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Question Number 106259 by qwerty111 last updated on 03/Aug/20

Commented by qwerty111 last updated on 03/Aug/20

(√(xy)) = ?

$$\sqrt{{xy}}\:=\:? \\ $$

Answered by john santu last updated on 03/Aug/20

y(√x) =((95)/8)−x = ((95−8x)/8) y=((95−8x)/(8(√x))) ...(1) x(√y) = ((93)/8)−y=((93−8y)/8) x=((93−8y)/(8(√y)))...(2) (1): (2)→(y/x)=((95−8x)/(8(√x))).((8(√y))/(93−8y)) (y/x)=(√(y/x))(((95−8x)/(93−8y))) ⇒((95−8x)/(93−8y))=((√y)/(√x)) (y/x)=(((95−8x)/(93−8y)))^2 ...

$$\mathrm{y}\sqrt{\mathrm{x}}\:=\frac{\mathrm{95}}{\mathrm{8}}−\mathrm{x}\:=\:\frac{\mathrm{95}−\mathrm{8x}}{\mathrm{8}} \\ $$$$\mathrm{y}=\frac{\mathrm{95}−\mathrm{8x}}{\mathrm{8}\sqrt{\mathrm{x}}}\:...\left(\mathrm{1}\right) \\ $$$$\mathrm{x}\sqrt{\mathrm{y}}\:=\:\frac{\mathrm{93}}{\mathrm{8}}−\mathrm{y}=\frac{\mathrm{93}−\mathrm{8y}}{\mathrm{8}} \\ $$$$\mathrm{x}=\frac{\mathrm{93}−\mathrm{8y}}{\mathrm{8}\sqrt{\mathrm{y}}}...\left(\mathrm{2}\right) \\ $$$$\left(\mathrm{1}\right):\:\left(\mathrm{2}\right)\rightarrow\frac{\mathrm{y}}{\mathrm{x}}=\frac{\mathrm{95}−\mathrm{8x}}{\mathrm{8}\sqrt{\mathrm{x}}}.\frac{\mathrm{8}\sqrt{\mathrm{y}}}{\mathrm{93}−\mathrm{8y}} \\ $$$$\frac{\mathrm{y}}{\mathrm{x}}=\sqrt{\frac{\mathrm{y}}{\mathrm{x}}}\left(\frac{\mathrm{95}−\mathrm{8x}}{\mathrm{93}−\mathrm{8y}}\right)\:\Rightarrow\frac{\mathrm{95}−\mathrm{8x}}{\mathrm{93}−\mathrm{8y}}=\frac{\sqrt{\mathrm{y}}}{\sqrt{\mathrm{x}}} \\ $$$$\frac{\mathrm{y}}{\mathrm{x}}=\left(\frac{\mathrm{95}−\mathrm{8x}}{\mathrm{93}−\mathrm{8y}}\right)^{\mathrm{2}} ... \\ $$$$ \\ $$

Answered by Her_Majesty last updated on 03/Aug/20

believing that x, y must be rational numbers and the denominator of both lhs must be 8 I guess x=(p^2 /4)∧y=(q^2 /4) with p, q natural ⇒ { ((2p^2 +pq^2 =95 ⇒ q^2 =((95−2p^2 )/p))),((p^2 q+2q^2 =93 ⇒ p^2 =((93−2q^2 )/q))) :} now it′s easy to try because p∣(95−2p^2 ) and q∣(93−2q^2 ) and both square naturals ⇒ p=5∧q=3 ⇒ x=((25)/4)∧q=(9/4) ⇒ (√(xy))=((15)/4)

$${believing}\:{that}\:{x},\:{y}\:{must}\:{be}\:{rational}\:{numbers} \\ $$$${and}\:{the}\:{denominator}\:{of}\:{both}\:{lhs}\:{must}\:{be}\:\mathrm{8} \\ $$$${I}\:{guess}\:{x}=\frac{{p}^{\mathrm{2}} }{\mathrm{4}}\wedge{y}=\frac{{q}^{\mathrm{2}} }{\mathrm{4}}\:{with}\:{p},\:{q}\:{natural} \\ $$$$\Rightarrow \\ $$$$\begin{cases}{\mathrm{2}{p}^{\mathrm{2}} +{pq}^{\mathrm{2}} =\mathrm{95}\:\Rightarrow\:{q}^{\mathrm{2}} =\frac{\mathrm{95}−\mathrm{2}{p}^{\mathrm{2}} }{{p}}}\\{{p}^{\mathrm{2}} {q}+\mathrm{2}{q}^{\mathrm{2}} =\mathrm{93}\:\Rightarrow\:{p}^{\mathrm{2}} =\frac{\mathrm{93}−\mathrm{2}{q}^{\mathrm{2}} }{{q}}}\end{cases} \\ $$$${now}\:{it}'{s}\:{easy}\:{to}\:{try}\:{because}\:{p}\mid\left(\mathrm{95}−\mathrm{2}{p}^{\mathrm{2}} \right)\:{and} \\ $$$${q}\mid\left(\mathrm{93}−\mathrm{2}{q}^{\mathrm{2}} \right)\:{and}\:{both}\:{square}\:{naturals} \\ $$$$\Rightarrow\:{p}=\mathrm{5}\wedge{q}=\mathrm{3} \\ $$$$\Rightarrow\:{x}=\frac{\mathrm{25}}{\mathrm{4}}\wedge{q}=\frac{\mathrm{9}}{\mathrm{4}} \\ $$$$\Rightarrow\:\sqrt{{xy}}=\frac{\mathrm{15}}{\mathrm{4}} \\ $$

Commented by Her_Majesty last updated on 03/Aug/20

why question 93???

$${why}\:{question}\:\mathrm{93}??? \\ $$

Commented by bemath last updated on 04/Aug/20

how do you do find p=5 and q=3

$$\mathrm{how}\:\mathrm{do}\:\mathrm{you}\:\mathrm{do}\:\mathrm{find}\:\mathrm{p}=\mathrm{5}\:\mathrm{and}\:\mathrm{q}=\mathrm{3} \\ $$

Commented by Her_Majesty last updated on 04/Aug/20

by trying

$${by}\:{trying} \\ $$

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