Solve-for-x-and-y-1-x-1-y-4-i-x-2-y-y-2-x-9-ii-

Question Number 58529 by Tawa1 last updated on 24/Apr/19

Solve for x and y

\frac{1}{x} + \frac{1}{y} = 4 \dots \dots . (i)

\frac{x^{2}}{y} + \frac{y^{2}}{x} = 9 \dots \dots . (ii)

Commented by MJS last updated on 24/Apr/19

it leads to an exactly solveable polynome

of 4^{th} degree . I will show later

Commented by Tawa1 last updated on 24/Apr/19

Waiting sir . God bless you

Commented by behi83417@gmail.com last updated on 25/Apr/19

x + y = 4 x y

x^{3} + y^{3} = 9 x y \Rightarrow [x + y = p, x y = q] \Rightarrow p^{3} - 3 p q = 9 q

\Rightarrow {\begin{cases} p = 4 q \\ p^{3} - 3 p q = 9 q \Rightarrow {(4 q)}^{3} - 3 (4 q) q = 9 q \end{cases}

\Rightarrow 64 q^{3} - 12 q^{2} - 9 q = 0 \Rightarrow q (64 q^{2} - 12 q - 9) = 0

\Rightarrow 1) q = 0 \Rightarrow (x = 0 \lor y = 0, n o t o k f o r o r i g i n a l

q u e s t i o n)

\Rightarrow 2) 64 q^{2} - 12 q - 9 = 0 \Rightarrow q = \frac{6 \pm \sqrt{36 + 9 \times 64}}{64}

\Rightarrow [q = xy = 3 \times \frac{1 \pm \sqrt{17}}{32} \Rightarrow p = x + y = 3 \times \frac{1 \pm \sqrt{17}}{8}]

z^{2} - \frac{3}{8} (1 \pm \sqrt{17}) z + \frac{3}{32} (1 \pm \sqrt{17}) = 0

z_{1, 2} = \frac{3}{16} [(1 + \sqrt{17}) \pm \sqrt{{(1 + \sqrt{17})}^{2} - (1 + \sqrt{17})}]

= \frac{3}{16} [(1 + \sqrt{17}) \pm \sqrt{17 + \sqrt{17}}]

z_{3, 4} = \frac{3}{16} [(1 - \sqrt{17}) \pm \sqrt{{(1 - \sqrt{17})}^{2} - (1 - \sqrt{17})}]

= \frac{3}{16} [(1 - \sqrt{17}) \pm \sqrt{17 - \sqrt{17}}]

z_{5, 6} = \frac{3}{16} [(1 + \sqrt{17}) \pm \sqrt{{(1 + \sqrt{17})}^{2} - (1 - \sqrt{17})}]

= \frac{3}{16} [(1 + \sqrt{17}) \pm \sqrt{17 + 3 \sqrt{17}}]

z_{7, 8} = \frac{3}{16} [(1 - \sqrt{17}) \pm \sqrt{{(1 - \sqrt{17})}^{2} - (1 + \sqrt{17})}]

= \frac{3}{16} [(1 - \sqrt{17}) \pm \sqrt{17 - 3 \sqrt{17}}]

Commented by Tawa1 last updated on 25/Apr/19

God bless you sir

Answered by MJS last updated on 25/Apr/19

(i) \Rightarrow y = \frac{x}{4 x - 1} \Rightarrow

\Rightarrow (ii) x^{4} - \frac{3}{4} x^{3} - \frac{33}{16} x^{2} + \frac{9}{8} x - \frac{9}{64} = 0

put x_{1, 2} = α \pm \sqrt{β}; x_{3, 4} = γ \pm \sqrt{δ}

(x - x_{1}) (x - x_{2}) (x - x_{3}) (x - x_{4}) = 0

x^{4} - 2 (α + γ) x^{3} + (α^{2} + 4 α γ - β + γ^{2} + δ) x^{2} - 2 (α γ (α + γ) + α δ - β γ) x + (α^{2} - β) (γ^{2} + δ) = 0

now compare factors :

(1) - 2 (α + γ) = - \frac{3}{4}

(2) α^{2} + 4 α γ - β + γ^{2} + δ = - \frac{33}{16}

(3) - 2 (α γ (α + γ) + α δ - β γ) = \frac{9}{8}

(4) (α^{2} - β) (γ^{2} + δ) = - \frac{9}{64}

now solve (1) for α, (2) for β, (3) for γ

α = \frac{3}{8} - γ

β = - 2 γ^{2} + \frac{3}{4} γ + δ + \frac{141}{64}

δ = \frac{32 γ^{3} - 18 γ^{2} - 33 γ + 9}{2 (16 γ - 3)}

\Rightarrow (4)

γ^{6} - \frac{9}{8} γ^{5} - \frac{39}{64} γ^{4} + \frac{369}{512} γ^{3} + \frac{423}{4096} γ^{2} - \frac{3051}{32768} γ + \frac{567}{65536} = 0

γ = r + \frac{3}{16}

r^{6} - \frac{291}{256} r^{4} + \frac{21267}{65536} r^{2} - \frac{23409}{16777216} = 0

r = \sqrt{s}

s^{3} - \frac{291}{256} s^{2} + \frac{21267}{65536} s - \frac{23409}{16777216} = 0

s = t + \frac{97}{256}

t^{3} - \frac{435}{4096} t + \frac{1673}{131072} = 0

trying factors of \frac{1673}{131072} \Rightarrow t_{1} = \frac{7}{32}

(we {don}^{'} t need t_{2, 3} = - \frac{7}{64} \pm \frac{12 \sqrt{2}}{64})

t = \frac{7}{32} \Rightarrow s = \frac{153}{256} \Rightarrow r = \frac{3 \sqrt{17}}{16} \Rightarrow γ = \frac{3}{16} + \frac{3 \sqrt{17}}{16}

now

α = \frac{3}{16} - \frac{3 \sqrt{17}}{16}

β = \frac{69}{128} + \frac{3 \sqrt{17}}{128}

γ = \frac{3}{16} + \frac{3 \sqrt{17}}{16}

δ = - \frac{69}{128} + \frac{3 \sqrt{17}}{16}

\Rightarrow

x_{1} = \frac{3}{16} - \frac{3 \sqrt{17}}{16} - \frac{\sqrt{6 (23 + \sqrt{17})}}{16}

x_{2} = \frac{3}{16} - \frac{3 \sqrt{17}}{16} + \frac{\sqrt{6 (23 + \sqrt{17})}}{16}

x_{3, 4} \notin R

y_{1} = x_{2}

y_{2} = x_{1}

Commented by MJS last updated on 25/Apr/19

I prepared these formulas, the rest is just

putting in the constants

x^{4} + {a x}^{3} + {b x}^{2} + c x + d = 0

(x - α - \sqrt{β}) (x - α + \sqrt{β}) (x - γ - \sqrt{δ}) (x - γ + \sqrt{δ}) = 0

(1) - 2 (α + γ) = a

(2) α^{2} + 4 α γ - β + γ^{2} - δ = b

(3) - 2 (α γ (α + γ) - α δ - β γ) = c

(4) (α^{2} - β) (γ^{2} - δ) = d

\Rightarrow

(1) α = - γ - \frac{a}{2}

(2) β = - 2 γ^{2} - a γ - δ + \frac{a^{2}}{4} - b

(3) δ = - γ^{2} - \frac{a}{2} γ + \frac{(a^{2} - 4 b) γ - 2 c}{2 (4 γ + a)}

\Rightarrow β = - γ^{2} - \frac{a}{2} γ + \frac{2 (a^{2} - 4 b) γ + a^{3} - 4 a b + 4 c}{4 (4 γ + a)}

(4)

γ^{6} + \frac{3 a}{2} γ^{5} + \frac{3 a^{2} + 2 b}{4} γ^{4} + \frac{a (a^{2} + 4 b)}{8} γ^{3} + \frac{2 a^{2} b + a c + b^{2} - 4 d}{16} γ^{2} + \frac{a (a c + b^{2} - 4 d)}{32} γ - \frac{a^{2} d - a b c + c^{2}}{64} = 0

γ = r - \frac{a}{4}

r^{6} - \frac{3 a^{2} - 8 b}{16} r^{4} + \frac{3 a^{4} - 16 (a^{2} b - a c - b^{2} + 4 d)}{256} r^{2} - \frac{{(a^{3} - 4 a b + 8 c)}^{2}}{4096} = 0

r = \sqrt{s}

s^{3} - \frac{3 a^{2} - 8 b}{16} s^{2} + \frac{3 a^{4} - 16 (a^{2} b - a c - b^{2} + 4 d)}{256} s - \frac{{(a^{3} - 4 a b + 8 c)}^{2}}{4096} = 0

s = t + \frac{3 a^{2} - 8 b}{48}

t^{3} + \frac{3 a c - b^{2} - 12 d}{48} t - \frac{2 b^{3} + 9 (3 a^{2} d - a b c - 8 b d + 3 c^{2})}{1728} = 0

p = \frac{3 a c - b^{2} - 12 d}{48}

q = - \frac{2 b^{3} + 9 (3 a^{2} d - a b c - 8 b d + 3 c^{2})}{1728}

now we have to

1 . try factors of q \Rightarrow solved

2 . calculate

D = \frac{p^{3}}{27} + \frac{q^{2}}{4} and decide which method to use

D < 0 \Rightarrow trigonometric solution

D ⩾ 0 \Rightarrow {Cardano}^{'} s solution

in most cases we {won}^{'} t get useable solutions

but we always get exact solutions

γ = \sqrt{t + \frac{3 a^{2} - 8 b}{48}} - \frac{a}{4}

if t {isn}^{'} t a “ {nice}^{″} number we cannot handle

α, β, γ, δ nor x = {\begin{cases} α \pm \sqrt{β} \\ γ \pm \sqrt{δ} \end{cases}

Commented by Tawa1 last updated on 24/Apr/19

God bless you sir, i appreciate your time sir

Commented by tanmay last updated on 25/Apr/19

w i t h o u t u s i n g c a l c u l a t o r h o w t o s o l v e t h i s p r o b l e m ?

i s i t f e a s i b l e t o c a c u l a t e s u c h b i g n u m b e r

65536, 16777216, 33768 e t c t o c a l c u l a t e b y s i m p l e

+, -, \times, / e t c

Commented by tanmay last updated on 25/Apr/19

o k s i r

Answered by tanmay last updated on 24/Apr/19

x + y = 4 x y

x^{3} + y^{3} = 9 x y

{(x + y)}^{3} - 3 x y (x + y) = 9 x y

64 x^{3} y^{3} - 12 x^{2} y^{2} - 9 x y = 0

x y \neq 0

64 x^{2} y^{2} - 12 x y - 9 = 0

x y = \frac{12 \pm \sqrt{144 + 64 \times 36}}{2 \times 64} = a

x + y = 4 a

x y = a

x (4 a - x) = a

- x^{2} + 4 a x - a = 0

x^{2} - 4 a x + a = 0

x = \frac{4 a \pm \sqrt{16 a^{2} - 4 a}}{2}

y = 4 a - (\frac{4 a \pm \sqrt{16 a^{2} - 4 a}}{2})

p l s c a l c u l a t e v a l u e o f a \dots i h a v e a p a t h y t o c a c u l a t e

Commented by Tawa1 last updated on 24/Apr/19

God bless you sir

Answered by Rasheed.Sindhi last updated on 24/Apr/19

(i) \Rightarrow \frac{x + y}{xy} = 4 \dots \dots . (iii)

(ii) \Rightarrow \frac{x^{3} + y^{3}}{xy} = 9 \dots \dots (iv)

(iv) \div (iii) : \frac{x^{3} + y^{3}}{x + y} = \frac{9}{4}

x^{2} + y^{2} - xy = \frac{9}{4}

{(x + y)}^{2} - 3 xy = \frac{9}{4}

From (iii) xy = \frac{x + y}{4}

So {(x + y)}^{2} - 3 (\frac{x + y}{4}) - \frac{9}{4} = 0

4 {(x + y)}^{2} - 3 (x + y) - 9 = 0

x + y = \frac{3 \pm \sqrt{9 - 4 (4) (- 9)}}{2 (4)}

= \frac{3 \pm 3 \sqrt{17}}{8}

C o n t i n u e

Commented by Tawa1 last updated on 24/Apr/19

God bless you sir

God bless you sir

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