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Question-201722

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Question Number 201722 by Calculusboy last updated on 11/Dec/23
Commented by Frix last updated on 11/Dec/23
There also should be complex solutions
$$\mathrm{There}\:\mathrm{also}\:\mathrm{should}\:\mathrm{be}\:\mathrm{complex}\:\mathrm{solutions} \\ $$
Answered by Sutrisno last updated on 11/Dec/23
x^2 +y^2 +xy=3 (x+y)^2 −xy=3 xy=(x+y)^2 −3 ...(1) ((x+y)+1)(4+(x+y)^2 −3+2(x+y))=27 ((x+y)+1)((x+y)^2 +2(x+y)+1)=27 ((x+y)+1)((x+y)+1)^2 =27 ((x+y)+1)^3 =27 x+y=2 ...(2) xy=2^2 −3→ xy=1 x(2−x)=1 x^2 −2x+1=0 x=1, y=1
$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{xy}=\mathrm{3} \\ $$$$\left({x}+{y}\right)^{\mathrm{2}} −{xy}=\mathrm{3} \\ $$$${xy}=\left({x}+{y}\right)^{\mathrm{2}} −\mathrm{3}\:…\left(\mathrm{1}\right) \\ $$$$\left(\left({x}+{y}\right)+\mathrm{1}\right)\left(\mathrm{4}+\left({x}+{y}\right)^{\mathrm{2}} −\mathrm{3}+\mathrm{2}\left({x}+{y}\right)\right)=\mathrm{27} \\ $$$$\left(\left({x}+{y}\right)+\mathrm{1}\right)\left(\left({x}+{y}\right)^{\mathrm{2}} +\mathrm{2}\left({x}+{y}\right)+\mathrm{1}\right)=\mathrm{27} \\ $$$$\left(\left({x}+{y}\right)+\mathrm{1}\right)\left(\left({x}+{y}\right)+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{27} \\ $$$$\left(\left({x}+{y}\right)+\mathrm{1}\right)^{\mathrm{3}} =\mathrm{27} \\ $$$${x}+{y}=\mathrm{2}\:…\left(\mathrm{2}\right) \\ $$$${xy}=\mathrm{2}^{\mathrm{2}} −\mathrm{3}\rightarrow\:{xy}=\mathrm{1} \\ $$$${x}\left(\mathrm{2}−{x}\right)=\mathrm{1} \\ $$$${x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}=\mathrm{0} \\ $$$${x}=\mathrm{1},\:{y}=\mathrm{1} \\ $$$$ \\ $$
Commented by Calculusboy last updated on 11/Dec/23
thanks sir
$$\boldsymbol{{thanks}}\:\boldsymbol{{sir}} \\ $$
Answered by esmaeil last updated on 12/Dec/23
x+y=s→S^2 −p=3 xy=p→(s+1)(4+p+2s)=27→ (s+1)((s^2 +1+2s)=27→ (s+1)=3→s=2→p=1 a^2 −2a+1=0→x=y=1
$${x}+{y}={s}\rightarrow{S}^{\mathrm{2}} −{p}=\mathrm{3} \\ $$$${xy}={p}\rightarrow\left({s}+\mathrm{1}\right)\left(\mathrm{4}+{p}+\mathrm{2}{s}\right)=\mathrm{27}\rightarrow \\ $$$$\left({s}+\mathrm{1}\right)\left(\left({s}^{\mathrm{2}} +\mathrm{1}+\mathrm{2}{s}\right)=\mathrm{27}\rightarrow\right. \\ $$$$\left({s}+\mathrm{1}\right)=\mathrm{3}\rightarrow{s}=\mathrm{2}\rightarrow{p}=\mathrm{1} \\ $$$${a}^{\mathrm{2}} −\mathrm{2}{a}+\mathrm{1}=\mathrm{0}\rightarrow{x}={y}=\mathrm{1} \\ $$$$ \\ $$$$ \\ $$
Answered by Rasheed.Sindhi last updated on 12/Dec/23
Slightly different way { ((x^2 +y^2 +xy=3.......(i))),(((x+y+1)(4+xy+2x+2y)=27...(ii))) :} (ii)⇒(x+y+1)(1+3+xy+2x+2y)=27 (x+y+1)(1+x^2 +y^2 +xy+xy+2x+2y)=27 (x+y+1)(1+x^2 +y^2 +2xy+2x+2y)=27 (x+y+1)( (x+y)^2 +2(x+y)+1 )=27 (x+y+1) (x+y+1)^2 =27 (x+y+1)^3 =27 x+y+1=3 x+y=2.....(iii) x^2 +y^2 +2xy=4 x^2 +y^2 +xy+xy=4 3+xy=4 xy=1.......(i) (iii) & (iv): x=1,y=1
$$\mathcal{S}{lightly}\:{different}\:{way} \\ $$$$\begin{cases}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{xy}=\mathrm{3}…….\left({i}\right)}\\{\left({x}+{y}+\mathrm{1}\right)\left(\mathrm{4}+{xy}+\mathrm{2}{x}+\mathrm{2}{y}\right)=\mathrm{27}…\left({ii}\right)}\end{cases}\:\: \\ $$$$\left({ii}\right)\Rightarrow\left({x}+{y}+\mathrm{1}\right)\left(\mathrm{1}+\mathrm{3}+{xy}+\mathrm{2}{x}+\mathrm{2}{y}\right)=\mathrm{27} \\ $$$$\:\:\:\:\:\:\:\:\:\:\left({x}+{y}+\mathrm{1}\right)\left(\mathrm{1}+{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{xy}+{xy}+\mathrm{2}{x}+\mathrm{2}{y}\right)=\mathrm{27} \\ $$$$\:\:\:\:\:\:\:\:\:\:\left({x}+{y}+\mathrm{1}\right)\left(\mathrm{1}+{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{2}{xy}+\mathrm{2}{x}+\mathrm{2}{y}\right)=\mathrm{27} \\ $$$$\:\:\:\:\:\:\:\:\:\:\left({x}+{y}+\mathrm{1}\right)\left(\:\left({x}+{y}\right)^{\mathrm{2}} +\mathrm{2}\left({x}+{y}\right)+\mathrm{1}\:\right)=\mathrm{27} \\ $$$$\:\:\:\:\:\:\:\:\:\left({x}+{y}+\mathrm{1}\right)\:\left({x}+{y}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{27} \\ $$$$\:\:\:\:\:\:\:\:\:\:\left({x}+{y}+\mathrm{1}\right)^{\mathrm{3}} =\mathrm{27} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{x}+{y}+\mathrm{1}=\mathrm{3} \\ $$$$\:\:\:\:\:\:\:\:\:\:{x}+{y}=\mathrm{2}…..\left({iii}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{2}{xy}=\mathrm{4} \\ $$$$\:\:\:\:\:\:\:\:\:\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{xy}+{xy}=\mathrm{4} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{3}+{xy}=\mathrm{4} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{xy}=\mathrm{1}…….\left({i}\right) \\ $$$$\left({iii}\right)\:\&\:\left({iv}\right):\:\:\:\:{x}=\mathrm{1},{y}=\mathrm{1} \\ $$
Commented by Calculusboy last updated on 15/Dec/23
thanks sir
$$\boldsymbol{{thanks}}\:\boldsymbol{{sir}} \\ $$
Answered by Rasheed.Sindhi last updated on 12/Dec/23
A way leading determinant (((ALL))) the solutions: { ((x^2 +y^2 +xy=3....(i))),(((x+y+1)(4+xy+2x+2y)=27...(ii))) :} (ii)⇒(x+y+1)(1+3+xy+2x+2y)=(3)^3 (x+y+1)(1+x^2 +y^2 +xy+xy+2x+2y)=(3)^3 (x+y+1)( (x+y)^2 +2(x+y)+1 )=(3)^3 (x+y+1)(x+y+1)^2 =(3)^3 (x+y+1)^3 =(3)^(3 ) (x+y+1)^3 −(3)^(3 ) =0 {(x+y+1)−3}{(x+y+1)^2 +3(x+y+1)+9}=0^(★★) x+y+1−3=0_(→real roots) ∣ (x+y+1)^2 +3(x+y+1)+9=0^★ _(→complex roots) x+y=2⇒y=2−x (i)⇒x^2 +(2−x)^2 +x(2−x)=3 x^2 −2x+1=0 (x−1)^2 =0 x=1^✓ ⇒y=2−x=2−1=1^✓ ^★ x+y+1=((−3±(√(9−4(9)(1))))/2)=((−3±3(√3))/2) x+y=((−3±3(√3))/2)−1=((−5±3(√3))/2) y=((−5±3(√3))/2)−x (i)⇒(x+y)^2 −xy=3⇒(((−5±3(√3))/2))^2 −x(((−5±3(√3))/2)−x)=3 let ((−5±3(√3))/2)=a a^2 −x(a−x)−3=0 x^2 −ax+a^2 −3=0 x=((a±(√(a^2 −4(a^2 −3))))/2)=((a±(√(12−3a^2 )))/2) y=a−x=a−((a±(√(12−3a^2 )))/2)=((a∓(√(12−3a^2 )))/2) where a= ((−5±3(√3))/2)
$$\boldsymbol{\mathrm{A}}\:\boldsymbol{\mathrm{way}}\:\boldsymbol{\mathrm{leading}}\:\begin{array}{|c|}{\mathbb{ALL}}\\\hline\end{array}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{solutions}}: \\ $$$$\begin{cases}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{xy}=\mathrm{3}….\left({i}\right)}\\{\left({x}+{y}+\mathrm{1}\right)\left(\mathrm{4}+{xy}+\mathrm{2}{x}+\mathrm{2}{y}\right)=\mathrm{27}…\left({ii}\right)}\end{cases}\: \\ $$$$\left({ii}\right)\Rightarrow\left({x}+{y}+\mathrm{1}\right)\left(\mathrm{1}+\mathrm{3}+{xy}+\mathrm{2}{x}+\mathrm{2}{y}\right)=\left(\mathrm{3}\right)^{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\left({x}+{y}+\mathrm{1}\right)\left(\mathrm{1}+{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{xy}+{xy}+\mathrm{2}{x}+\mathrm{2}{y}\right)=\left(\mathrm{3}\right)^{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\left({x}+{y}+\mathrm{1}\right)\left(\:\left({x}+{y}\right)^{\mathrm{2}} +\mathrm{2}\left({x}+{y}\right)+\mathrm{1}\:\right)=\left(\mathrm{3}\right)^{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\left({x}+{y}+\mathrm{1}\right)\left({x}+{y}+\mathrm{1}\right)^{\mathrm{2}} =\left(\mathrm{3}\right)^{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\left({x}+{y}+\mathrm{1}\right)^{\mathrm{3}} =\left(\mathrm{3}\right)^{\mathrm{3}\:} \\ $$$$\:\:\:\:\:\:\left({x}+{y}+\mathrm{1}\right)^{\mathrm{3}} −\left(\mathrm{3}\right)^{\mathrm{3}\:} =\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\left\{\left({x}+{y}+\mathrm{1}\right)−\mathrm{3}\right\}\left\{\left({x}+{y}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{3}\left({x}+{y}+\mathrm{1}\right)+\mathrm{9}\right\}=\mathrm{0}^{\bigstar\bigstar} \\ $$$$\:\:\:\:\:\underset{\rightarrow{real}\:{roots}} {{x}+{y}+\mathrm{1}−\mathrm{3}=\mathrm{0}}\:\mid\:\left(\underset{\rightarrow{complex}\:{roots}} {{x}+{y}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{3}\left({x}+{y}+\mathrm{1}\right)+\mathrm{9}=\mathrm{0}^{\bigstar} } \\ $$$$\:\:\:\:\:{x}+{y}=\mathrm{2}\Rightarrow{y}=\mathrm{2}−{x} \\ $$$$\left({i}\right)\Rightarrow{x}^{\mathrm{2}} +\left(\mathrm{2}−{x}\right)^{\mathrm{2}} +{x}\left(\mathrm{2}−{x}\right)=\mathrm{3} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\left({x}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}=\mathrm{1}^{\checkmark} \Rightarrow{y}=\mathrm{2}−{x}=\mathrm{2}−\mathrm{1}=\mathrm{1}^{\checkmark} \\ $$$$\: \\ $$$$\:^{\bigstar} {x}+{y}+\mathrm{1}=\frac{−\mathrm{3}\pm\sqrt{\mathrm{9}−\mathrm{4}\left(\mathrm{9}\right)\left(\mathrm{1}\right)}}{\mathrm{2}}=\frac{−\mathrm{3}\pm\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\:\:\:\:\:{x}+{y}=\frac{−\mathrm{3}\pm\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}}−\mathrm{1}=\frac{−\mathrm{5}\pm\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:{y}=\frac{−\mathrm{5}\pm\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}}−{x} \\ $$$$\left({i}\right)\Rightarrow\left({x}+{y}\right)^{\mathrm{2}} −{xy}=\mathrm{3}\Rightarrow\left(\frac{−\mathrm{5}\pm\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} −{x}\left(\frac{−\mathrm{5}\pm\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}}−{x}\right)=\mathrm{3} \\ $$$${let}\:\frac{−\mathrm{5}\pm\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}}={a} \\ $$$${a}^{\mathrm{2}} −{x}\left({a}−{x}\right)−\mathrm{3}=\mathrm{0} \\ $$$$\:{x}^{\mathrm{2}} −{ax}+{a}^{\mathrm{2}} −\mathrm{3}=\mathrm{0} \\ $$$${x}=\frac{{a}\pm\sqrt{{a}^{\mathrm{2}} −\mathrm{4}\left({a}^{\mathrm{2}} −\mathrm{3}\right)}}{\mathrm{2}}=\frac{{a}\pm\sqrt{\mathrm{12}−\mathrm{3}{a}^{\mathrm{2}} }}{\mathrm{2}}\: \\ $$$${y}={a}−{x}={a}−\frac{{a}\pm\sqrt{\mathrm{12}−\mathrm{3}{a}^{\mathrm{2}} }}{\mathrm{2}}=\frac{{a}\mp\sqrt{\mathrm{12}−\mathrm{3}{a}^{\mathrm{2}} }}{\mathrm{2}}\:\: \\ $$$${where}\:{a}=\:\frac{−\mathrm{5}\pm\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$ \\ $$
Commented by Rasheed.Sindhi last updated on 12/Dec/23
^(★★) x+y+1−3=0 ∣ (x+y+1)^2 +3(x+y+1)+9=0 x+y+1= 3 , 3ω , 3ω^2 x+y=2 , 3ω−1 , 3ω^2 −1 (i)⇒(x+y)^2 −xy=3⇒xy=(x+y)^2 −3 xy=4−3 , (3ω−1)^2 −3 , (3ω^2 −1)^2 −3 ...... ....
$$\:^{\bigstar\bigstar} \:\:\:{x}+{y}+\mathrm{1}−\mathrm{3}=\mathrm{0}\:\mid\:\left({x}+{y}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{3}\left({x}+{y}+\mathrm{1}\right)+\mathrm{9}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:{x}+{y}+\mathrm{1}=\:\mathrm{3}\:,\:\mathrm{3}\omega\:,\:\mathrm{3}\omega^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:{x}+{y}=\mathrm{2}\:,\:\mathrm{3}\omega−\mathrm{1}\:,\:\mathrm{3}\omega^{\mathrm{2}} −\mathrm{1} \\ $$$$\left({i}\right)\Rightarrow\left({x}+{y}\right)^{\mathrm{2}} −{xy}=\mathrm{3}\Rightarrow{xy}=\left({x}+{y}\right)^{\mathrm{2}} −\mathrm{3} \\ $$$$\:\:\:\:\:\:\:\:\:{xy}=\mathrm{4}−\mathrm{3}\:,\:\left(\mathrm{3}\omega−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{3}\:,\:\left(\mathrm{3}\omega^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} −\mathrm{3} \\ $$$$\:\:\:\:\:…… \\ $$$$\:\:\:\:\:…. \\ $$
Commented by Calculusboy last updated on 15/Dec/23
nice solution sir
$$\boldsymbol{{nice}}\:\boldsymbol{{solution}}\:\boldsymbol{{sir}} \\ $$
Answered by ajfour last updated on 12/Dec/23
x+y=s , xy=m s^2 =3+m (s+1)(2s+m+4)=27 (s+1)(s^2 +2s+1)=27 s=−1+3 , −1+3ω, −1+3ω^2 x, y=((s±(√(3(4−s^2 ))))/2) say if s=2 x, y=(s/2) = 1
$${x}+{y}={s}\:\:\:,\:\:{xy}={m} \\ $$$${s}^{\mathrm{2}} =\mathrm{3}+{m} \\ $$$$\left({s}+\mathrm{1}\right)\left(\mathrm{2}{s}+{m}+\mathrm{4}\right)=\mathrm{27} \\ $$$$\left({s}+\mathrm{1}\right)\left({s}^{\mathrm{2}} +\mathrm{2}{s}+\mathrm{1}\right)=\mathrm{27} \\ $$$${s}=−\mathrm{1}+\mathrm{3}\:,\:−\mathrm{1}+\mathrm{3}\omega,\:−\mathrm{1}+\mathrm{3}\omega^{\mathrm{2}} \\ $$$${x},\:{y}=\frac{{s}\pm\sqrt{\mathrm{3}\left(\mathrm{4}−{s}^{\mathrm{2}} \right)}}{\mathrm{2}} \\ $$$${say}\:{if}\:{s}=\mathrm{2} \\ $$$${x},\:{y}=\frac{{s}}{\mathrm{2}}\:=\:\mathrm{1} \\ $$
Commented by Calculusboy last updated on 15/Dec/23
thanks sir
$$\boldsymbol{{thanks}}\:\boldsymbol{{sir}} \\ $$

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