Solve in real numbers the system of equations { (((3x+y)(x+3y)(√(xy)) =14)),(((x+y)(x^2 +14xy+y^2 )= 36)) :}


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Question Number 119795 by bobhans last updated on 27/Oct/20
$Solve in real numbers the system of equations { (((3x+y)(x+3y)(√(xy)) =14)),(((x+y)(x^2 +14xy+y^2 )= 36)) :}$
$S o l v e i n r e a l n u m b e r s t h e s y s t e m o f$ $e q u a t i o n s {\begin{cases} (3 x + y) (x + 3 y) \sqrt{x y} = 14 \\ (x + y) (x^{2} + 14 x y + y^{2}) = 36 \end{cases}$
	Answered by 1549442205PVT last updated on 27/Oct/20
	$Put x=ky we get (((3k+1)(k+3)(√k) )/((k+1)(k^2 +14k+1)))=(7/(18)) ⇔18(3k+1)(k+3)(√k) =7(k+1)(k^2 +14k+1) { (((3x+y)(x+3y)(√(xy)) =14 (1))),(((x+y)(x^2 +14xy+y^2 )= 36 (2))) :} (2)⇔x^4 +15x^2 y+15xy^2 +y^3 −36=0(3) ⇔324k(3k^2 +10k+3)^2 =49(k^3 +15k^2 +15k+1)^2 ⇔324k(9k^4 +118k^2 +60k^3 +60k+9) =49(k^6 +30k^5 +255k^4 +452k^3 +255k^2 +30k+1) ⇔49k^6 −1446k^5 −6945k^4 −16084k^3 −6945k^2 −1446k+49i ⇔49(k^3 +(1/k^3 ))−1446(k^2 +(1/k^2 ))−6945(k+(1/k))−16084=0(∗) Put k+(1/k)=t⇒k^2 +(1/k^2 )=t^2 −2 k^3 +(1/k^3 )=t^3 −3t.Replace into we get 49(t^3 −3t)−1446(t^2 −2)−6945t−16084=0 ⇔49t^3 −1446t^2 −7092t−13192=0 49(t−34)(49t^2 +220t+338)=0 ⇒t=34⇔k+(1/k)=34⇔k^2 −34k+1=0 Δ′=17^2 −1=288=12^2 .2 k=17±12(√2) i)k=17+12(√2) ⇒x=(17+12(√2))y Replace into (3)we get (3)⇔37/y^3 =(k^3 +15k^2 +15k+1) =k(k^2 +1)+15(k^2 +1)+14k−14 =k(34k)+15.34k+14k−14 =34(k^2 +1)+524k−48=34.34k+524k−48 =1680k−48=1680(17+12(√2))−48=36/y^3 ⇔28512+20160(√2)=36/y^3 y^3 =((36)/(28512+20160(√2))) ⇒y=^3 (√(1/(792+560(√2)))) =^3 (√(1/((6+4(√2))^3 )))=(1/(6+4(√2)))=((3−2(√2))/2) x=(17+12(√2)).((3−2(√2))/2)=((3+2(√2))/2) ii)k=17−12(√2) .Similarly,we get y=((3+2(√2))/2),x=((3−2(√2))/2) Thus the given system has two solutions (x,y)∈{(((3−2(√2))/2),((3+2(√2))/2)),(((3+2(√2))/2),((3−2(√2))/2))}$
	$Put x = ky we get$ $\frac{(3 k + 1) (k + 3) \sqrt{k}}{(k + 1) (k^{2} + 14 k + 1)} = \frac{7}{18}$ $\Leftrightarrow 18 (3 k + 1) (k + 3) \sqrt{k} = 7 (k + 1) (k^{2} + 14 k + 1)$ ${\begin{cases} (3 x + y) (x + 3 y) \sqrt{x y} = 14 (1) \\ (x + y) (x^{2} + 14 x y + y^{2}) = 36 (2) \end{cases}$ $(2) \Leftrightarrow x^{4} + {15 x}^{2} y + {15 xy}^{2} + y^{3} - 36 = 0 (3)$ $\Leftrightarrow 324 k {({3 k}^{2} + 10 k + 3)}^{2} = 49 {(k^{3} + {15 k}^{2} + 15 k + 1)}^{2}$ $\Leftrightarrow 324 k ({9 k}^{4} + {118 k}^{2} + {60 k}^{3} + 60 k + 9)$ $= 49 (k^{6} + {30 k}^{5} + {255 k}^{4} + {452 k}^{3} + {255 k}^{2} + 30 k + 1)$ $\Leftrightarrow {49 k}^{6} - {1446 k}^{5} - {6945 k}^{4} - {16084 k}^{3} - {6945 k}^{2} - 1446 k + 49 i$ $\Leftrightarrow 49 (k^{3} + \frac{1}{k^{3}}) - 1446 (k^{2} + \frac{1}{k^{2}}) - 6945 (k + \frac{1}{k}) - 16084 = 0 (*)$ $Put k + \frac{1}{k} = t \Rightarrow k^{2} + \frac{1}{k^{2}} = t^{2} - 2$ $k^{3} + \frac{1}{k^{3}} = t^{3} - 3 t . Replace into we get$ $49 (t^{3} - 3 t) - 1446 (t^{2} - 2) - 6945 t - 16084 = 0$ $\Leftrightarrow {49 t}^{3} - {1446 t}^{2} - 7092 t - 13192 = 0$ $49 (t - 34) ({49 t}^{2} + 220 t + 338) = 0$ $\Rightarrow t = 34 \Leftrightarrow k + \frac{1}{k} = 34 \Leftrightarrow k^{2} - 34 k + 1 = 0$ $Δ^{'} = 17^{2} - 1 = 288 = 12^{2} . 2$ $k = 17 \pm 12 \sqrt{2}$ $i) k = 17 + 12 \sqrt{2} \Rightarrow x = (17 + 12 \sqrt{2}) y$ $Replace into (3) we get$ $(3) \Leftrightarrow 37 / y^{3} = (k^{3} + {15 k}^{2} + 15 k + 1)$ $= k (k^{2} + 1) + 15 (k^{2} + 1) + 14 k - 14$ $= k (34 k) + 15 . 34 k + 14 k - 14$ $= 34 (k^{2} + 1) + 524 k - 48 = 34 . 34 k + 524 k - 48$ $= 1680 k - 48 = 1680 (17 + 12 \sqrt{2}) - 48 = 36 / y^{3}$ $\Leftrightarrow 28512 + 20160 \sqrt{2} = 36 / y^{3}$ $y^{3} = \frac{36}{28512 + 20160 \sqrt{2}} \Rightarrow y =^{3} \sqrt{\frac{1}{792 + 560 \sqrt{2}}}$ $=^{3} \sqrt{\frac{1}{{(6 + 4 \sqrt{2})}^{3}}} = \frac{1}{6 + 4 \sqrt{2}} = \frac{3 - 2 \sqrt{2}}{2}$ $x = (17 + 12 \sqrt{2}) . \frac{3 - 2 \sqrt{2}}{2} = \frac{3 + 2 \sqrt{2}}{2}$ $ii) k = 17 - 12 \sqrt{2} . Similarly, we get$ $y = \frac{3 + 2 \sqrt{2}}{2}, x = \frac{3 - 2 \sqrt{2}}{2}$ $Thus the given system has two solutions$ $(x, y) \in {(\frac{3 - 2 \sqrt{2}}{2}, \frac{3 + 2 \sqrt{2}}{2}), (\frac{3 + 2 \sqrt{2}}{2}, \frac{3 - 2 \sqrt{2}}{2})}$
	Answered by MJS_new last updated on 27/Oct/20
	$in this case it′s easier to let x=p−q∧y=p+q we get { ((4(4p^2 −q^2 )(√(p^2 −q^2 ))=14)),((8p(4p^2 −3q^2 )=36 ⇒ q^2 =((8p^3 −9)/(6p)))) :} insert in first equation (((16p^3 +9)(√(54−12p^3 )))/(9(√p^3 )))=14 squaring and transforming leads to p^9 −((27)/8)p^6 +((27)/(64))p^3 −((729)/(512))=0 (p^3 −((27)/8))(p^6 +((27)/(64)))=0 ⇒ p=(3/2) ⇒ q=±(√2) ⇒ x=(3/2)∓(√2)∧y=(3/2)±(√2)$
	$in this case {it}^{'} s easier to let$ $x = p - q \land y = p + q$ $we get$ ${\begin{cases} 4 (4 p^{2} - q^{2}) \sqrt{p^{2} - q^{2}} = 14 \\ 8 p (4 p^{2} - 3 q^{2}) = 36 \Rightarrow q^{2} = \frac{8 p^{3} - 9}{6 p} \end{cases}$ $insert in first equation$ $\frac{(16 p^{3} + 9) \sqrt{54 - 12 p^{3}}}{9 \sqrt{p^{3}}} = 14$ $squaring and transforming leads to$ $p^{9} - \frac{27}{8} p^{6} + \frac{27}{64} p^{3} - \frac{729}{512} = 0$ $(p^{3} - \frac{27}{8}) (p^{6} + \frac{27}{64}) = 0 \Rightarrow p = \frac{3}{2} \Rightarrow q = \pm \sqrt{2}$ $\Rightarrow x = \frac{3}{2} \mp \sqrt{2} \land y = \frac{3}{2} \pm \sqrt{2}$
	Answered by benjo_mathlover last updated on 27/Oct/20
	$by substituting { ((u=(√x))),((v=(√y))) :} we obtain equivalent form { ((uv(3u^4 +10u^2 v^2 +3v^4 )=14)),((u^6 +15u^4 v^2 +15u^2 v^4 +v^6 =36)) :} Here we should recognize elements of binomial with exponent equal to 6. { ((36+2.14=u^6 +6u^5 v+15u^4 v^2 +20u^3 v^3 +15u^2 v^4 +6uv^5 +v^6 )),((36−2.14=u^6 −6u^5 v+15u^4 v^2 −20u^3 v^3 +15u^2 v^4 −6uv^5 +v^6 )) :} therefore { (((u+v)^6 =64)),(((u−v)^6 =8)) :} which implies { ((u+v=2)),((u−v=(√2))) :} since u,v have to be +ve so { ((u=1+((√2)/2))),((v=1−((√2)/2))) :} ⇔ { (((x,y)=((3/2)+(√2), (3/2)−(√2)))),(((x,y)=((3/2)−(√2), (3/2)+(√2)) )) :}$
	$b y s u b s t i t u t i n g {\begin{cases} u = \sqrt{x} \\ v = \sqrt{y} \end{cases}$ $w e o b t a i n e q u i v a l e n t f o r m$ ${\begin{cases} u v (3 u^{4} + 10 u^{2} v^{2} + 3 v^{4}) = 14 \\ u^{6} + 15 u^{4} v^{2} + 15 u^{2} v^{4} + v^{6} = 36 \end{cases}$ $H e r e w e s h o u l d r e c o g n i z e e l e m e n t s$ $o f b i n o m i a l w i t h e x p o n e n t e q u a l t o 6 .$ ${\begin{cases} 36 + 2.14 = u^{6} + 6 u^{5} v + 15 u^{4} v^{2} + 20 u^{3} v^{3} + 15 u^{2} v^{4} + 6 {u v}^{5} + v^{6} \\ 36 - 2.14 = u^{6} - 6 u^{5} v + 15 u^{4} v^{2} - 20 u^{3} v^{3} + 15 u^{2} v^{4} - 6 {u v}^{5} + v^{6} \end{cases}$ $t h e r e f o r e {\begin{cases} {(u + v)}^{6} = 64 \\ {(u - v)}^{6} = 8 \end{cases}$ $w h i c h i m p l i e s {\begin{cases} u + v = 2 \\ u - v = \sqrt{2} \end{cases} s i n c e u, v h a v e t o b e + v e$ $s o {\begin{cases} u = 1 + \frac{\sqrt{2}}{2} \\ v = 1 - \frac{\sqrt{2}}{2} \end{cases} \Leftrightarrow {\begin{cases} (x, y) = (\frac{3}{2} + \sqrt{2}, \frac{3}{2} - \sqrt{2}) \\ (x, y) = (\frac{3}{2} - \sqrt{2}, \frac{3}{2} + \sqrt{2}) \end{cases}$
	Answered by behi83417@gmail.com last updated on 27/Oct/20
	${ (((3x^2 +3y^2 +10xy)(√(xy))=14)),(((x+y)(x^2 +14xy+y^2 )=36)) :} ⇒_(xy=q^2 ) ^(x+y=p) { (([3(x+y)^2 +4xy](√(xy))=14)),(((x+y)[(x+y)^2 +12xy]=36)) :} ⇒ { ((q.(3p^2 +4q^2 )=14)),((p.(p^2 +12q^2 )=36)) :}⇒(q/p).((3p^2 +4q^2 )/(p^2 +12q^2 ))=(7/(18)) ⇒^((p/q)=t) (1/t).((3t^2 +4)/(t^2 +12))=(7/(18))⇒54t^2 +72=7t^3 +84t ⇒7t^3 −54t^2 +84t−72=0 ⇒(t−6)(7t^2 −12t+12)=0 ⇒ { ((t=(p/q)=6)),((t=((6±(√(36−84)))/7)=((6±4i(√3))/7))) :} (p/q)=6⇒6q^2 (36q^2 +12q^2 )=36⇒q^3 =(1/8) ⇒q=(1/2),p=6q=3⇒ { ((x+y=3)),((xy=q^2 =(1/4))) :} ⇒z^2 −3z+(1/4)=0⇒z=x∨y=((3±(√(9−1)))/2) ⇒x∨y=((3±2(√2))/2)=(3/2)±(√2) .■$
	${\begin{cases} ({3 x}^{2} + {3 y}^{2} + 10 xy) \sqrt{xy} = 14 \\ (x + y) (x^{2} + 14 xy + y^{2}) = 36 \end{cases}$ $\underset{xy = q^{2}}{\overset{x + y = p}{\Rightarrow}} {\begin{cases} [3 {(x + y)}^{2} + 4 xy] \sqrt{xy} = 14 \\ (x + y) [{(x + y)}^{2} + 12 xy] = 36 \end{cases}$ $\Rightarrow {\begin{cases} q . ({3 p}^{2} + {4 q}^{2}) = 14 \\ p . (p^{2} + {12 q}^{2}) = 36 \end{cases} \Rightarrow \frac{q}{p} . \frac{{3 p}^{2} + {4 q}^{2}}{p^{2} + {12 q}^{2}} = \frac{7}{18}$ $\overset{\frac{p}{q} = t}{\Rightarrow} \frac{1}{t} . \frac{{3 t}^{2} + 4}{t^{2} + 12} = \frac{7}{18} \Rightarrow {54 t}^{2} + 72 = {7 t}^{3} + 84 t$ $\Rightarrow {7 t}^{3} - {54 t}^{2} + 84 t - 72 = 0$ $\Rightarrow (t - 6) ({7 t}^{2} - 12 t + 12) = 0$ $\Rightarrow {\begin{cases} t = \frac{p}{q} = 6 \\ t = \frac{6 \pm \sqrt{36 - 84}}{7} = \frac{6 \pm 4 i \sqrt{3}}{7} \end{cases}$ $\frac{p}{q} = 6 \Rightarrow {6 q}^{2} ({36 q}^{2} + {12 q}^{2}) = 36 \Rightarrow q^{3} = \frac{1}{8}$ $\Rightarrow q = \frac{1}{2}, p = 6 q = 3 \Rightarrow {\begin{cases} x + y = 3 \\ xy = q^{2} = \frac{1}{4} \end{cases}$ $\Rightarrow z^{2} - 3 z + \frac{1}{4} = 0 \Rightarrow z = x \lor y = \frac{3 \pm \sqrt{9 - 1}}{2}$ $\Rightarrow x \lor y = \frac{3 \pm 2 \sqrt{2}}{2} = \frac{3}{2} \pm \sqrt{2} . ◼$
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