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Question Number 135486 by EDWIN88 last updated on 13/Mar/21
(((√(2x+1))+(√(x−1)))/( (√(2x+1))−(√(x−1)))) = (3/( (√(x+2))))
$$\frac{\sqrt{\mathrm{2x}+\mathrm{1}}+\sqrt{\mathrm{x}−\mathrm{1}}}{\:\sqrt{\mathrm{2x}+\mathrm{1}}−\sqrt{\mathrm{x}−\mathrm{1}}}\:=\:\frac{\mathrm{3}}{\:\sqrt{\mathrm{x}+\mathrm{2}}}\: \\ $$
Commented by benjo_mathlover last updated on 13/Mar/21
wrong
$${wrong} \\ $$$$ \\ $$
Commented by benjo_mathlover last updated on 13/Mar/21
(((√(2.1+1)) +0)/( (√(2.1+1)) −0)) = (1/1) = 1(LHS) RHS ⇒(3/( (√3))) = (√3) how (√3) = 1?
$$\frac{\sqrt{\mathrm{2}.\mathrm{1}+\mathrm{1}}\:+\mathrm{0}}{\:\sqrt{\mathrm{2}.\mathrm{1}+\mathrm{1}}\:−\mathrm{0}}\:=\:\frac{\mathrm{1}}{\mathrm{1}}\:=\:\mathrm{1}\left({LHS}\right) \\ $$$${RHS}\:\Rightarrow\frac{\mathrm{3}}{\:\sqrt{\mathrm{3}}}\:=\:\sqrt{\mathrm{3}}\: \\ $$$${how}\:\sqrt{\mathrm{3}}\:=\:\mathrm{1}? \\ $$
Commented by Dwaipayan Shikari last updated on 13/Mar/21
Oopps ! Sorry
$${Oopps}\:!\:{Sorry} \\ $$
Answered by benjo_mathlover last updated on 13/Mar/21
Answered by mr W last updated on 13/Mar/21
x≥1 ((((√(2x+1))+(√(x−1)))^2 )/(x+2))=(3/( (√(x+2)))) 2x+1+x−1+2(√((2x+1)(x−1)))=3(√(x+2)) 2(√((2x+1)(x−1)))=3((√(x+2))−x) 4(2x+1)(x−1)=9(x+2+x^2 −2x(√(x+2))) x^2 +13x+22=18x(√(x+2)) x^4 −298x^3 −435x^2 +572x+484=0 (x+2)(x^3 −200x^2 +165x+242)=0 ⇒x^3 −300x^2 +165x+242=0 ⇒x≈1.2175
$${x}\geqslant\mathrm{1} \\ $$$$\frac{\left(\sqrt{\mathrm{2}{x}+\mathrm{1}}+\sqrt{{x}−\mathrm{1}}\right)^{\mathrm{2}} }{{x}+\mathrm{2}}=\frac{\mathrm{3}}{\:\sqrt{{x}+\mathrm{2}}} \\ $$$$\mathrm{2}{x}+\mathrm{1}+{x}−\mathrm{1}+\mathrm{2}\sqrt{\left(\mathrm{2}{x}+\mathrm{1}\right)\left({x}−\mathrm{1}\right)}=\mathrm{3}\sqrt{{x}+\mathrm{2}} \\ $$$$\mathrm{2}\sqrt{\left(\mathrm{2}{x}+\mathrm{1}\right)\left({x}−\mathrm{1}\right)}=\mathrm{3}\left(\sqrt{{x}+\mathrm{2}}−{x}\right) \\ $$$$\mathrm{4}\left(\mathrm{2}{x}+\mathrm{1}\right)\left({x}−\mathrm{1}\right)=\mathrm{9}\left({x}+\mathrm{2}+{x}^{\mathrm{2}} −\mathrm{2}{x}\sqrt{{x}+\mathrm{2}}\right) \\ $$$${x}^{\mathrm{2}} +\mathrm{13}{x}+\mathrm{22}=\mathrm{18}{x}\sqrt{{x}+\mathrm{2}} \\ $$$${x}^{\mathrm{4}} −\mathrm{298}{x}^{\mathrm{3}} −\mathrm{435}{x}^{\mathrm{2}} +\mathrm{572}{x}+\mathrm{484}=\mathrm{0} \\ $$$$\left({x}+\mathrm{2}\right)\left({x}^{\mathrm{3}} −\mathrm{200}{x}^{\mathrm{2}} +\mathrm{165}{x}+\mathrm{242}\right)=\mathrm{0} \\ $$$$\Rightarrow{x}^{\mathrm{3}} −\mathrm{300}{x}^{\mathrm{2}} +\mathrm{165}{x}+\mathrm{242}=\mathrm{0} \\ $$$$\Rightarrow{x}\approx\mathrm{1}.\mathrm{2175} \\ $$
Commented by MJS_new last updated on 13/Mar/21
you read my mind or I read yours? anyway you type faster than me...
$$\mathrm{you}\:\mathrm{read}\:\mathrm{my}\:\mathrm{mind}\:\mathrm{or}\:\mathrm{I}\:\mathrm{read}\:\mathrm{yours}?\:\mathrm{anyway} \\ $$$$\mathrm{you}\:\mathrm{type}\:\mathrm{faster}\:\mathrm{than}\:\mathrm{me}… \\ $$
Commented by mr W last updated on 13/Mar/21
haha, it′s telepathy :)
$$\left.{haha},\:{it}'{s}\:{telepathy}\::\right) \\ $$
Answered by MJS_new last updated on 13/Mar/21
((((√(2x+1))+(√(x−1)))^2 )/(x+2))=((3(√(x+2)))/(x+2)) 3x+2(√((x−1)(2x+1)))=3(√(x+2)) 3x=3(√(x+2))−2(√((x−1)(2x+1))) 9x^2 =9(x+2)+4(x−1)(2x+1)−12(√((x−1)(x+2)(2x+1))) 14+5x−x^2 =12(√((x−1)(x+2)(2x+1))) (x^2 −5x−14)^2 =144(x−1)(x+2)(2x+1) x^4 −298x^3 −435x^2 +572x+484=0 (x+2)(x^3 −300x^2 +165x+242)=0 x=−2 impossible of the solutions of the 3^(rd) degree only one solves the given equation: x≈1.21750059
$$\frac{\left(\sqrt{\mathrm{2}{x}+\mathrm{1}}+\sqrt{{x}−\mathrm{1}}\right)^{\mathrm{2}} }{{x}+\mathrm{2}}=\frac{\mathrm{3}\sqrt{{x}+\mathrm{2}}}{{x}+\mathrm{2}} \\ $$$$\mathrm{3}{x}+\mathrm{2}\sqrt{\left({x}−\mathrm{1}\right)\left(\mathrm{2}{x}+\mathrm{1}\right)}=\mathrm{3}\sqrt{{x}+\mathrm{2}} \\ $$$$\mathrm{3}{x}=\mathrm{3}\sqrt{{x}+\mathrm{2}}−\mathrm{2}\sqrt{\left({x}−\mathrm{1}\right)\left(\mathrm{2}{x}+\mathrm{1}\right)} \\ $$$$\mathrm{9}{x}^{\mathrm{2}} =\mathrm{9}\left({x}+\mathrm{2}\right)+\mathrm{4}\left({x}−\mathrm{1}\right)\left(\mathrm{2}{x}+\mathrm{1}\right)−\mathrm{12}\sqrt{\left({x}−\mathrm{1}\right)\left({x}+\mathrm{2}\right)\left(\mathrm{2}{x}+\mathrm{1}\right)} \\ $$$$\mathrm{14}+\mathrm{5}{x}−{x}^{\mathrm{2}} =\mathrm{12}\sqrt{\left({x}−\mathrm{1}\right)\left({x}+\mathrm{2}\right)\left(\mathrm{2}{x}+\mathrm{1}\right)} \\ $$$$\left({x}^{\mathrm{2}} −\mathrm{5}{x}−\mathrm{14}\right)^{\mathrm{2}} =\mathrm{144}\left({x}−\mathrm{1}\right)\left({x}+\mathrm{2}\right)\left(\mathrm{2}{x}+\mathrm{1}\right) \\ $$$${x}^{\mathrm{4}} −\mathrm{298}{x}^{\mathrm{3}} −\mathrm{435}{x}^{\mathrm{2}} +\mathrm{572}{x}+\mathrm{484}=\mathrm{0} \\ $$$$\left({x}+\mathrm{2}\right)\left({x}^{\mathrm{3}} −\mathrm{300}{x}^{\mathrm{2}} +\mathrm{165}{x}+\mathrm{242}\right)=\mathrm{0} \\ $$$${x}=−\mathrm{2}\:\mathrm{impossible} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{solutions}\:\mathrm{of}\:\mathrm{the}\:\mathrm{3}^{\mathrm{rd}} \:\mathrm{degree}\:\mathrm{only}\:\mathrm{one} \\ $$$$\mathrm{solves}\:\mathrm{the}\:\mathrm{given}\:\mathrm{equation}: \\ $$$${x}\approx\mathrm{1}.\mathrm{21750059} \\ $$
Answered by Dwaipayan Shikari last updated on 13/Mar/21
((√(2x+1))/( (√(x−1))))=((3+(√(x+2)))/(3−(√(x+2)))) ⇒((√(2x+1))/( (√(x−1))))=(((3+(√(x+2))))/(3−(√(x+2))))⇒((2x+1)/(x−1))=(((3+(√(x+2)))^2 )/((3−(√(x+2)))^2 )) ⇒((3x)/(x+2))=(((3+(√(x+2)))^2 +(3−(√(x+2)))^2 )/((3+(√(x+2)))^2 −(3−(√(x+2)))^2 )) ⇒((3x)/(x+2))=((2(9+x+2))/(12(√(x+2))))⇒((18x)/( (√(x+2))))=11+x⇒((324x^2 )/(x+2))=121+x^2 +22x ⇒324x^2 =121x+x^3 +22x^2 +242+2x^2 +44x ⇒x^3 −300x^2 +165x+242=0 Which gives x∼1.2175
$$\frac{\sqrt{\mathrm{2}{x}+\mathrm{1}}}{\:\sqrt{{x}−\mathrm{1}}}=\frac{\mathrm{3}+\sqrt{{x}+\mathrm{2}}}{\mathrm{3}−\sqrt{{x}+\mathrm{2}}} \\ $$$$\Rightarrow\frac{\sqrt{\mathrm{2}{x}+\mathrm{1}}}{\:\sqrt{{x}−\mathrm{1}}}=\frac{\left(\mathrm{3}+\sqrt{{x}+\mathrm{2}}\right)}{\mathrm{3}−\sqrt{{x}+\mathrm{2}}}\Rightarrow\frac{\mathrm{2}{x}+\mathrm{1}}{{x}−\mathrm{1}}=\frac{\left(\mathrm{3}+\sqrt{{x}+\mathrm{2}}\right)^{\mathrm{2}} }{\left(\mathrm{3}−\sqrt{{x}+\mathrm{2}}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow\frac{\mathrm{3}{x}}{{x}+\mathrm{2}}=\frac{\left(\mathrm{3}+\sqrt{{x}+\mathrm{2}}\right)^{\mathrm{2}} +\left(\mathrm{3}−\sqrt{{x}+\mathrm{2}}\right)^{\mathrm{2}} }{\left(\mathrm{3}+\sqrt{{x}+\mathrm{2}}\right)^{\mathrm{2}} −\left(\mathrm{3}−\sqrt{{x}+\mathrm{2}}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow\frac{\mathrm{3}{x}}{{x}+\mathrm{2}}=\frac{\mathrm{2}\left(\mathrm{9}+{x}+\mathrm{2}\right)}{\mathrm{12}\sqrt{{x}+\mathrm{2}}}\Rightarrow\frac{\mathrm{18}{x}}{\:\sqrt{{x}+\mathrm{2}}}=\mathrm{11}+{x}\Rightarrow\frac{\mathrm{324}{x}^{\mathrm{2}} }{{x}+\mathrm{2}}=\mathrm{121}+{x}^{\mathrm{2}} +\mathrm{22}{x} \\ $$$$\Rightarrow\mathrm{324}{x}^{\mathrm{2}} =\mathrm{121}{x}+{x}^{\mathrm{3}} +\mathrm{22}{x}^{\mathrm{2}} +\mathrm{242}+\mathrm{2}{x}^{\mathrm{2}} +\mathrm{44}{x} \\ $$$$\Rightarrow{x}^{\mathrm{3}} −\mathrm{300}{x}^{\mathrm{2}} +\mathrm{165}{x}+\mathrm{242}=\mathrm{0} \\ $$$${Which}\:{gives}\:{x}\sim\mathrm{1}.\mathrm{2175}\:\:\: \\ $$

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