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Question Number 171199 by mr W last updated on 09/Jun/22
solve x^2 +y^2 −xy=9 y^2 +z^2 −yz=30 z^2 +x^2 −zx=50
$${solve} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} −{xy}=\mathrm{9} \\ $$$${y}^{\mathrm{2}} +{z}^{\mathrm{2}} −{yz}=\mathrm{30} \\ $$$${z}^{\mathrm{2}} +{x}^{\mathrm{2}} −{zx}=\mathrm{50} \\ $$
Commented by MJS_new last updated on 10/Jun/22
this has no “nice” solution...
$$\mathrm{this}\:\mathrm{has}\:\mathrm{no}\:“\mathrm{nice}''\:\mathrm{solution}… \\ $$
Answered by aleks041103 last updated on 10/Jun/22
a=((x+y)/2) b=((y+z)/2) c=((x+z)/2) ⇒a+b+c=x+y+z x=a+b+c−2b=a−b+c y=a+b+c−2c=a+b−c z=a+b+c−2a=−a+b+c ⇒x^2 =a^2 +b^2 +c^2 +2ac−2ab−2bc y^2 =a^2 +b^2 +c^2 +2ab−2ac−2bc z^2 =a^2 +b^2 +c^2 +2bc−2ab−2ac xy=(a−b+c)(a+b−c)=a^2 −(b−c)^2 = =a^2 −b^2 −c^2 +2bc yz=(a+b−c)(−a+b+c)=b^2 −(a−c)^2 = =b^2 −a^2 −c^2 +2ac xz=(a−b+c)(−a+b+c)=c^2 −(a−b)^2 = =c^2 −a^2 −b^2 +2ab x^2 +y^2 −xy= =a^2 +b^2 +c^2 +2ac−2ab−2bc+ +a^2 +b^2 +c^2 +2ab−2ac−2bc− −a^2 +b^2 +c^2 −2bc= =a^2 +3b^2 +3c^2 −6bc=a^2 +3(b−c)^2 =9 y^2 +z^2 −yz=...=b^2 +3(a−c)^2 =30 x^2 +z^2 −xz=...=c^2 +3(a−b)^2 =50 (b−a)(a+b)+3(a+b−2c)(a−b)=21 (a−b)(3a+3b−6c−a−b)=21 2(a−b)(a+b−3c)=21 .... no idea for now
$${a}=\frac{{x}+{y}}{\mathrm{2}} \\ $$$${b}=\frac{{y}+{z}}{\mathrm{2}} \\ $$$${c}=\frac{{x}+{z}}{\mathrm{2}} \\ $$$$\Rightarrow{a}+{b}+{c}={x}+{y}+{z} \\ $$$${x}={a}+{b}+{c}−\mathrm{2}{b}={a}−{b}+{c} \\ $$$${y}={a}+{b}+{c}−\mathrm{2}{c}={a}+{b}−{c} \\ $$$${z}={a}+{b}+{c}−\mathrm{2}{a}=−{a}+{b}+{c} \\ $$$$\Rightarrow{x}^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{2}{ac}−\mathrm{2}{ab}−\mathrm{2}{bc} \\ $$$${y}^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{2}{ab}−\mathrm{2}{ac}−\mathrm{2}{bc} \\ $$$${z}^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{2}{bc}−\mathrm{2}{ab}−\mathrm{2}{ac} \\ $$$${xy}=\left({a}−{b}+{c}\right)\left({a}+{b}−{c}\right)={a}^{\mathrm{2}} −\left({b}−{c}\right)^{\mathrm{2}} = \\ $$$$={a}^{\mathrm{2}} −{b}^{\mathrm{2}} −{c}^{\mathrm{2}} +\mathrm{2}{bc} \\ $$$${yz}=\left({a}+{b}−{c}\right)\left(−{a}+{b}+{c}\right)={b}^{\mathrm{2}} −\left({a}−{c}\right)^{\mathrm{2}} = \\ $$$$={b}^{\mathrm{2}} −{a}^{\mathrm{2}} −{c}^{\mathrm{2}} +\mathrm{2}{ac} \\ $$$${xz}=\left({a}−{b}+{c}\right)\left(−{a}+{b}+{c}\right)={c}^{\mathrm{2}} −\left({a}−{b}\right)^{\mathrm{2}} = \\ $$$$={c}^{\mathrm{2}} −{a}^{\mathrm{2}} −{b}^{\mathrm{2}} +\mathrm{2}{ab} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} −{xy}= \\ $$$$={a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{2}{ac}−\mathrm{2}{ab}−\mathrm{2}{bc}+ \\ $$$$+{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{2}{ab}−\mathrm{2}{ac}−\mathrm{2}{bc}− \\ $$$$−{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −\mathrm{2}{bc}= \\ $$$$={a}^{\mathrm{2}} +\mathrm{3}{b}^{\mathrm{2}} +\mathrm{3}{c}^{\mathrm{2}} −\mathrm{6}{bc}={a}^{\mathrm{2}} +\mathrm{3}\left({b}−{c}\right)^{\mathrm{2}} =\mathrm{9} \\ $$$${y}^{\mathrm{2}} +{z}^{\mathrm{2}} −{yz}=…={b}^{\mathrm{2}} +\mathrm{3}\left({a}−{c}\right)^{\mathrm{2}} =\mathrm{30} \\ $$$${x}^{\mathrm{2}} +{z}^{\mathrm{2}} −{xz}=…={c}^{\mathrm{2}} +\mathrm{3}\left({a}−{b}\right)^{\mathrm{2}} =\mathrm{50} \\ $$$$\left({b}−{a}\right)\left({a}+{b}\right)+\mathrm{3}\left({a}+{b}−\mathrm{2}{c}\right)\left({a}−{b}\right)=\mathrm{21} \\ $$$$\left({a}−{b}\right)\left(\mathrm{3}{a}+\mathrm{3}{b}−\mathrm{6}{c}−{a}−{b}\right)=\mathrm{21} \\ $$$$\mathrm{2}\left({a}−{b}\right)\left({a}+{b}−\mathrm{3}{c}\right)=\mathrm{21} \\ $$$$…. \\ $$$${no}\:{idea}\:{for}\:{now} \\ $$
Commented by Tawa11 last updated on 11/Jun/22
Nice try sir. Weldone.
$$\mathrm{Nice}\:\mathrm{try}\:\mathrm{sir}.\:\mathrm{Weldone}. \\ $$
Answered by MJS_new last updated on 12/Jun/22
let y=px∧z=qx ⇒ x^2 =(9/(p^2 −p+1))=((30)/(p^2 −pq+q^2 ))=((50)/(q^2 −q+1)) (I) (II) (III) from I=II∧II=III we get q=((29p^2 −20p+11)/(9(p−1))) inserting this leads to p^4 −((71)/(391))p^3 −((240)/(391))p^2 +((469)/(391))p−((149)/(391))=0 this has no “nice”solution p_1 ≈−1.25724339 ⇒ q_1 ≈−4.03560259 p_2 ≈.383450177 ⇒ q_2 ≈−1.36872485 p_(3, 4) ≈.527689447±.715546104i ⇒ q_(3, 4) ≈1.27249620±.142489563i ⇒ x_1 ≈1.53134901 y_1 ≈−1.92527842 z_1 ≈−6.17991602 x_2 ≈3.43315083 y_2 ≈1.31644230 z_2 ≈−4.69903886 x_(3, 4) ≈6.07742719±.500897689i y_(3, 4) ≈3.56540959±4.08436093i z_(3, 4) ≈7.80487569±.228579540i plus all triplets with opposite signs (x_5 =−x_1 ∧y_5 =−y_1 ∧z_5 =−z_1 etc.)
$$\mathrm{let}\:{y}={px}\wedge{z}={qx}\:\Rightarrow \\ $$$${x}^{\mathrm{2}} =\frac{\mathrm{9}}{{p}^{\mathrm{2}} −{p}+\mathrm{1}}=\frac{\mathrm{30}}{{p}^{\mathrm{2}} −{pq}+{q}^{\mathrm{2}} }=\frac{\mathrm{50}}{{q}^{\mathrm{2}} −{q}+\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left({I}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left({II}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left({III}\right) \\ $$$$\mathrm{from}\:{I}={II}\wedge{II}={III}\:\mathrm{we}\:\mathrm{get} \\ $$$${q}=\frac{\mathrm{29}{p}^{\mathrm{2}} −\mathrm{20}{p}+\mathrm{11}}{\mathrm{9}\left({p}−\mathrm{1}\right)} \\ $$$$\mathrm{inserting}\:\mathrm{this}\:\mathrm{leads}\:\mathrm{to} \\ $$$${p}^{\mathrm{4}} −\frac{\mathrm{71}}{\mathrm{391}}{p}^{\mathrm{3}} −\frac{\mathrm{240}}{\mathrm{391}}{p}^{\mathrm{2}} +\frac{\mathrm{469}}{\mathrm{391}}{p}−\frac{\mathrm{149}}{\mathrm{391}}=\mathrm{0} \\ $$$$\mathrm{this}\:\mathrm{has}\:\mathrm{no}\:“\mathrm{nice}''\mathrm{solution} \\ $$$${p}_{\mathrm{1}} \approx−\mathrm{1}.\mathrm{25724339}\:\Rightarrow\:{q}_{\mathrm{1}} \approx−\mathrm{4}.\mathrm{03560259} \\ $$$${p}_{\mathrm{2}} \approx.\mathrm{383450177}\:\Rightarrow\:{q}_{\mathrm{2}} \approx−\mathrm{1}.\mathrm{36872485} \\ $$$${p}_{\mathrm{3},\:\mathrm{4}} \approx.\mathrm{527689447}\pm.\mathrm{715546104i} \\ $$$$\:\:\:\:\:\Rightarrow\:{q}_{\mathrm{3},\:\mathrm{4}} \approx\mathrm{1}.\mathrm{27249620}\pm.\mathrm{142489563i} \\ $$$$\Rightarrow \\ $$$${x}_{\mathrm{1}} \approx\mathrm{1}.\mathrm{53134901} \\ $$$${y}_{\mathrm{1}} \approx−\mathrm{1}.\mathrm{92527842} \\ $$$${z}_{\mathrm{1}} \approx−\mathrm{6}.\mathrm{17991602} \\ $$$$ \\ $$$${x}_{\mathrm{2}} \approx\mathrm{3}.\mathrm{43315083} \\ $$$${y}_{\mathrm{2}} \approx\mathrm{1}.\mathrm{31644230} \\ $$$${z}_{\mathrm{2}} \approx−\mathrm{4}.\mathrm{69903886} \\ $$$$ \\ $$$${x}_{\mathrm{3},\:\mathrm{4}} \approx\mathrm{6}.\mathrm{07742719}\pm.\mathrm{500897689i} \\ $$$${y}_{\mathrm{3},\:\mathrm{4}} \approx\mathrm{3}.\mathrm{56540959}\pm\mathrm{4}.\mathrm{08436093i} \\ $$$${z}_{\mathrm{3},\:\mathrm{4}} \approx\mathrm{7}.\mathrm{80487569}\pm.\mathrm{228579540i} \\ $$$$ \\ $$$$\mathrm{plus}\:\mathrm{all}\:\mathrm{triplets}\:\mathrm{with}\:\mathrm{opposite}\:\mathrm{signs} \\ $$$$\left({x}_{\mathrm{5}} =−{x}_{\mathrm{1}} \wedge{y}_{\mathrm{5}} =−{y}_{\mathrm{1}} \wedge{z}_{\mathrm{5}} =−{z}_{\mathrm{1}} \:{etc}.\right) \\ $$
Commented by Tawa11 last updated on 12/Jun/22
Great sir.
$$\mathrm{Great}\:\mathrm{sir}. \\ $$

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