find-x-1-x-1-x-2-x-2-x-x-1-

Question Number 53262 by Tawa1 last updated on 19/Jan/19

$$\mathrm{find}\:\mathrm{x}:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{x}\:+\:\mathrm{1}\:+\:\sqrt{\mathrm{x}}}}\:\:−\:\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{x}\:−\:\mathrm{2}\:+\:\sqrt{\mathrm{x}}}}\:\:=\:\:\sqrt{\mathrm{x}\:−\:\mathrm{1}} \\ $$

Commented by mr W last updated on 20/Jan/19

there is no real solution! LHS: say x+(√x)=u >2 (√(u+1))>(√(u−2)) >0 ⇒(1/( (√(u+1))))<(1/( (√(u−2))))<(2/( (√(u−2)))) ⇒LHS=(1/( (√(x+1+(√x)))))−(2/( (√(x−2+(√x)))))=(1/( (√(u+1))))−(2/( (√(u−2))))<0 RHS: (√(x−1))≥0 i.e. LHS is always negative, but RHS is always positive or zero. LHS<0 ≠ RHS≥0 ⇒no solution for LHS=RHS!

$${there}\:{is}\:{no}\:{real}\:{solution}! \\ $$$${LHS}: \\ $$$${say}\:{x}+\sqrt{{x}}={u}\:>\mathrm{2} \\ $$$$\sqrt{{u}+\mathrm{1}}>\sqrt{{u}−\mathrm{2}}\:>\mathrm{0} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\:\sqrt{{u}+\mathrm{1}}}<\frac{\mathrm{1}}{\:\sqrt{{u}−\mathrm{2}}}<\frac{\mathrm{2}}{\:\sqrt{{u}−\mathrm{2}}} \\ $$$$\Rightarrow{LHS}=\frac{\mathrm{1}}{\:\sqrt{{x}+\mathrm{1}+\sqrt{{x}}}}−\frac{\mathrm{2}}{\:\sqrt{{x}−\mathrm{2}+\sqrt{{x}}}}=\frac{\mathrm{1}}{\:\sqrt{{u}+\mathrm{1}}}−\frac{\mathrm{2}}{\:\sqrt{{u}−\mathrm{2}}}<\mathrm{0} \\ $$$$ \\ $$$${RHS}: \\ $$$$\sqrt{{x}−\mathrm{1}}\geqslant\mathrm{0} \\ $$$$ \\ $$$${i}.{e}.\:{LHS}\:{is}\:{always}\:{negative},\:{but}\:{RHS} \\ $$$${is}\:{always}\:{positive}\:{or}\:{zero}. \\ $$$${LHS}<\mathrm{0}\:\neq\:{RHS}\geqslant\mathrm{0} \\ $$$$ \\ $$$$\Rightarrow{no}\:{solution}\:{for}\:{LHS}={RHS}! \\ $$

Commented by Tawa1 last updated on 20/Jan/19

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 20/Jan/19

condition... 1)x>0 2)(x−1)>0 i,e x>1 3)x−2+(√x) >0 if x=1 then x−2+(√x) =0 but value of x+(√(x )) should be greater than 2 so x should be x>1 4)x+1+(√x) >0 when x>1 so x should be greater than 1 wait... f(x)=(1/( (√(x+(√x)++1)))) g(x)=(√(x−1)) +(1/( (√(x+(√x) −2)))) from graph it is confirmed that f(x) never intersect g(x) hence no solution... however... condition of two curve intesect/touch is shortest distance zero..

$${condition}… \\ $$$$\left.\mathrm{1}\right){x}>\mathrm{0} \\ $$$$\left.\mathrm{2}\right)\left({x}−\mathrm{1}\right)>\mathrm{0}\:{i},{e}\:{x}>\mathrm{1} \\ $$$$\left.\mathrm{3}\right){x}−\mathrm{2}+\sqrt{{x}}\:>\mathrm{0} \\ $$$${if}\:{x}=\mathrm{1}\:\:{then}\:{x}−\mathrm{2}+\sqrt{{x}}\:=\mathrm{0} \\ $$$${but}\:{value}\:{of}\:{x}+\sqrt{{x}\:}\:{should}\:{be}\:{greater}\:{than}\:\mathrm{2} \\ $$$${so}\:{x}\:{should}\:{be}\:{x}>\mathrm{1} \\ $$$$\left.\mathrm{4}\right){x}+\mathrm{1}+\sqrt{{x}}\:>\mathrm{0}\:{when}\:{x}>\mathrm{1} \\ $$$${so}\:{x}\:{should}\:{be}\:{greater}\:{than}\:\mathrm{1} \\ $$$${wait}… \\ $$$${f}\left({x}\right)=\frac{\mathrm{1}}{\:\sqrt{{x}+\sqrt{{x}}++\mathrm{1}}} \\ $$$${g}\left({x}\right)=\sqrt{{x}−\mathrm{1}}\:+\frac{\mathrm{1}}{\:\sqrt{{x}+\sqrt{{x}}\:−\mathrm{2}}} \\ $$$${from}\:{graph}\:{it}\:{is}\:{confirmed}\:{that}\:{f}\left({x}\right)\:{never} \\ $$$${intersect}\:{g}\left({x}\right)\:\:\:{hence}\:\:{no}\:{solution}… \\ $$$${however}… \\ $$$${condition}\:{of}\:{two}\:{curve}\:{intesect}/{touch}\:{is}\:{shortest} \\ $$$${distance}\:{zero}.. \\ $$$$ \\ $$

Commented by Tawa1 last updated on 20/Jan/19