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Question Number 109922 by Ar Brandon last updated on 26/Aug/20

	Answered by 1549442205PVT last updated on 26/Aug/20
	$3x^2 +12xy+6y=0⇔x^2 +4xy+2y=0(1) i)Case x=0⇒y=0 substituting into first eqn. we see it satisfy,so we get y=0⇒(x,y)=(0,0)is a root ii)Case x≠0.From (1)we get y=((−x^2 )/(4x+2))(2) .Replace into first eqn. 3x^2 +6x.((−x^2 )/(4x+2))+6.(x^4 /(16x^2 +16x+4))+10x=0 ⇔6x^4 +48x^4 +48x^3 +12x^2 +160x^3 +160x^2 +40x−24x^4 −12x^3 =0 ⇔30x^4 +196x^3 +172x^2 +40x=0 ⇔15x^3 +98x^2 +86x+20=0.Multiplying two sides by 15^2 and put 15x=t we get t^3 +98t^2 +1290t+4500=0.Set t=u−((98)/3) After some simple transformation we obtain u^3 −((5734)/3)u+((866104)/(27))=0(∗) Putting u=(2/3)(√(5734))×z replace into (∗)we get ((8.5734(√(5734)))/(27))z^3 −((2.5734.(√(5734)))/9)z+((866104)/(27))=0 Multiplying two sides by ((27)/(2.5734(√(5734)))) we get 4z^3 −3z+((216526)/(2867(√(5734))))=0(∗∗).On the other hands,we have 4cos^3 v−3cosv−cos3v=0 .Hence Putting cos3v=((−216526)/(2867(√(5734)))) (3) we infer cosv are roots of (∗∗) (3)⇔3v=cos^(−1) (((−216526)/(2867(√(5734)))))+2kπ ⇔v=(1/3)[cos^(−1) (((−216526)/(2867(√(5734)))))+2kπ](k=0,1,2) Thus z=cosv=cos{(1/3)[cos^(−1) (((−216526)/(2867(√(5734)))))+2kπ]} are three different roots of (∗∗).Hence u=(2/3)(√(5734)) z=(2/3)(√(5734))cos{(1/3)[cos^(−1) (((−216526)/(2867(√(5734)))))+2kπ]}(k=0..2^(−) ) are three roots of (∗).From that we get x=((u−98/3)/(15))=(((2/3)(√(5734 )) cos{(1/3)[cos^(−1) (((−216526)/(2867(√(5734)))))+2kπ]}−((98)/3))/(15)) are three different roots of equation 15x^3 +98x^2 +86x+20=0 Replace into (2) we get corresponding values of y and obtain roots of given system$
	${3 x}^{2} + 12 xy + 6 y = 0 \Leftrightarrow x^{2} + 4 xy + 2 y = 0 (1)$ $i) Case x = 0 \Rightarrow y = 0 substituting into first eqn .$ $we see it satisfy, so$ $we get y = 0 \Rightarrow (x, y) = (0, 0) is a root$ $ii) Case x \neq 0 . From (1) we get y = \frac{- x^{2}}{4 x + 2} (2)$ $. Replace into first eqn .$ ${3 x}^{2} + 6 x . \frac{- x^{2}}{4 x + 2} + 6 . \frac{x^{4}}{{16 x}^{2} + 16 x + 4} + 10 x = 0$ $\Leftrightarrow {6 x}^{4} + {48 x}^{4} + {48 x}^{3} + {12 x}^{2} + {160 x}^{3} + {160 x}^{2} + 40 x - {24 x}^{4} - {12 x}^{3} = 0$ $\Leftrightarrow {30 x}^{4} + {196 x}^{3} + {172 x}^{2} + 40 x = 0$ $\Leftrightarrow {15 x}^{3} + {98 x}^{2} + 86 x + 20 = 0 . Multiplying$ $two sides by 15^{2} and put 15 x = t we get$ $t^{3} + {98 t}^{2} + 1290 t + 4500 = 0 . Set t = u - \frac{98}{3}$ $After some simple transformation we$ $obtain u^{3} - \frac{5734}{3} u + \frac{866104}{27} = 0 ()$ $Putting u = \frac{2}{3} \sqrt{5734} \times z replace into$ $() we get \frac{8.5734 \sqrt{5734}}{27} z^{3} - \frac{2.5734 . \sqrt{5734}}{9} z + \frac{866104}{27} = 0$ $Multiplying two sides by \frac{27}{2.5734 \sqrt{5734}}$ $we get {4 z}^{3} - 3 z + \frac{216526}{2867 \sqrt{5734}} = 0 (* ) . On the$ $other hands, we have$ ${4 \cos}^{3} v - 3 cosv - \cos 3 v = 0 . Hence$ $Putting \cos 3 v = \frac{- 216526}{2867 \sqrt{5734}} (3) we infer$ $cosv are roots of ( )$ $(3) \Leftrightarrow 3 v = \cos^{- 1} (\frac{- 216526}{2867 \sqrt{5734}}) + 2 k π$ $\Leftrightarrow v = \frac{1}{3} [\cos^{- 1} (\frac{- 216526}{2867 \sqrt{5734}}) + 2 k π] (k = 0, 1, 2)$ $Thus z = cosv = \cos {\frac{1}{3} [\cos^{- 1} (\frac{- 216526}{2867 \sqrt{5734}}) + 2 k π]}$ $are three different roots of ( ) . Hence$ $u = \frac{2}{3} \sqrt{5734} z = \frac{2}{3} \sqrt{5734} \cos {\frac{1}{3} [\cos^{- 1} (\frac{- 216526}{2867 \sqrt{5734}}) + 2 k π]} (k = \overset{―}{0 . . 2})$ $are three roots of () . From that we$ $get x = \frac{u - 98 / 3}{15} = \frac{\frac{2}{3} \sqrt{5734} \cos {\frac{1}{3} [\cos^{- 1} (\frac{- 216526}{2867 \sqrt{5734}}) + 2 k π]} - \frac{98}{3}}{15}$ $are three different roots of equation$ $15 x^{3} + 98 x^{2} + 86 x + 20 = 0$ $Replace into (2) we get corresponding values$ $of y and obtain roots of given system$
	Commented by Ar Brandon last updated on 27/Aug/20
	Thanks so much, Sir.
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