1-x-1-y-34-1-x-1-y-23-1-xy-find-the-solution-

Question Number 80702 by john santu last updated on 05/Feb/20

{\begin{cases} \frac{1}{x} + \frac{1}{y} = 34 \\ \frac{1}{\sqrt{x}} + \frac{1}{\sqrt{y}} = 23 - \frac{1}{\sqrt{x y}} \end{cases}

f i n d t h e s o l u t i o n .

Commented by mind is power last updated on 05/Feb/20

l e t \frac{1}{\sqrt{x}} + \frac{1}{\sqrt{y}} = u

\frac{1}{\sqrt{x y}} .

\Rightarrow u^{2} - 2 v = 34

u = 23 - v

\Rightarrow {(23 - v)}^{2} - 2 v = 34

v^{2} - 48 v + 495 = 0

v \in {15, 28}

u ⩾ 0 \Rightarrow v = 15 \Rightarrow u = 8

{\begin{cases} \frac{1}{\sqrt{x}} + \frac{1}{\sqrt{y}} = 8 \\ \frac{1}{\sqrt{x}} . \frac{1}{\sqrt{y}} = 15 \end{cases}

\frac{1}{\sqrt{x}}, \frac{1}{\sqrt{y}} r o o t o f X^{2} - 8 X + 15

X \in {3, 5}

\Rightarrow (x, y) \in {(\frac{1}{9}, \frac{1}{25}); (\frac{1}{25}, \frac{1}{9})}

Commented by john santu last updated on 05/Feb/20

l e t \frac{1}{\sqrt{x}} + \frac{1}{\sqrt{y}} = u, \frac{1}{\sqrt{x y}} = v

u^{2} - 2 v = 34, u + v = 23

u^{2} + 2 u - 80 = 0 \Rightarrow {\begin{cases} u = 8 \\ u = - 10 \end{cases}

f o r u = 8, v = 15

\frac{1}{\sqrt{x}} + \frac{1}{\sqrt{y}} = 8 \land \frac{1}{\sqrt{x y}} = 15

\frac{1}{\sqrt{x}} = 3, \frac{1}{\sqrt{x}} = 5 \Rightarrow x = \frac{1}{9}, x = \frac{1}{25}

∴ (\frac{1}{9}, \frac{1}{25}), (\frac{1}{25}, \frac{1}{9})

Commented by john santu last updated on 05/Feb/20

g r e a t s i r

Commented by MJS last updated on 06/Feb/20

\frac{1}{\sqrt{x}} = u - v; \frac{1}{\sqrt{y}} = u + v; u - v > 0 \land u + v > 0

{\begin{cases} v^{2} = 17 - u^{2} \\ u^{2} + u - 20 = 0 \end{cases}

{\begin{cases} v_{1} = \pm 2 \sqrt{2} i; v_{2} = \pm 1 \\ u_{1} = - 5; u_{2} = 4 \end{cases} [u_{1}, v_{1} not useable]

\frac{1}{\sqrt{x}} = 5 \land \frac{1}{\sqrt{y}} = 3 \lor \frac{1}{\sqrt{x}} = 3 \land \frac{1}{\sqrt{y}} = 5

x = \frac{1}{25} \land y = \frac{1}{9} \lor x = \frac{1}{9} \land y = \frac{1}{25}

Answered by behi83417@gmail.com last updated on 05/Feb/20

\frac{1}{x} = t^{2}, \frac{1}{y} = s^{2}

\Rightarrow {\begin{cases} t^{2} + s^{2} = 34 \\ t + s = 23 - ts \end{cases} \underset{ts = q}{\overset{t + s = p}{\Rightarrow}} {\begin{cases} {(t + s)}^{2} - 2 ts = 34 \\ t + s + ts = 23 \end{cases}

\Rightarrow {\begin{cases} p^{2} - 2 q = 34 \\ p + q = 23 \end{cases} \Rightarrow p^{2} - 2 (23 - p) - 34 = 0

\Rightarrow p^{2} + 2 p + 1 = 34 + 46 + 1 \Rightarrow {(p + 1)}^{2} = 81

\Rightarrow p + 1 = \pm 9 \Rightarrow {\begin{cases} p = 8 \Rightarrow q = 15 \\ p = - 10 \Rightarrow q = 33 \end{cases}

\Rightarrow {\begin{cases} [p = 8, q = 15] \Rightarrow [\frac{1}{\sqrt{x}} = 3, 5, \frac{1}{\sqrt{y}} = 5, 3] \\ \Rightarrow (x, y) = (\frac{1}{9}, \frac{1}{25}), (\frac{1}{25}, \frac{1}{9}) \end{cases}

\Rightarrow {\begin{cases} [p = - 10, q = 33] \Rightarrow z^{2} + 10 z + 33 = 0 \\ z = - 5 \pm \sqrt{25 - 4 \times 33} = - 5 \pm \sqrt{107} i \end{cases}

\Rightarrow \frac{1}{\sqrt{x}} = - 5 \pm \sqrt{107} i \Rightarrow {\begin{cases} x = \frac{1}{{(- 5 - \sqrt{107} i)}^{2}} \\ y = \frac{1}{{(- 5 + \sqrt{107} i)}^{2}} \end{cases}