Solve-simultaneously-1-u-1-v-1-3-equation-i-u-2-v-v-2-u-12-equation-ii-

Question Number 6824 by Tawakalitu. last updated on 30/Jul/16

S o l v e s i m u l t a n e o u s l y

\frac{1}{u} + \frac{1}{v} = \frac{1}{3} \dots \dots \dots . e q u a t i o n (i)

\frac{u^{2}}{v} + \frac{v^{2}}{u} = 12 \dots \dots . . e q u a t i o n (i i)

Commented by sou1618 last updated on 30/Jul/16

(i) \frac{v + u}{u v} = \frac{1}{3} \Leftrightarrow 3 (u + v) = u v

(i i) \frac{u^{3} + v^{3}}{u v} = 12 \Leftrightarrow (u + v) (u^{2} - u v + v^{2}) = 12 u v

\Leftrightarrow (u + v) {{(u + v)}^{2} - 3 u v} = 12 u v

s e t {\begin{cases} u + v = x \\ u v = y \end{cases}

(i), (i i) \Rightarrow {\begin{cases} (1) . 3 x = y \\ (2) . x^{3} - 3 x y = 12 y \end{cases}

(2) \Rightarrow x^{3} - 3 x (3 x) = 12 (3 x) ((1))

x^{3} - 9 x^{2} - 36 x = 0

x (x - 12) (x + 3) = 0

x = 0, 12, - 3

(x, y) = (0, 0) (12, 36) (- 3, - 9)

(u + v, u v) = (0, 0) (12, 36) (- 3, - 9)

w h e n u + v = 0, u v = 0

(i) i s n o t d e f i n e d

(α + β, α β) \Rightarrow z^{2} - (α + β) z + α β = 0^{'} s s o l u t i o n

z^{2} - 12 z + 36 = 0 \Rightarrow z = 6

z^{2} + 3 z - 9 = 0 \Rightarrow z = \frac{- 3 \pm 3 \sqrt{5}}{2}

\Rightarrow (u, v) = (6, 6) (\frac{- 3 \pm 3 \sqrt{5}}{2}, \frac{- 3 \mp 3 \sqrt{3}}{2})

∴ (u, v) = (6, 6) (\frac{- 3 \pm 3 \sqrt{5}}{2}, \frac{- 3 \mp 3 \sqrt{3}}{2}) .

Commented by Tawakalitu. last updated on 30/Jul/16

T h a n k s f o r y o u r h e l p . i a p p r e c i a t e .