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Question Number 166391 by mnjuly1970 last updated on 19/Feb/22

    solve in  R     (√(x.⌊x⌋)) − (√(⌊x⌋)) = 1    −−−−−−−

$$ \\ $$$$\:\:{solve}\:{in}\:\:\mathbb{R} \\ $$$$\:\:\:\sqrt{{x}.\lfloor{x}\rfloor}\:−\:\sqrt{\lfloor{x}\rfloor}\:=\:\mathrm{1} \\ $$$$\:\:−−−−−−− \\ $$

Answered by mr W last updated on 19/Feb/22

say x=n+f  1=(√(n(n+f)))−(√n)≥n−(√n)  ((√n))^2 −(√n)−1≤0  (√n)≤((1+(√5))/2) ⇒n≤((3+(√5))/2)  ⇒n≤2   ...(i)  1=(√(n(n+f)))−(√n)<(√(n(n+1)))−(√n)  ⇒n≥2   ...(ii)  ⇒n=2  (√(2x))−(√2)=1  2x=(1+(√2))^2 =3+2(√2)  ⇒x=(3/2)+(√2)

$${say}\:{x}={n}+{f} \\ $$$$\mathrm{1}=\sqrt{{n}\left({n}+{f}\right)}−\sqrt{{n}}\geqslant{n}−\sqrt{{n}} \\ $$$$\left(\sqrt{{n}}\right)^{\mathrm{2}} −\sqrt{{n}}−\mathrm{1}\leqslant\mathrm{0} \\ $$$$\sqrt{{n}}\leqslant\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\:\Rightarrow{n}\leqslant\frac{\mathrm{3}+\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\Rightarrow{n}\leqslant\mathrm{2}\:\:\:...\left({i}\right) \\ $$$$\mathrm{1}=\sqrt{{n}\left({n}+{f}\right)}−\sqrt{{n}}<\sqrt{{n}\left({n}+\mathrm{1}\right)}−\sqrt{{n}} \\ $$$$\Rightarrow{n}\geqslant\mathrm{2}\:\:\:...\left({ii}\right) \\ $$$$\Rightarrow{n}=\mathrm{2} \\ $$$$\sqrt{\mathrm{2}{x}}−\sqrt{\mathrm{2}}=\mathrm{1} \\ $$$$\mathrm{2}{x}=\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)^{\mathrm{2}} =\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}} \\ $$$$\Rightarrow{x}=\frac{\mathrm{3}}{\mathrm{2}}+\sqrt{\mathrm{2}} \\ $$

Commented by mnjuly1970 last updated on 19/Feb/22

bravo sir W

$${bravo}\:{sir}\:{W} \\ $$

Answered by mnjuly1970 last updated on 19/Feb/22

  (√(⌊x⌋)) ((√x) −1)=1       D=(1,∞)      { (√x) −1 = (1/( (√(⌊x⌋)) )) }    ⌊x ⌋=n        (√x) ≥(√(⌊x⌋))       (√n)−1 ≤(1/( (√n))) ⇒ n−(√n) ≤1            n^( 2)  −2n+1≤n           n^( 2) −3n +1≤0            (n−(3/2))^( 2) ≤(5/4)              n ≤(((√5) +3)/2) ⇒ n=2               (√(x )) = 1+(1/( (√2))) ⇒x= ((3+2(√2))/2)

$$\:\:\sqrt{\lfloor{x}\rfloor}\:\left(\sqrt{{x}}\:−\mathrm{1}\right)=\mathrm{1}\:\:\:\:\:\:\:\mathrm{D}=\left(\mathrm{1},\infty\right) \\ $$$$\:\:\:\:\left\{\:\sqrt{{x}}\:−\mathrm{1}\:=\:\frac{\mathrm{1}}{\:\sqrt{\lfloor{x}\rfloor}\:}\:\right\}\:\:\:\:\lfloor{x}\:\rfloor={n} \\ $$$$\:\:\:\:\:\:\sqrt{{x}}\:\geqslant\sqrt{\lfloor{x}\rfloor} \\ $$$$\:\:\:\:\:\sqrt{{n}}−\mathrm{1}\:\leqslant\frac{\mathrm{1}}{\:\sqrt{{n}}}\:\Rightarrow\:{n}−\sqrt{{n}}\:\leqslant\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:{n}^{\:\mathrm{2}} \:−\mathrm{2}{n}+\mathrm{1}\leqslant{n} \\ $$$$\:\:\:\:\:\:\:\:\:{n}^{\:\mathrm{2}} −\mathrm{3}{n}\:+\mathrm{1}\leqslant\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\left({n}−\frac{\mathrm{3}}{\mathrm{2}}\right)^{\:\mathrm{2}} \leqslant\frac{\mathrm{5}}{\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{n}\:\leqslant\frac{\sqrt{\mathrm{5}}\:+\mathrm{3}}{\mathrm{2}}\:\Rightarrow\:{n}=\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\sqrt{{x}\:}\:=\:\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\Rightarrow{x}=\:\frac{\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$$\:\:\:\:\: \\ $$

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