} \\ $$$$\begin{cases}{\:{a}+{b}={x}}\\{{a}−{b}=\frac{{x}−\mathrm{1}}{{x}}=\mathrm{1}−\frac{\mathrm{1}}{{x}}}\end{cases}\:\: \\ $$$$\:\Rightarrow\begin{cases}{\mathrm{2}{a}={x}+\frac{{x}−\mathrm{1}}{{x}}={x}−\frac{\mathrm{1}}{{x}}+\mathrm{1}={a}^{\mathrm{2}} +\mathrm{1}}\\{\mathrm{2}{b}={x}−\frac{{x}−\mathrm{1}}{{x}}={x}+\frac{\mathrm{1}}{{x}}−\mathrm{1}={x}−\left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right)={x}−{b}^{\mathrm{2}} }\end{cases} \\ $$$$\Rightarrow\begin{cases}{{a}^{\mathrm{2}} −\mathrm{2}{a}+\mathrm{1}=\mathrm{0}\Rightarrow\left({a}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{0}\Rightarrow{a}=\mathrm{1}\Rightarrow{b}^{\mathrm{2}} +{b}={a}=\mathrm{1}}\\{{b}^{\mathrm{2}} +\mathrm{2}{b}={x}}\end{cases} \\ $$$$\Rightarrow\sqrt{{x}−\frac{\mathrm{1}}{{x}}}\:=\mathrm{1}\Rightarrow{x}−\frac{\mathrm{1}}{{x}}=\mathrm{1}\Rightarrow{x}^{\mathrm{2}} −{x}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{x}_{\geqslant\mathrm{0}} =\frac{\mathrm{1}+\sqrt{\mathrm{1}+\mathrm{4}}}{\mathrm{2}}=\frac{\mathrm{1}+\sqrt{\mathrm{5}}\:}{\mathrm{2}} \\ $$$$\:\:\:\:\: \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com