Find x and y x^2 + y^2 = 25 ...... (i) x^3 + y^3 = 91 ....... (ii)


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Question Number 51431 by Tawa1 last updated on 26/Dec/18

$Find x and y$ $x^{2} + y^{2} = 25 . . . . . . (i)$ $x^{3} + y^{3} = 91 . . . . . . . (ii)$
Commented by Tawa1 last updated on 27/Dec/18

$God bless you sir .$
Commented by Abdo msup. last updated on 26/Dec/18

$l e t x + y = s a n d x y = p w e h a v e s^{2} = x^{2} + y^{2} + 2 x y$ $= 25 + 2 p s l s o s^{3} = x^{3} + y^{3} + 3 x y (x + y)$ $= 91 + 3 s p \Rightarrow s^{2} - 2 p = 25 a n d s^{3} - 3 s p = 91 \Rightarrow$ $s^{3} - 2 p s = 25 s \Rightarrow s^{3} - 3 p s + p s = 25 s \Rightarrow$ $91 + p s = 25 s \Rightarrow (25 - p) s = 91 \Rightarrow$ ${(25 - p)}^{2} s^{2} = 91^{2} \Rightarrow {(25 - p)}^{2} (2 p + 25) = 91^{2} \Rightarrow$ $(p^{2} - 50 p + 25^{2}) (2 p + 25) = 91^{2} \Rightarrow$ $2 p^{3} + 25 p^{2} - 100 p^{2} - 50.25 p + 2 . 25^{2} p + 25^{3} = 91^{2} \Rightarrow$ $2 p^{3} - 75 p^{2} + 25^{3} - 91^{2} = 0 \Rightarrow$ $p^{3} - \frac{75}{2} p^{2} + \frac{25^{3} - 91^{2}}{2} = 0 w e m u s t s o l v e t h e$ $e q u a t i o n x^{3} + {a x}^{2} + b = 0 l e t u s e t h e c h a n g e m e n t$ $x = t - \frac{a}{3}$ $(e) \Leftrightarrow t^{3} - 3 t^{2} . \frac{a}{3} + 3 t . \frac{a^{2}}{9} - \frac{a^{3}}{27} + a (t^{2} - \frac{2 a t}{3} + \frac{a^{2}}{9}) + b = 0$ $\Rightarrow t^{3} - {a t}^{2} + \frac{a^{2} t}{3} - \frac{a^{3}}{27} + {a t}^{2} - 2 \frac{a^{2} t}{3} + \frac{a^{3}}{9} + b = 0 \Rightarrow$ $t^{3} - \frac{a^{2}}{3} t + \frac{2 a^{3}}{9} + b = 0 a f t e r w e u s e t h e c h s n g e m e n t$ $t = u + v . . . . b e c o n t i n u e d . . .$
Commented by Abdo msup. last updated on 26/Dec/18

$a f t e r f i n d i n g s a n d p x a n d y a r e r o o t s o f$ $X^{2} - s X + p = 0$
	Answered by mr W last updated on 26/Dec/18

	$l e t x + y = a, x y = b$ $x^{3} + y^{3} = (x + y) (x^{2} + y^{2} - x y)$ $\Rightarrow 91 = (x + y) (25 - x y)$ $\Rightarrow 91 = a (25 - b) . . . (1)$ $x^{2} + y^{2} = {(x + y)}^{2} - 2 x y$ $\Rightarrow 25 = {(x + y)}^{2} - 2 x y$ $\Rightarrow 25 = a^{2} - 2 b . . . (2)$ $f r o m (2) :$ $\Rightarrow b = \frac{a^{2} - 25}{2}$ $91 = a (25 - \frac{a^{2} - 25}{2})$ $a^{3} - 75 a + 182 = 0$ $(a - 7) (a^{2} + 7 a - 26) = 0$ $\Rightarrow a = 7$ $\Rightarrow a = \frac{- 7 \pm \sqrt{153}}{2}$ $w i t h a = 7 :$ $b = \frac{49 - 25}{2} = 12$ $x, y a r e r o o t s o f$ $t^{2} - 7 t + 12 = 0$ $(t - 3) (t - 4) = 0$ $\Rightarrow x, y = 3, 4$ $w i t h a = \frac{- 7 - \sqrt{153}}{2}$ $\Rightarrow b = \frac{1 - 7 a}{2} = \frac{51 + 7 \sqrt{153}}{4}$ $x, y a r e r o o t s o f$ $t^{2} + \frac{7 + \sqrt{153}}{2} t + \frac{51 + 7 \sqrt{153}}{4} = 0$ $\Rightarrow x, y = \frac{1}{4} (- 7 - \sqrt{153} \pm i \sqrt{2 + 14 \sqrt{153}})$ $w i t h a = \frac{- 7 + \sqrt{153}}{2}$ $\Rightarrow b = \frac{1 - 7 a}{2} = \frac{51 - 7 \sqrt{153}}{4}$ $x, y a r e r o o t s o f$ $t^{2} + \frac{7 - \sqrt{153}}{2} t + \frac{51 - 7 \sqrt{153}}{4} = 0$ $\Rightarrow x, y = \frac{1}{4} (- 7 + \sqrt{153} \pm \sqrt{- 2 + 14 \sqrt{153}})$ $\Rightarrow x, y \approx - 1.928, 4.613$
	Commented by Tawa1 last updated on 27/Dec/18

	$God bless you sir$
	Answered by peter frank last updated on 26/Dec/18

	$l e t$ $a = x^{2}$ $b = y^{2}$ $a + b = 25 . . . . (i) \times x$ $a x + b y = 91 . . . (i i)$ $a x + b x = 25 x$ $a x + b y = 91$ $b (x - y) = 25 x - 91 . . . (i i i)$ $a + b = 25 . . . . (i) \times y$ $a x + b y = 91 . . . (i i)$ $a y + b y = 25 y$ $a x + b y = 91$ $a (y - x) = 25 y - 91 . . . . (i v)$ $b (x - y) = 25 x - 91 . . . (i i i)$ $- a (x - y) = - 25 y + 91 . . . . (i v)$ $(x - y) [- a + b] = 25 (x - y)$ $b - a = 25 . . . . . . v$ $. . . .$
	Commented by Tawa1 last updated on 27/Dec/18

	$God bless you sir$
	Answered by behi83417@gmail.com last updated on 27/Dec/18
	$x+y=p,xy=q ⇒ { ((p^2 −2q=25)),((p(p^2 −3q)=91)) :}⇒ { ((3p^3 −6pq=75p)),((2p^3 −6pq=182)) :} ⇒p^3 −75p+182=0 ⇒(p−7)(p^2 +7p−26)=0 ⇒p=7,[((−7±(√(49+4×26)))/2)=2.68,−9.68] p=7⇒49−2q=25⇒q=12 p=2.68⇒7.18−2q=25⇒q=−8.91 p=−9.68⇒93.7−2q=25⇒q=34.35 1.[p=7,q=12]⇒z^2 −7z+12=0 ⇒[z_1 =x=3,z_2 =y=4]∨[x=4,y=3] 2.[p=2.68,q=−8.91]⇒z^2 −2.68z−8.91=0 ⇒[z_(1,2) =x∨y=((2.68±(√(2.68^2 +4×8.91)))/2)] ⇒[z_(1,2) =x∨y= 4.61 ∨−1.93] 3.[p=−9.69,q=34.35]⇒z^2 +9.69z+34.35=0 ⇒[z_(1,2) =((−9.69±(√(9.69^2 −4×34.35)))/2)] ⇒[z_(1,2) =x∨y=−4.85±3.3i] .■$
	$x + y = p, x y = q$ $\Rightarrow {\begin{cases} p^{2} - 2 q = 25 \\ p (p^{2} - 3 q) = 91 \end{cases} \Rightarrow {\begin{cases} 3 p^{3} - 6 p q = 75 p \\ 2 p^{3} - 6 p q = 182 \end{cases}$ $\Rightarrow p^{3} - 75 p + 182 = 0$ $\Rightarrow (p - 7) (p^{2} + 7 p - 26) = 0$ $\Rightarrow p = 7, [\frac{- 7 \pm \sqrt{49 + 4 \times 26}}{2} = 2.68, - 9.68]$ $p = 7 \Rightarrow 49 - 2 q = 25 \Rightarrow q = 12$ $p = 2.68 \Rightarrow 7.18 - 2 q = 25 \Rightarrow q = - 8.91$ $p = - 9.68 \Rightarrow 93.7 - 2 q = 25 \Rightarrow q = 34.35$ $1 . [p = 7, q = 12] \Rightarrow z^{2} - 7 z + 12 = 0$ $\Rightarrow [z_{1} = x = 3, z_{2} = y = 4] \lor [x = 4, y = 3]$ $2 . [p = 2.68, q = - 8.91] \Rightarrow z^{2} - 2.68 z - 8.91 = 0$ $\Rightarrow [z_{1, 2} = x \lor y = \frac{2.68 \pm \sqrt{2 . 68^{2} + 4 \times 8.91}}{2}]$ $\Rightarrow [z_{1, 2} = x \lor y = 4.61 \lor - 1.93]$ $3 . [p = - 9.69, q = 34.35] \Rightarrow z^{2} + 9.69 z + 34.35 = 0$ $\Rightarrow [z_{1, 2} = \frac{- 9.69 \pm \sqrt{9 . 69^{2} - 4 \times 34.35}}{2}]$ $\Rightarrow [z_{1, 2} = x \lor y = - 4.85 \pm 3.3 i] . ◼$
	Commented by Tawa1 last updated on 27/Dec/18

	$God bless you sir$
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