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Question Number 98761 by bemath last updated on 16/Jun/20
(√(x+(√x) )) −(√(x−(√x))) = m(√(x/(x+(√x)))) m is a real parameter
$$\sqrt{\mathrm{x}+\sqrt{\mathrm{x}}\:}\:−\sqrt{\mathrm{x}−\sqrt{\mathrm{x}}}\:=\:\mathrm{m}\sqrt{\frac{\mathrm{x}}{\mathrm{x}+\sqrt{\mathrm{x}}}} \\ $$$$\mathrm{m}\:\mathrm{is}\:\mathrm{a}\:\mathrm{real}\:\mathrm{parameter} \\ $$
Commented by MJS last updated on 16/Jun/20
if x∈R ⇒ x>0∧x−(√x)≥0 ⇒ x≥1 (√(x+(√x)))−(√(x−(√x)))=m(√(x/(x+(√x)))) (√((1+(√x))(√x)))−(√((−1+(√x))(√x)))=m(√((√x)/(1+(√x)))) x≥1 ⇒ we are allowed to divide by (√x) (√(1+(√x)))−(√(−1+(√x)))=(m/( (√(1+(√x))))) 1+(√x)−(√(−1+(√x)))×(√(1+(√x)))=m −1+(√x)≥0∧1+(√x)≥0 1+(√x)−(√(x−1))=m ⇒ 1<m≤2 (√x)−(√(x−1))=m−1 squaring (both sides >0) 2x−1−2(√(x(x−1)))=(m−1)^2 2(√(x(x−1)))=2x−1−(m−1)^2 to prove: 2x−1−(m−1)^2 ≥0 2x−1≥(m−1)^2 x≥1 ⇒ 2x−1≥1 (√(2x−1))≥m−1 (√(2x−1))+1≥m (√(2x−1))+1≥2∧m≤2 ⇒ true squaring (both sides >0) 4x(x−1)=(2x−1−(m−1)^2 )^2 4(m−1)^2 x=(m^2 −2m+2)^2 x=(((m^2 −2m+2)^2 )/(4(m−1)^2 ))∧1<m≤2
$$\mathrm{if}\:{x}\in\mathbb{R} \\ $$$$\Rightarrow\:{x}>\mathrm{0}\wedge{x}−\sqrt{{x}}\geqslant\mathrm{0}\:\Rightarrow\:{x}\geqslant\mathrm{1} \\ $$$$\sqrt{{x}+\sqrt{{x}}}−\sqrt{{x}−\sqrt{{x}}}={m}\sqrt{\frac{{x}}{{x}+\sqrt{{x}}}} \\ $$$$\sqrt{\left(\mathrm{1}+\sqrt{{x}}\right)\sqrt{{x}}}−\sqrt{\left(−\mathrm{1}+\sqrt{{x}}\right)\sqrt{{x}}}={m}\sqrt{\frac{\sqrt{{x}}}{\mathrm{1}+\sqrt{{x}}}} \\ $$$${x}\geqslant\mathrm{1}\:\Rightarrow\:\mathrm{we}\:\mathrm{are}\:\mathrm{allowed}\:\mathrm{to}\:\mathrm{divide}\:\mathrm{by}\:\sqrt{{x}} \\ $$$$\sqrt{\mathrm{1}+\sqrt{{x}}}−\sqrt{−\mathrm{1}+\sqrt{{x}}}=\frac{{m}}{\:\sqrt{\mathrm{1}+\sqrt{{x}}}} \\ $$$$\mathrm{1}+\sqrt{{x}}−\sqrt{−\mathrm{1}+\sqrt{{x}}}×\sqrt{\mathrm{1}+\sqrt{{x}}}={m} \\ $$$$−\mathrm{1}+\sqrt{{x}}\geqslant\mathrm{0}\wedge\mathrm{1}+\sqrt{{x}}\geqslant\mathrm{0} \\ $$$$\mathrm{1}+\sqrt{{x}}−\sqrt{{x}−\mathrm{1}}={m} \\ $$$$\Rightarrow\:\mathrm{1}<{m}\leqslant\mathrm{2} \\ $$$$\sqrt{{x}}−\sqrt{{x}−\mathrm{1}}={m}−\mathrm{1} \\ $$$$\mathrm{squaring}\:\left(\mathrm{both}\:\mathrm{sides}\:>\mathrm{0}\right) \\ $$$$\mathrm{2}{x}−\mathrm{1}−\mathrm{2}\sqrt{{x}\left({x}−\mathrm{1}\right)}=\left({m}−\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\mathrm{2}\sqrt{{x}\left({x}−\mathrm{1}\right)}=\mathrm{2}{x}−\mathrm{1}−\left({m}−\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\mathrm{to}\:\mathrm{prove}:\:\mathrm{2}{x}−\mathrm{1}−\left({m}−\mathrm{1}\right)^{\mathrm{2}} \geqslant\mathrm{0} \\ $$$$\:\:\:\:\:\mathrm{2}{x}−\mathrm{1}\geqslant\left({m}−\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:{x}\geqslant\mathrm{1}\:\Rightarrow\:\mathrm{2}{x}−\mathrm{1}\geqslant\mathrm{1} \\ $$$$\:\:\:\:\:\sqrt{\mathrm{2}{x}−\mathrm{1}}\geqslant{m}−\mathrm{1} \\ $$$$\:\:\:\:\:\sqrt{\mathrm{2}{x}−\mathrm{1}}+\mathrm{1}\geqslant{m} \\ $$$$\:\:\:\:\:\sqrt{\mathrm{2}{x}−\mathrm{1}}+\mathrm{1}\geqslant\mathrm{2}\wedge{m}\leqslant\mathrm{2}\:\Rightarrow\:\mathrm{true} \\ $$$$\mathrm{squaring}\:\left(\mathrm{both}\:\mathrm{sides}\:>\mathrm{0}\right) \\ $$$$\mathrm{4}{x}\left({x}−\mathrm{1}\right)=\left(\mathrm{2}{x}−\mathrm{1}−\left({m}−\mathrm{1}\right)^{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$$\mathrm{4}\left({m}−\mathrm{1}\right)^{\mathrm{2}} {x}=\left({m}^{\mathrm{2}} −\mathrm{2}{m}+\mathrm{2}\right)^{\mathrm{2}} \\ $$$${x}=\frac{\left({m}^{\mathrm{2}} −\mathrm{2}{m}+\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{4}\left({m}−\mathrm{1}\right)^{\mathrm{2}} }\wedge\mathrm{1}<{m}\leqslant\mathrm{2} \\ $$
Commented by john santu last updated on 16/Jun/20
(√(x+(√x))) {(√(x+(√x)))−(√(x−(√x)))} = m(√x) x+(√x)−(√(x^2 −x)) = m(√x) x+(1−m)(√x) = (√(x^2 −x)) squaring x^2 +2x(√x)(1−m)+(1−m)^2 x=x^2 −x 2x(√x)(1−m)+(1−m)^2 x+x=0 x { 2(√x) (1−m)+2−2m+m^2 } =0 2(√x) = ((m^2 −2m+2)/(m−1)) ⇒(√x) = ((m^2 −2m+2)/(2m−2)) x = (((m^2 −2m+2)/(2m−2)))^2
$$\sqrt{\mathrm{x}+\sqrt{\mathrm{x}}}\:\left\{\sqrt{\mathrm{x}+\sqrt{\mathrm{x}}}−\sqrt{\mathrm{x}−\sqrt{\mathrm{x}}}\right\}\:=\:\mathrm{m}\sqrt{\mathrm{x}} \\ $$$$\mathrm{x}+\sqrt{\mathrm{x}}−\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{x}}\:=\:\mathrm{m}\sqrt{\mathrm{x}}\: \\ $$$$\mathrm{x}+\left(\mathrm{1}−\mathrm{m}\right)\sqrt{\mathrm{x}}\:=\:\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{x}} \\ $$$$\mathrm{squaring}\: \\ $$$$\mathrm{x}^{\mathrm{2}} +\mathrm{2x}\sqrt{\mathrm{x}}\left(\mathrm{1}−\mathrm{m}\right)+\left(\mathrm{1}−\mathrm{m}\right)^{\mathrm{2}} \mathrm{x}=\mathrm{x}^{\mathrm{2}} −\mathrm{x} \\ $$$$\mathrm{2x}\sqrt{\mathrm{x}}\left(\mathrm{1}−\mathrm{m}\right)+\left(\mathrm{1}−\mathrm{m}\right)^{\mathrm{2}} \mathrm{x}+\mathrm{x}=\mathrm{0} \\ $$$$\mathrm{x}\:\left\{\:\mathrm{2}\sqrt{\mathrm{x}}\:\left(\mathrm{1}−\mathrm{m}\right)+\mathrm{2}−\mathrm{2m}+\mathrm{m}^{\mathrm{2}} \:\right\}\:=\mathrm{0} \\ $$$$\mathrm{2}\sqrt{\mathrm{x}}\:=\:\frac{\mathrm{m}^{\mathrm{2}} −\mathrm{2m}+\mathrm{2}}{\mathrm{m}−\mathrm{1}}\:\Rightarrow\sqrt{\mathrm{x}}\:=\:\frac{\mathrm{m}^{\mathrm{2}} −\mathrm{2m}+\mathrm{2}}{\mathrm{2m}−\mathrm{2}} \\ $$$$\mathrm{x}\:=\:\left(\frac{\mathrm{m}^{\mathrm{2}} −\mathrm{2m}+\mathrm{2}}{\mathrm{2m}−\mathrm{2}}\right)^{\mathrm{2}} \\ $$
Commented by MJS last updated on 16/Jun/20
your answer is only true for m=2
$$\mathrm{your}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{only}\:\mathrm{true}\:\mathrm{for}\:{m}=\mathrm{2} \\ $$
Commented by john santu last updated on 16/Jun/20
typo sir. your answer and my answer is same
$$\mathrm{typo}\:\mathrm{sir}.\:\mathrm{your}\:\mathrm{answer}\:\mathrm{and}\:\mathrm{my}\:\mathrm{answer}\: \\ $$$$\mathrm{is}\:\mathrm{same} \\ $$
Commented by MJS last updated on 16/Jun/20
I see. but we need to restrict m like I showed
$$\mathrm{I}\:\mathrm{see}.\:\mathrm{but}\:\mathrm{we}\:\mathrm{need}\:\mathrm{to}\:\mathrm{restrict}\:{m}\:\mathrm{like}\:\mathrm{I}\:\mathrm{showed} \\ $$

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