u-2-v-v-2-u-12-1-u-1-v-1-3-find-u-and-v-

Question Number 96821 by bobhans last updated on 05/Jun/20

{\begin{cases} \frac{u^{2}}{v} + \frac{v^{2}}{u} = 12 \\ \frac{1}{u} + \frac{1}{v} = \frac{1}{3} \end{cases} . find u and v ?

Answered by john santu last updated on 05/Jun/20

Commented by bobhans last updated on 05/Jun/20

thank you

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

{\begin{cases} u^{3} + v^{3} = 12 uv \\ 3 (u + v) = uv \end{cases}

\underset{uv = q}{\overset{u + v = p}{\Rightarrow}} {\begin{cases} p (p^{2} - 3 q) = 12 q \\ 3 p = q \end{cases} \Rightarrow {\begin{cases} p^{3} - 3 pq = 12 q \\ 3 p = q \end{cases}

\Rightarrow p^{3} - {9 p}^{2} - 36 p = 0 \Rightarrow p (p^{2} - 9 p - 36) = 0

\Rightarrow {\begin{cases} p = 0 \\ p^{2} - 9 p - 36 = 0 \Rightarrow p = \frac{9 \pm \sqrt{81 + 144}}{2} = 12, - 3 \end{cases}

1 . {\begin{cases} p = 0 \Rightarrow q = 0 \Rightarrow u = v = 0 (not ok) \\ p = 12, - 3 \Rightarrow q = 36, - 9 \end{cases}

2 . {\begin{cases} u + v = 12 \\ uv = 36 \end{cases} \Rightarrow u = v = 6

3 . {\begin{cases} u + v = - 3 \\ uv = - 9 \end{cases} z^{2} + 3 z - 9 = 0 \Rightarrow u \lor v = - \frac{3}{2} (1 \pm \sqrt{5})

Commented by bobhans last updated on 05/Jun/20

thank you

Answered by 1549442205 last updated on 05/Jun/20

Condition for the system defined as u, v \neq 0

\Rightarrow uv \neq 0, u + v \neq 0 . Then

the system of eqs is equivalent to

{\begin{cases} u^{3} + v^{3} = 12 uv (1) \\ 3 (u + v) = uv (2) \end{cases} \Leftrightarrow {\begin{cases} {(u + v)}^{3} - 3 uv (u + v) = 12 uv \\ 9 {(u + v)}^{2} = 3 uv (u + v) \end{cases}

Subtructing two equations we get :

Missing \left or extra \right

\Leftrightarrow (x + 3) (x - 12) = 0 \Rightarrow x \in {- 3; 12}

a / if x = - 3 then putting (2) we get

α / {\begin{cases} u + v = - 3 \\ uv = - 9 \end{cases} \Rightarrow {(u - v)}^{2} = {(u + v)}^{2} - 4 uv = 9 + 36 = 45

\Rightarrow u - v = \pm 3 \sqrt{5} . We have :

{\begin{cases} u + v = - 3 \\ u - v = 3 \sqrt{5} \end{cases} \Leftrightarrow {\begin{cases} u = \frac{- 3 + \sqrt{5}}{2} \\ v = \frac{- 3 - \sqrt{5}}{2} \end{cases}

β / {\begin{cases} u + v = - 3 \\ u - v = - 3 \sqrt{5} \end{cases} \Leftrightarrow {\begin{cases} u = \frac{- 3 - 3 \sqrt{5}}{2} \\ v = \frac{- 3 + 3 \sqrt{5}}{2} \end{cases}

b / if x = u + v = 12 then putting (2) we get :

uv = 36 \Rightarrow {(u - v)}^{2} = {(u + v)}^{2} - 4 uv = 144 - 144 = 0

\Rightarrow u = v \Rightarrow u = v = 6 . Thus, the system has three roots :

(u; v) \in {(6; 6); (\frac{- 3 + \sqrt{5}}{2}; \frac{- 3 - \sqrt{5}}{2}); (\frac{- 3 - \sqrt{5}}{2}; \frac{- 3 + \sqrt{5}}{2})}