x-2-y-2-5-1-3x-2-xy-y-2-1-2-please-help-find-x-and-y-

Question Number 13647 by chux last updated on 22/May/17

x^{2} + y^{2} = 5 \dots . . (1)

{3 x}^{2} + xy + y^{2} = 1 \dots . . (2)

please help find x and y

Commented by ajfour last updated on 22/May/17

2 x^{2} + x y + (x^{2} + y^{2}) = 1

2 x^{2} + x y + 4 = 0 \Rightarrow y = - (2 x + \frac{4}{x}) . . (i)

s q u a r i n g,

x^{2} y^{2} = {(2 x^{2} + 4)}^{2}

x^{2} (5 - x^{2}) = 4 {(x^{2} + 2)}^{2}

- {(x^{2} + 2)}^{2} + 9 (x^{2} + 2) - 14 = 4 {(x^{2} + 2)}^{2}

5 {(x^{2} + 2)}^{2} - 9 (x^{2} + 2) + 14 = 0

x^{2} + 2 = \frac{9 \pm \sqrt{81 - 280}}{10}

x^{2} = - \frac{11}{10} \pm \sqrt{{(\frac{1}{10})}^{2} - 2} \dots . (i i)

y = - (2 x + \frac{4}{x}) \dots . . (i)

d i f f i c u l t f o r m e b e y o n d t h i s p o i n t . .

Commented by prakash jain last updated on 22/May/17

3 x^{2} + x y + y^{2} = 1

2 x^{2} + x y + 5 = 1

2 x^{2} + x y = - 4 \dots . (A)

x^{2} + y^{2} = 5 \Rightarrow x = \sqrt{5} \cos θ \Rightarrow y = \sqrt{5} \sin θ

s u b s t i t u t i n g i n (A)

{10 \cos}^{2} θ + 5 \cos θ \sin θ = - 4

\cos^{2} θ = (\cos 2 θ + 1) / 2, 2 \sin θ \cos θ = \sin 2 θ

5 (\cos 2 θ + 1) + \frac{5}{2} \sin 2 θ = - 4

10 \cos 2 θ + 10 + 5 \sin 2 θ = - 8

10 \cos 2 θ + 5 \sin 2 θ = - 18

2 \cos 2 θ + \sin 2 θ = \frac{- 18}{5}

\frac{2}{\sqrt{5}} \cos 2 θ + \frac{1}{\sqrt{5}} \sin 2 θ = - \frac{18}{5 \sqrt{5}}

\sin (2 θ + \sin^{- 1} \frac{2}{\sqrt{5}}) = - \frac{18}{5 \sqrt{5}} > 1

5 \sqrt{5} \approx 11.18 < 18

n o r e a l s o l u t i o n .

Commented by ajfour last updated on 22/May/17

You can't use 'macro parameter character #' in math mode

- 8 - 50 \neq - 42

Commented by prakash jain last updated on 22/May/17

o k . w i l l c o r r e c t

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 22/May/17

\frac{x}{y} = t

{\begin{cases} t^{2} + 1 = \frac{5}{y^{2}} (i) \\ 3 t^{2} + t + 1 = \frac{1}{y^{2}} (i i) \end{cases}

(i) - 5 (i i) \Rightarrow t^{2} + 1 - 15 t^{2} - 5 t - 5 = 0

\Rightarrow 14 t^{2} + 5 t + 4 = 0

t = \frac{- 5 \pm \sqrt{25 - 14 \times 16}}{28} = \frac{- 5 \pm i \sqrt{199}}{28}

\Rightarrow \frac{x}{y} = - 0.18 \pm 0.5 i (i^{2} = - 1)

n o r e a l a n s w e r s .

Commented by tawa tawa last updated on 22/May/17

God bless you sir