Given-xy-1-x-1-y-9-x-y-20-where-x-gt-y-find-the-value-of-x-y-y-x-

Question Number 106156 by bemath last updated on 03/Aug/20

Given {\begin{cases} \sqrt{xy} + \frac{1}{\sqrt{x}} + \frac{1}{\sqrt{y}} = 9 \\ \sqrt{x} + \sqrt{y} = 20 \end{cases}

where x > y . find the value of

x \sqrt{y} - y \sqrt{x} .

Commented by bemath last updated on 03/Aug/20

let \sqrt{x} = a & \sqrt{y} = ♭

\to {\begin{cases} a ♭ + \frac{1}{a} + \frac{1}{♭} = 9 \\ a + ♭ = 20 \to ♭ = 20 - a \end{cases}

eq (1) : a ♭ + \frac{a + ♭}{a ♭} = 9

\to {(a ♭)}^{2} + 20 = 9 (a ♭)

\to {(a ♭ - 5) (a ♭ - 4)} = 0

\to {\begin{cases} a ♭ = 5 \to a^{2} - 20 + 5 = 0 \\ a ♭ = 4 \to a^{2} - 20 a + 4 = 0 \end{cases}

\to {\begin{cases} a = \frac{20 \pm \sqrt{380}}{2} = \frac{20 \pm 2 \sqrt{95}}{2} \\ a = \frac{20 \pm \sqrt{384}}{2} = \frac{20 \pm 8 \sqrt{6}}{2} \end{cases}

\to {\begin{cases} a = {\begin{cases} 10 + \sqrt{95} \\ 10 - \sqrt{95} \end{cases} \Rightarrow {\begin{cases} b = 10 - \sqrt{95} \\ b = 10 + \sqrt{95} \end{cases} \\ a = {\begin{cases} 10 + 4 \sqrt{6} \\ 10 - 4 \sqrt{6} \end{cases} \to {\begin{cases} b = 10 - 4 \sqrt{6} \\ b = 10 + 4 \sqrt{6} \end{cases} \end{cases}

the solution {a, b} = {(10 + 4 \sqrt{6}, 10 - 4 \sqrt{6})} or

{(10 + \sqrt{95}, 10 - \sqrt{95})}

then a^{2} ♭ - a ♭^{2} = a ♭ (a - ♭)

= (100 - 96) (8 \sqrt{6})

= 32 \sqrt{6} or 10 \sqrt{95}

Answered by 1549442205PVT last updated on 03/Aug/20

Set \sqrt{xy} = t . From first eqn . of the system

and \sqrt{x} + \sqrt{y} = 20 we get : t + \frac{20}{t} = 9

\Leftrightarrow t^{2} - 9 t + 20 = 0 \Leftrightarrow (t - 4) (t - 5) = 0

\Leftrightarrow t \in {4; 5}

a) For t = 5 \Leftrightarrow \sqrt{xy} = 4, we have {(\sqrt{x} - \sqrt{y})}^{2} = {(\sqrt{x} + \sqrt{y})}^{2} - 4 \sqrt{xy}

= 20^{2} - 4.5 = 380 \Rightarrow \sqrt{x} - \sqrt{y} = 2 \sqrt{95} (as x > y), so

M = x \sqrt{y} - y \sqrt{x} = \sqrt{xy} (\sqrt{x} - \sqrt{y}) = 5 \times 2 \sqrt{95} = \pm 10 \sqrt{95}

\Rightarrow M = 10 \sqrt{95}

b) Fo r t = 4 \Leftrightarrow \sqrt{xy} = 4, we have {(\sqrt{x} - \sqrt{y})}^{2} = {(\sqrt{x} + \sqrt{y})}^{2} - 4 \sqrt{xy}

= 20^{2} - 4.4 = 384 = 6.64 \Rightarrow \sqrt{x} - \sqrt{y} = 8 \sqrt{6} (as x > y)

\Rightarrow M = \sqrt{xy} (\sqrt{x} - \sqrt{y}) = 4 \times 8 \sqrt{6} = 32 \sqrt{6}

\Rightarrow M = 32 \sqrt{6}

Thus, M = x \sqrt{y} - y \sqrt{x} \in {10 \sqrt{95}; 32 \sqrt{6}}

Answered by Rasheed.Sindhi last updated on 03/Aug/20

Without determining x, y

Given {\begin{cases} \sqrt{xy} + \frac{1}{\sqrt{x}} + \frac{1}{\sqrt{y}} = 9 \\ \sqrt{x} + \sqrt{y} = 20 \end{cases}

(i) \Rightarrow xy + \sqrt{x} + \sqrt{y} = 9 \sqrt{xy}

{(\sqrt{xy})}^{2} - 9 \sqrt{xy} + 20 = 0

\sqrt{xy} = \frac{9 \pm \sqrt{81 - 80}}{2} = 5, 4

xy = 25, 16 \dots \dots \dots \dots \dots \dots . A

- - - -

(i) \times 20, (ii) \times 9

{\begin{cases} 20 \sqrt{xy} + \frac{20}{\sqrt{x}} + \frac{20}{\sqrt{y}} = 180 \\ 9 \sqrt{x} + 9 \sqrt{y} = 180 \end{cases}

20 \sqrt{xy} + \frac{20}{\sqrt{x}} + \frac{20}{\sqrt{y}} = 9 \sqrt{x} + 9 \sqrt{y}

M u l t i p l y i n g b y \sqrt{x} \sqrt{y} :

20 xy + 20 \sqrt{y} + 20 \sqrt{x} = 9 x \sqrt{y} + 9 y \sqrt{x}

\frac{20}{9} (xy + \sqrt{x} + \sqrt{y}) = x \sqrt{y} + y \sqrt{x}

x \sqrt{y} + y \sqrt{x} = \frac{20}{9} (xy + 20) \dots \dots \dots . . B

F r o m A & B :

{\begin{cases} x \sqrt{y} + y \sqrt{x} = \frac{20}{9} (25 + 20) = 100 \\ x \sqrt{y} + y \sqrt{x} = \frac{20}{9} (16 + 20) = 80 \end{cases}

{\begin{cases} (x \sqrt{y}) (y \sqrt{x}) = xy \sqrt{xy} = 25 \times 5 = 125 \\ xy \sqrt{xy} = 16 \times 4 = 64 \end{cases}

{(x \sqrt{y} - y \sqrt{x})}^{2} = {(x \sqrt{y} + y \sqrt{x})}^{2} - 4 (x \sqrt{y}) (y \sqrt{x})

{\begin{cases} {(x \sqrt{y} - y \sqrt{x})}^{2} = 100^{2} - 4 (125) \\ {(x \sqrt{y} - y \sqrt{x})}^{2} = 80^{2} - 4 (64) \end{cases}

{\begin{cases} x \sqrt{y} - y \sqrt{x} = 10 \sqrt{95} \\ x \sqrt{y} - y \sqrt{x} = 32 \sqrt{6} \end{cases}

Commented by Rasheed.Sindhi last updated on 03/Aug/20

P l e a s e s i r b o b h a n s s e e n o w . I^{'} v e

c a l c u l a t e d x \sqrt{y} - y \sqrt{x}

Commented by bobhans last updated on 03/Aug/20

sorry sir . the original question is x \sqrt{y} - y \sqrt{x}

sir

Commented by bobhans last updated on 03/Aug/20

yes sir . your answer is same .

Commented by bemath last updated on 03/Aug/20

thank you sir

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