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Question Number 106156 by bemath last updated on 03/Aug/20
Given { (((√(xy)) +(1/( (√x))) +(1/( (√y))) =9)),(((√x)+(√y) = 20)) :} where x > y. find the value of x(√y) −y(√x) .
$$\mathrm{Given}\:\begin{cases}{\sqrt{\mathrm{xy}}\:+\frac{\mathrm{1}}{\:\sqrt{\mathrm{x}}}\:+\frac{\mathrm{1}}{\:\sqrt{\mathrm{y}}}\:=\mathrm{9}}\\{\sqrt{\mathrm{x}}+\sqrt{\mathrm{y}}\:=\:\mathrm{20}}\end{cases} \\ $$$$\mathrm{where}\:\mathrm{x}\:>\:\mathrm{y}.\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of} \\ $$$$\mathrm{x}\sqrt{\mathrm{y}}\:−\mathrm{y}\sqrt{\mathrm{x}}\:. \\ $$
Commented by bemath last updated on 03/Aug/20
let (√x) = a & (√y) = ♭ → { ((a♭ + (1/a)+(1/♭) = 9)),((a+♭ = 20 →♭=20−a)) :} eq(1) : a♭ +((a+♭)/(a♭)) = 9 → (a♭)^2 +20=9(a♭) →{(a♭−5)(a♭−4)}=0 → { ((a♭=5→a^2 −20+5=0)),((a♭=4 →a^2 −20a+4=0)) :} → { ((a=((20±(√(380)))/2)=((20±2(√(95)))/2))),((a=((20±(√(384)))/2)=((20±8(√6))/2))) :} → { ((a= { ((10+(√(95)))),((10−(√(95)))) :} ⇒ { ((b=10−(√(95)))),((b=10+(√(95)))) :})),((a= { ((10+4(√6))),((10−4(√6))) :}→ { ((b=10−4(√6))),((b=10+4(√6))) :})) :} the solution {a,b}={(10+4(√6) ,10−4(√6)) }or {(10+(√(95)) ,10−(√(95)))} then a^2 ♭−a♭^2 =a♭(a−♭) = (100−96)(8(√6)) = 32(√6) or 10(√(95))
$$\mathrm{let}\:\sqrt{\mathrm{x}}\:=\:\mathrm{a}\:\&\:\sqrt{\mathrm{y}}\:=\:\flat\: \\ $$$$\rightarrow\begin{cases}{\mathrm{a}\flat\:+\:\frac{\mathrm{1}}{\mathrm{a}}+\frac{\mathrm{1}}{\flat}\:=\:\mathrm{9}}\\{\mathrm{a}+\flat\:=\:\mathrm{20}\:\rightarrow\flat=\mathrm{20}−\mathrm{a}}\end{cases} \\ $$$$\mathrm{eq}\left(\mathrm{1}\right)\::\:\mathrm{a}\flat\:+\frac{\mathrm{a}+\flat}{\mathrm{a}\flat}\:=\:\mathrm{9}\: \\ $$$$\rightarrow\:\left(\mathrm{a}\flat\right)^{\mathrm{2}} \:+\mathrm{20}=\mathrm{9}\left(\mathrm{a}\flat\right) \\ $$$$\rightarrow\left\{\left(\mathrm{a}\flat−\mathrm{5}\right)\left(\mathrm{a}\flat−\mathrm{4}\right)\right\}=\mathrm{0} \\ $$$$\rightarrow\begin{cases}{\mathrm{a}\flat=\mathrm{5}\rightarrow\mathrm{a}^{\mathrm{2}} −\mathrm{20}+\mathrm{5}=\mathrm{0}}\\{\mathrm{a}\flat=\mathrm{4}\:\rightarrow\mathrm{a}^{\mathrm{2}} −\mathrm{20a}+\mathrm{4}=\mathrm{0}}\end{cases} \\ $$$$\rightarrow\begin{cases}{\mathrm{a}=\frac{\mathrm{20}\pm\sqrt{\mathrm{380}}}{\mathrm{2}}=\frac{\mathrm{20}\pm\mathrm{2}\sqrt{\mathrm{95}}}{\mathrm{2}}}\\{\mathrm{a}=\frac{\mathrm{20}\pm\sqrt{\mathrm{384}}}{\mathrm{2}}=\frac{\mathrm{20}\pm\mathrm{8}\sqrt{\mathrm{6}}}{\mathrm{2}}}\end{cases} \\ $$$$\rightarrow\begin{cases}{\mathrm{a}=\begin{cases}{\mathrm{10}+\sqrt{\mathrm{95}}}\\{\mathrm{10}−\sqrt{\mathrm{95}}}\end{cases}\:\Rightarrow\begin{cases}{\mathrm{b}=\mathrm{10}−\sqrt{\mathrm{95}}}\\{\mathrm{b}=\mathrm{10}+\sqrt{\mathrm{95}}}\end{cases}}\\{\mathrm{a}=\begin{cases}{\mathrm{10}+\mathrm{4}\sqrt{\mathrm{6}}}\\{\mathrm{10}−\mathrm{4}\sqrt{\mathrm{6}}}\end{cases}\rightarrow\begin{cases}{\mathrm{b}=\mathrm{10}−\mathrm{4}\sqrt{\mathrm{6}}}\\{\mathrm{b}=\mathrm{10}+\mathrm{4}\sqrt{\mathrm{6}}}\end{cases}}\end{cases} \\ $$$$\mathrm{the}\:\mathrm{solution}\:\left\{\mathrm{a},\mathrm{b}\right\}=\left\{\left(\mathrm{10}+\mathrm{4}\sqrt{\mathrm{6}}\:,\mathrm{10}−\mathrm{4}\sqrt{\mathrm{6}}\right)\:\right\}\mathrm{or} \\ $$$$\left\{\left(\mathrm{10}+\sqrt{\mathrm{95}}\:,\mathrm{10}−\sqrt{\mathrm{95}}\right)\right\} \\ $$$$\mathrm{then}\:\mathrm{a}^{\mathrm{2}} \flat−\mathrm{a}\flat^{\mathrm{2}} =\mathrm{a}\flat\left(\mathrm{a}−\flat\right) \\ $$$$=\:\left(\mathrm{100}−\mathrm{96}\right)\left(\mathrm{8}\sqrt{\mathrm{6}}\right) \\ $$$$=\:\mathrm{32}\sqrt{\mathrm{6}}\:\mathrm{or}\:\mathrm{10}\sqrt{\mathrm{95}} \\ $$
Answered by 1549442205PVT last updated on 03/Aug/20
Set (√(xy))=t.From first eqn.of the system and (√x)+(√y)=20 we get: t+((20)/t)=9 ⇔t^2 −9t+20=0⇔(t−4)(t−5)=0 ⇔t∈{4;5} a)For t=5⇔(√(xy))=4, we have ((√x)−(√y))^2 =((√x)+(√y))^2 −4(√(xy)) =20^2 −4.5=380⇒(√x)−(√y)=2(√(95)) (as x>y) ,so M=x(√y)−y(√(x )) =(√(xy))((√x)−(√y))=5×2(√(95))=±10(√(95)) ⇒M=10(√(95)) b)For t=4⇔(√(xy))=4,we have ((√x)−(√y))^2 =((√x)+(√y))^2 −4(√(xy)) =20^2 −4.4=384=6.64⇒(√x)−(√y)=8(√6) (as x>y) ⇒M=(√(xy))((√x)−(√y))=4×8(√6)=32(√6) ⇒M=32(√6) Thus ,M=x(√y)−y(√x) ∈{10(√(95)) ;32(√6) }
$$\mathrm{Set}\:\sqrt{\mathrm{xy}}=\mathrm{t}.\mathrm{From}\:\mathrm{first}\:\mathrm{eqn}.\mathrm{of}\:\mathrm{the}\:\mathrm{system}\: \\ $$$$\mathrm{and}\:\sqrt{\mathrm{x}}+\sqrt{\mathrm{y}}=\mathrm{20}\:\mathrm{we}\:\mathrm{get}:\:\mathrm{t}+\frac{\mathrm{20}}{\mathrm{t}}=\mathrm{9} \\ $$$$\Leftrightarrow\mathrm{t}^{\mathrm{2}} −\mathrm{9t}+\mathrm{20}=\mathrm{0}\Leftrightarrow\left(\mathrm{t}−\mathrm{4}\right)\left(\mathrm{t}−\mathrm{5}\right)=\mathrm{0} \\ $$$$\Leftrightarrow\mathrm{t}\in\left\{\mathrm{4};\mathrm{5}\right\} \\ $$$$\left.\mathrm{a}\right)\mathrm{For}\:\mathrm{t}=\mathrm{5}\Leftrightarrow\sqrt{\mathrm{xy}}=\mathrm{4},\:\mathrm{we}\:\mathrm{have}\:\left(\sqrt{\mathrm{x}}−\sqrt{\mathrm{y}}\right)^{\mathrm{2}} =\left(\sqrt{\mathrm{x}}+\sqrt{\mathrm{y}}\right)^{\mathrm{2}} −\mathrm{4}\sqrt{\mathrm{xy}} \\ $$$$=\mathrm{20}^{\mathrm{2}} −\mathrm{4}.\mathrm{5}=\mathrm{380}\Rightarrow\sqrt{\mathrm{x}}−\sqrt{\mathrm{y}}=\mathrm{2}\sqrt{\mathrm{95}}\:\left(\mathrm{as}\:\mathrm{x}>\mathrm{y}\right)\:,\mathrm{so} \\ $$$$\mathrm{M}=\mathrm{x}\sqrt{\mathrm{y}}−\mathrm{y}\sqrt{\mathrm{x}\:}\:=\sqrt{\mathrm{xy}}\left(\sqrt{\mathrm{x}}−\sqrt{\mathrm{y}}\right)=\mathrm{5}×\mathrm{2}\sqrt{\mathrm{95}}=\pm\mathrm{10}\sqrt{\mathrm{95}} \\ $$$$\Rightarrow\boldsymbol{\mathrm{M}}=\mathrm{10}\sqrt{\mathrm{95}}\: \\ $$$$\left.\mathrm{b}\right)\mathrm{Fo}\boldsymbol{\mathrm{r}}\:\boldsymbol{\mathrm{t}}=\mathrm{4}\Leftrightarrow\sqrt{\boldsymbol{\mathrm{xy}}}=\mathrm{4},\boldsymbol{\mathrm{we}}\:\boldsymbol{\mathrm{have}}\:\left(\sqrt{\boldsymbol{\mathrm{x}}}−\sqrt{\boldsymbol{\mathrm{y}}}\right)^{\mathrm{2}} =\left(\sqrt{\mathrm{x}}+\sqrt{\mathrm{y}}\right)^{\mathrm{2}} −\mathrm{4}\sqrt{\mathrm{xy}} \\ $$$$=\mathrm{20}^{\mathrm{2}} −\mathrm{4}.\mathrm{4}=\mathrm{384}=\mathrm{6}.\mathrm{64}\Rightarrow\sqrt{\mathrm{x}}−\sqrt{\mathrm{y}}=\mathrm{8}\sqrt{\mathrm{6}}\:\left(\mathrm{as}\:\mathrm{x}>\mathrm{y}\right) \\ $$$$\Rightarrow\mathrm{M}=\sqrt{\mathrm{xy}}\left(\sqrt{\mathrm{x}}−\sqrt{\mathrm{y}}\right)=\mathrm{4}×\mathrm{8}\sqrt{\mathrm{6}}=\mathrm{32}\sqrt{\mathrm{6}} \\ $$$$\Rightarrow\boldsymbol{\mathrm{M}}=\mathrm{32}\sqrt{\mathrm{6}}\: \\ $$$$\boldsymbol{\mathrm{Thus}}\:,\boldsymbol{\mathrm{M}}=\boldsymbol{\mathrm{x}}\sqrt{\boldsymbol{\mathrm{y}}}−\boldsymbol{\mathrm{y}}\sqrt{\boldsymbol{\mathrm{x}}}\:\in\left\{\mathrm{10}\sqrt{\mathrm{95}}\:;\mathrm{32}\sqrt{\mathrm{6}}\:\right\} \\ $$$$ \\ $$
Answered by Rasheed.Sindhi last updated on 03/Aug/20
Without determining x,y Given { (((√(xy)) +(1/( (√x))) +(1/( (√y))) =9)),(((√x)+(√y) = 20)) :} (i)⇒xy+(√x)+(√y)=9(√(xy)) ((√(xy)))^2 −9(√(xy))+20=0 (√(xy))=((9±(√(81−80)))/2)=5,4 xy=25,16...................A −−−− (i)×20 , (ii)×9 { ((20(√(xy)) +((20)/( (√x))) +((20)/( (√y))) =180)),((9(√x)+9(√y) = 180)) :} 20(√(xy)) +((20)/( (√x))) +((20)/( (√y)))=9(√x)+9(√y) Multiplying by (√x)(√y) : 20xy+20(√y)+20(√x)=9x(√y)+9y(√x) ((20)/9)(xy+(√x)+(√y))=x(√y)+y(√x) x(√y)+y(√x)=((20)/9)(xy+20)...........B From A & B: { ((x(√y)+y(√x)=((20)/9)(25+20)=100)),((x(√y)+y(√x)=((20)/9)(16+20)=80)) :} { (((x(√y))(y(√x))=xy(√(xy))=25×5=125)),((xy(√(xy))=16×4=64)) :} (x(√y)−y(√x))^2 =(x(√y)+y(√x))^2 −4(x(√y))(y(√x)) { (((x(√y)−y(√x))^2 =100^2 −4(125))),(((x(√y)−y(√x))^2 =80^2 −4(64))) :} { ((x(√y)−y(√x)=10(√(95)))),((x(√y)−y(√x)=32(√6))) :}
$$\mathrm{Without}\:\mathrm{determining}\:\mathrm{x},\mathrm{y} \\ $$$$\mathrm{Given}\:\begin{cases}{\sqrt{\mathrm{xy}}\:+\frac{\mathrm{1}}{\:\sqrt{\mathrm{x}}}\:+\frac{\mathrm{1}}{\:\sqrt{\mathrm{y}}}\:=\mathrm{9}}\\{\sqrt{\mathrm{x}}+\sqrt{\mathrm{y}}\:=\:\mathrm{20}}\end{cases} \\ $$$$\left(\mathrm{i}\right)\Rightarrow\mathrm{xy}+\sqrt{\mathrm{x}}+\sqrt{\mathrm{y}}=\mathrm{9}\sqrt{\mathrm{xy}} \\ $$$$\:\:\:\:\:\:\:\:\left(\sqrt{\mathrm{xy}}\right)^{\mathrm{2}} −\mathrm{9}\sqrt{\mathrm{xy}}+\mathrm{20}=\mathrm{0} \\ $$$$\:\:\:\:\:\sqrt{\mathrm{xy}}=\frac{\mathrm{9}\pm\sqrt{\mathrm{81}−\mathrm{80}}}{\mathrm{2}}=\mathrm{5},\mathrm{4} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{xy}=\mathrm{25},\mathrm{16}……………….\mathrm{A} \\ $$$$−−−− \\ $$$$\left(\mathrm{i}\right)×\mathrm{20}\:,\:\left(\mathrm{ii}\right)×\mathrm{9} \\ $$$$\:\begin{cases}{\mathrm{20}\sqrt{\mathrm{xy}}\:+\frac{\mathrm{20}}{\:\sqrt{\mathrm{x}}}\:+\frac{\mathrm{20}}{\:\sqrt{\mathrm{y}}}\:=\mathrm{180}}\\{\mathrm{9}\sqrt{\mathrm{x}}+\mathrm{9}\sqrt{\mathrm{y}}\:=\:\mathrm{180}}\end{cases} \\ $$$$\mathrm{20}\sqrt{\mathrm{xy}}\:+\frac{\mathrm{20}}{\:\sqrt{\mathrm{x}}}\:+\frac{\mathrm{20}}{\:\sqrt{\mathrm{y}}}=\mathrm{9}\sqrt{\mathrm{x}}+\mathrm{9}\sqrt{\mathrm{y}}\: \\ $$$${Multiplying}\:{by}\:\sqrt{\mathrm{x}}\sqrt{\mathrm{y}}\:: \\ $$$$\mathrm{20xy}+\mathrm{20}\sqrt{\mathrm{y}}+\mathrm{20}\sqrt{\mathrm{x}}=\mathrm{9x}\sqrt{\mathrm{y}}+\mathrm{9y}\sqrt{\mathrm{x}} \\ $$$$\frac{\mathrm{20}}{\mathrm{9}}\left(\mathrm{xy}+\sqrt{\mathrm{x}}+\sqrt{\mathrm{y}}\right)=\mathrm{x}\sqrt{\mathrm{y}}+\mathrm{y}\sqrt{\mathrm{x}} \\ $$$$\mathrm{x}\sqrt{\mathrm{y}}+\mathrm{y}\sqrt{\mathrm{x}}=\frac{\mathrm{20}}{\mathrm{9}}\left(\mathrm{xy}+\mathrm{20}\right)………..\mathrm{B} \\ $$$${From}\:\mathrm{A}\:\&\:\mathrm{B}: \\ $$$$\begin{cases}{\mathrm{x}\sqrt{\mathrm{y}}+\mathrm{y}\sqrt{\mathrm{x}}=\frac{\mathrm{20}}{\mathrm{9}}\left(\mathrm{25}+\mathrm{20}\right)=\mathrm{100}}\\{\mathrm{x}\sqrt{\mathrm{y}}+\mathrm{y}\sqrt{\mathrm{x}}=\frac{\mathrm{20}}{\mathrm{9}}\left(\mathrm{16}+\mathrm{20}\right)=\mathrm{80}}\end{cases} \\ $$$$\begin{cases}{\left(\mathrm{x}\sqrt{\mathrm{y}}\right)\left(\mathrm{y}\sqrt{\mathrm{x}}\right)=\mathrm{xy}\sqrt{\mathrm{xy}}=\mathrm{25}×\mathrm{5}=\mathrm{125}}\\{\mathrm{xy}\sqrt{\mathrm{xy}}=\mathrm{16}×\mathrm{4}=\mathrm{64}}\end{cases} \\ $$$$\left(\mathrm{x}\sqrt{\mathrm{y}}−\mathrm{y}\sqrt{\mathrm{x}}\right)^{\mathrm{2}} =\left(\mathrm{x}\sqrt{\mathrm{y}}+\mathrm{y}\sqrt{\mathrm{x}}\right)^{\mathrm{2}} −\mathrm{4}\left(\mathrm{x}\sqrt{\mathrm{y}}\right)\left(\mathrm{y}\sqrt{\mathrm{x}}\right) \\ $$$$\begin{cases}{\left(\mathrm{x}\sqrt{\mathrm{y}}−\mathrm{y}\sqrt{\mathrm{x}}\right)^{\mathrm{2}} =\mathrm{100}^{\mathrm{2}} −\mathrm{4}\left(\mathrm{125}\right)}\\{\left(\mathrm{x}\sqrt{\mathrm{y}}−\mathrm{y}\sqrt{\mathrm{x}}\right)^{\mathrm{2}} =\mathrm{80}^{\mathrm{2}} −\mathrm{4}\left(\mathrm{64}\right)}\end{cases}\:\: \\ $$$$\begin{cases}{\mathrm{x}\sqrt{\mathrm{y}}−\mathrm{y}\sqrt{\mathrm{x}}=\mathrm{10}\sqrt{\mathrm{95}}}\\{\mathrm{x}\sqrt{\mathrm{y}}−\mathrm{y}\sqrt{\mathrm{x}}=\mathrm{32}\sqrt{\mathrm{6}}}\end{cases}\:\: \\ $$
Commented by Rasheed.Sindhi last updated on 03/Aug/20
Please sir bobhans see now.I′ve calculated x(√y)−y(√x)
$${Please}\:{sir}\:{bobhans}\:{see}\:{now}.{I}'{ve} \\ $$$${calculated}\:\:\mathrm{x}\sqrt{\mathrm{y}}−\mathrm{y}\sqrt{\mathrm{x}} \\ $$
Commented by bobhans last updated on 03/Aug/20
sorry sir. the original question is x(√y)−y(√x) sir
$$\mathrm{sorry}\:\mathrm{sir}.\:\mathrm{the}\:\mathrm{original}\:\mathrm{question}\:\mathrm{is}\:\mathrm{x}\sqrt{\mathrm{y}}−\mathrm{y}\sqrt{\mathrm{x}} \\ $$$$\mathrm{sir} \\ $$
Commented by bobhans last updated on 03/Aug/20
yes sir. your answer is same.
$$\mathrm{yes}\:\mathrm{sir}.\:\mathrm{your}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{same}.\: \\ $$
Commented by bemath last updated on 03/Aug/20
thank you sir
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

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