Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 36692 by math2018 last updated on 04/Jun/18

x^3 +y^3 =5 x^2 +y^2 =3

$${x}^{\mathrm{3}} +{y}^{\mathrm{3}} =\mathrm{5} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{3} \\ $$

Commented by behi83417@gmail.com last updated on 04/Jun/18

x+y=s,xy=t (x+y)(x^2 −xy+y^2 )=5⇒s(3−t)=5 x^2 +y^2 =(x+y)^2 −2xy=s^2 −2t=3 ⇒ { ((3s−st=5)),((s^2 −2t=3)) :}⇒ { ((−6s+2st=−10)),((s^3 −2st=3s)) :}⇒ s^3 −9s+10=0⇒(s−2)(s^2 +2s−5)=0 ⇒s=2,−1±(√6) 1)s=2⇒6−2t=5⇒t=(1/2) ⇒ { ((x+y=2)),((xy=(1/2))) :}⇒z^2 −2z+(1/2)=0⇒z=1±((√2)/2) ⇒(x,y)=(1+((√2)/2),1−((√2)/2)),(1+((√2)/2),1−((√2)/2)). 2)s=−1+(√6)⇒t=3−(5/s)=3−(5/((√6)−1))= =3−((√6)+1)=2−(√6) ⇒ { ((x+y=(√6)−1)),((xy=2−(√6))) :}⇒z^2 −((√6)−1)z+(2−(√6))=0 ⇒z=((((√6)−1)±(√(((√6)−1)^2 −4(2−(√6)))))/2)= =((((√6)−1)±(√(6+1−2(√6)−8+4(√6))))/2)⇒ ⇒ { ((x=(((√6)+(√(2(√6)−1))−1)/2))),((y=(((√6)−(√(2(√6)−1))−1)/2))) :} 3)s=−1−(√6)⇒t=3−(5/s)=3+(5/((√6)+1))= =3+(√6)−1=2+(√6) ⇒ { ((x+y=−1−(√6))),((xy=2+(√6))) :}⇒z^2 +((√6)+1)z+(2+(√6))=0 ⇒z=((−((√6)+1)±(√(((√6)+1)^2 −4(2+(√6)))))/2)= =((−((√6)+1)±(√(−1−2(√6))))/2) ⇒ { ((x=((−((√6)+1)+i(√(2(√6)+1)))/2))),((y=((−((√6)+1)−i(√(2(√6)+1)))/2))) :} .■

$${x}+{y}={s},{xy}={t} \\ $$$$\left({x}+{y}\right)\left({x}^{\mathrm{2}} −{xy}+{y}^{\mathrm{2}} \right)=\mathrm{5}\Rightarrow{s}\left(\mathrm{3}−{t}\right)=\mathrm{5} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\left({x}+{y}\right)^{\mathrm{2}} −\mathrm{2}{xy}={s}^{\mathrm{2}} −\mathrm{2}{t}=\mathrm{3} \\ $$$$\Rightarrow\begin{cases}{\mathrm{3}{s}−{st}=\mathrm{5}}\\{{s}^{\mathrm{2}} −\mathrm{2}{t}=\mathrm{3}}\end{cases}\Rightarrow\begin{cases}{−\mathrm{6}{s}+\mathrm{2}{st}=−\mathrm{10}}\\{{s}^{\mathrm{3}} −\mathrm{2}{st}=\mathrm{3}{s}}\end{cases}\Rightarrow \\ $$$${s}^{\mathrm{3}} −\mathrm{9}{s}+\mathrm{10}=\mathrm{0}\Rightarrow\left({s}−\mathrm{2}\right)\left({s}^{\mathrm{2}} +\mathrm{2}{s}−\mathrm{5}\right)=\mathrm{0} \\ $$$$\Rightarrow{s}=\mathrm{2},−\mathrm{1}\pm\sqrt{\mathrm{6}} \\ $$$$\left.\mathrm{1}\right){s}=\mathrm{2}\Rightarrow\mathrm{6}−\mathrm{2}{t}=\mathrm{5}\Rightarrow{t}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\begin{cases}{{x}+{y}=\mathrm{2}}\\{{xy}=\frac{\mathrm{1}}{\mathrm{2}}}\end{cases}\Rightarrow{z}^{\mathrm{2}} −\mathrm{2}{z}+\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{0}\Rightarrow{z}=\mathrm{1}\pm\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$$\Rightarrow\left({x},{y}\right)=\left(\mathrm{1}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}},\mathrm{1}−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right),\left(\mathrm{1}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}},\mathrm{1}−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right). \\ $$$$\left.\mathrm{2}\right){s}=−\mathrm{1}+\sqrt{\mathrm{6}}\Rightarrow{t}=\mathrm{3}−\frac{\mathrm{5}}{{s}}=\mathrm{3}−\frac{\mathrm{5}}{\sqrt{\mathrm{6}}−\mathrm{1}}= \\ $$$$=\mathrm{3}−\left(\sqrt{\mathrm{6}}+\mathrm{1}\right)=\mathrm{2}−\sqrt{\mathrm{6}} \\ $$$$\Rightarrow\begin{cases}{{x}+{y}=\sqrt{\mathrm{6}}−\mathrm{1}}\\{{xy}=\mathrm{2}−\sqrt{\mathrm{6}}}\end{cases}\Rightarrow{z}^{\mathrm{2}} −\left(\sqrt{\mathrm{6}}−\mathrm{1}\right){z}+\left(\mathrm{2}−\sqrt{\mathrm{6}}\right)=\mathrm{0} \\ $$$$\Rightarrow{z}=\frac{\left(\sqrt{\mathrm{6}}−\mathrm{1}\right)\pm\sqrt{\left(\sqrt{\mathrm{6}}−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{4}\left(\mathrm{2}−\sqrt{\mathrm{6}}\right)}}{\mathrm{2}}= \\ $$$$=\frac{\left(\sqrt{\mathrm{6}}−\mathrm{1}\right)\pm\sqrt{\mathrm{6}+\mathrm{1}−\mathrm{2}\sqrt{\mathrm{6}}−\mathrm{8}+\mathrm{4}\sqrt{\mathrm{6}}}}{\mathrm{2}}\Rightarrow \\ $$$$\Rightarrow\begin{cases}{{x}=\frac{\sqrt{\mathrm{6}}+\sqrt{\mathrm{2}\sqrt{\mathrm{6}}−\mathrm{1}}−\mathrm{1}}{\mathrm{2}}}\\{{y}=\frac{\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}\sqrt{\mathrm{6}}−\mathrm{1}}−\mathrm{1}}{\mathrm{2}}}\end{cases} \\ $$$$\left.\mathrm{3}\right){s}=−\mathrm{1}−\sqrt{\mathrm{6}}\Rightarrow{t}=\mathrm{3}−\frac{\mathrm{5}}{{s}}=\mathrm{3}+\frac{\mathrm{5}}{\sqrt{\mathrm{6}}+\mathrm{1}}= \\ $$$$=\mathrm{3}+\sqrt{\mathrm{6}}−\mathrm{1}=\mathrm{2}+\sqrt{\mathrm{6}} \\ $$$$\Rightarrow\begin{cases}{{x}+{y}=−\mathrm{1}−\sqrt{\mathrm{6}}}\\{{xy}=\mathrm{2}+\sqrt{\mathrm{6}}}\end{cases}\Rightarrow{z}^{\mathrm{2}} +\left(\sqrt{\mathrm{6}}+\mathrm{1}\right){z}+\left(\mathrm{2}+\sqrt{\mathrm{6}}\right)=\mathrm{0} \\ $$$$\Rightarrow{z}=\frac{\left.−\left(\sqrt{\mathrm{6}}+\mathrm{1}\right)\pm\sqrt{\left(\sqrt{\mathrm{6}}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{4}\left(\mathrm{2}+\sqrt{\mathrm{6}}\right.}\right)}{\mathrm{2}}= \\ $$$$=\frac{−\left(\sqrt{\mathrm{6}}+\mathrm{1}\right)\pm\sqrt{−\mathrm{1}−\mathrm{2}\sqrt{\mathrm{6}}}}{\mathrm{2}} \\ $$$$\Rightarrow\begin{cases}{{x}=\frac{−\left(\sqrt{\mathrm{6}}+\mathrm{1}\right)+\boldsymbol{{i}}\sqrt{\mathrm{2}\sqrt{\mathrm{6}}+\mathrm{1}}}{\mathrm{2}}}\\{\boldsymbol{{y}}=\frac{−\left(\sqrt{\mathrm{6}}+\mathrm{1}\right)−\boldsymbol{{i}}\sqrt{\mathrm{2}\sqrt{\mathrm{6}}+\mathrm{1}}}{\mathrm{2}}}\end{cases}\:\:\:.\blacksquare \\ $$

Answered by Joel579 last updated on 04/Jun/18

a = x + y b = xy a^2 − 2b = 3 → b = ((a^2 − 3)/2) a(3 − b) = 5 a(((9 − a^2 )/2)) = 5 9a − a^3 = 10 a^3 − 9a + 10 = 0 (a − 2)(a^2 + 2a − 5) = 0 • a − 2 = 0 → a = 2 → b = (1/2) { ((x + y = 2)),((xy = (1/2))) :} • a^2 + 2a − 5 = 0 a = −1 + (√6) → b = 2 − (√6) { ((x + y = −1 + (√6))),((xy = 2 − (√6))) :} a = −1 − (√6) → b = 2 + (√6) { ((x + y = −1 − (√6))),((xy = 2 + (√6))) :}

$${a}\:=\:{x}\:+\:{y} \\ $$$${b}\:=\:{xy} \\ $$$$ \\ $$$${a}^{\mathrm{2}} \:−\:\mathrm{2}{b}\:=\:\mathrm{3}\:\:\rightarrow\:\:{b}\:=\:\frac{{a}^{\mathrm{2}} \:−\:\mathrm{3}}{\mathrm{2}} \\ $$$$ \\ $$$${a}\left(\mathrm{3}\:−\:{b}\right)\:=\:\mathrm{5} \\ $$$${a}\left(\frac{\mathrm{9}\:−\:{a}^{\mathrm{2}} }{\mathrm{2}}\right)\:=\:\mathrm{5} \\ $$$$\mathrm{9}{a}\:−\:{a}^{\mathrm{3}} \:=\:\mathrm{10} \\ $$$${a}^{\mathrm{3}} \:−\:\mathrm{9}{a}\:+\:\mathrm{10}\:=\:\mathrm{0} \\ $$$$\left({a}\:−\:\mathrm{2}\right)\left({a}^{\mathrm{2}} \:+\:\mathrm{2}{a}\:−\:\mathrm{5}\right)\:=\:\mathrm{0} \\ $$$$ \\ $$$$\bullet\:{a}\:−\:\mathrm{2}\:=\:\mathrm{0}\:\rightarrow\:{a}\:=\:\mathrm{2}\:\:\rightarrow\:\:{b}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:\:\:\begin{cases}{{x}\:+\:{y}\:=\:\mathrm{2}}\\{{xy}\:=\:\frac{\mathrm{1}}{\mathrm{2}}}\end{cases} \\ $$$$ \\ $$$$\bullet\:{a}^{\mathrm{2}} \:+\:\mathrm{2}{a}\:−\:\mathrm{5}\:=\:\mathrm{0} \\ $$$$\:\:\:{a}\:=\:−\mathrm{1}\:+\:\sqrt{\mathrm{6}}\:\:\rightarrow\:\:{b}\:=\:\mathrm{2}\:−\:\sqrt{\mathrm{6}} \\ $$$$\:\:\:\begin{cases}{{x}\:+\:{y}\:=\:−\mathrm{1}\:+\:\sqrt{\mathrm{6}}}\\{{xy}\:=\:\mathrm{2}\:−\:\sqrt{\mathrm{6}}}\end{cases} \\ $$$$ \\ $$$$\:\:\:{a}\:=\:−\mathrm{1}\:−\:\sqrt{\mathrm{6}}\:\:\rightarrow\:\:{b}\:=\:\mathrm{2}\:+\:\sqrt{\mathrm{6}} \\ $$$$\:\:\:\begin{cases}{{x}\:+\:{y}\:=\:−\mathrm{1}\:−\:\sqrt{\mathrm{6}}}\\{{xy}\:=\:\mathrm{2}\:+\:\sqrt{\mathrm{6}}}\end{cases} \\ $$

Commented by Joel579 last updated on 04/Jun/18

Now it′s easier to find (x,y)

$$\mathrm{Now}\:\mathrm{it}'\mathrm{s}\:\mathrm{easier}\:\mathrm{to}\:\mathrm{find}\:\left({x},{y}\right) \\ $$

Commented by math2018 last updated on 04/Jun/18

Thanks Sir

$${Thanks}\:{Sir} \\ $$

Answered by MJS last updated on 04/Jun/18

looking for solutions of the form (because I found no integer solutions and otherwise we can′t exactly solve): x=a(√b)+c(√d) ∧ y=a(√b)−c(√d) ⇒ ⇒ x=p(1+q) ∧ y=p(1−q) (1) p^2 (1+q)^2 +p^2 (1−q)^2 =3 (2) p^3 (1+q)^3 +p^3 (1−q)^3 =5 (1) p^2 (2q^2 +2)=3 ⇒ q^2 =(3/(2p^2 ))−1 (2) p^3 (6q^2 +2)=5 ⇒ q^2 =(5/(6p^3 ))−(1/3) (3/(2p^2 ))−1=(5/(6p^3 ))−(1/3) ⇒ p^3 −(9/4)p+(5/4)=0 ⇒ ⇒ (p−1)(p^2 +p−(5/4))=0 p_1 =1 ⇒ q_1 =±((√2)/2) p_2 =−(1/2)−((√6)/2) ⇒ q_2 =±((√(−17+12(√6)))/5)i p_3 =−(1/2)+((√6)/2) ⇒ q_3 =±((√(17+12(√6)))/5) x_1 =1±((√2)/2); y_1 =1∓((√2)/2) x_2 =−(1/2)−((√6)/2)±((√(1+2(√6)))/2)i; y_2 =−(1/2)−((√6)/2)∓((√(1+2(√6)))/2)i x_3 =−(1/2)+(√6)±((√(−1+2(√6)))/2); y_3 =−(1/2)+(√6)∓((√(−1+2(√6)))/2)

$$\mathrm{looking}\:\mathrm{for}\:\mathrm{solutions}\:\mathrm{of}\:\mathrm{the}\:\mathrm{form}\:\left(\mathrm{because}\right. \\ $$$$\mathrm{I}\:\mathrm{found}\:\mathrm{no}\:\mathrm{integer}\:\mathrm{solutions}\:\mathrm{and}\:\mathrm{otherwise} \\ $$$$\left.\mathrm{we}\:\mathrm{can}'\mathrm{t}\:\mathrm{exactly}\:\mathrm{solve}\right): \\ $$$${x}={a}\sqrt{{b}}+{c}\sqrt{{d}}\:\wedge\:{y}={a}\sqrt{{b}}−{c}\sqrt{{d}}\:\Rightarrow \\ $$$$\Rightarrow\:{x}={p}\left(\mathrm{1}+{q}\right)\:\wedge\:{y}={p}\left(\mathrm{1}−{q}\right) \\ $$$$ \\ $$$$\left(\mathrm{1}\right)\:{p}^{\mathrm{2}} \left(\mathrm{1}+{q}\right)^{\mathrm{2}} +{p}^{\mathrm{2}} \left(\mathrm{1}−{q}\right)^{\mathrm{2}} =\mathrm{3} \\ $$$$\left(\mathrm{2}\right)\:{p}^{\mathrm{3}} \left(\mathrm{1}+{q}\right)^{\mathrm{3}} +{p}^{\mathrm{3}} \left(\mathrm{1}−{q}\right)^{\mathrm{3}} =\mathrm{5} \\ $$$$ \\ $$$$\left(\mathrm{1}\right)\:{p}^{\mathrm{2}} \left(\mathrm{2}{q}^{\mathrm{2}} +\mathrm{2}\right)=\mathrm{3}\:\Rightarrow\:{q}^{\mathrm{2}} =\frac{\mathrm{3}}{\mathrm{2}{p}^{\mathrm{2}} }−\mathrm{1} \\ $$$$\left(\mathrm{2}\right)\:{p}^{\mathrm{3}} \left(\mathrm{6}{q}^{\mathrm{2}} +\mathrm{2}\right)=\mathrm{5}\:\Rightarrow\:{q}^{\mathrm{2}} =\frac{\mathrm{5}}{\mathrm{6}{p}^{\mathrm{3}} }−\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$ \\ $$$$\frac{\mathrm{3}}{\mathrm{2}{p}^{\mathrm{2}} }−\mathrm{1}=\frac{\mathrm{5}}{\mathrm{6}{p}^{\mathrm{3}} }−\frac{\mathrm{1}}{\mathrm{3}}\:\Rightarrow\:{p}^{\mathrm{3}} −\frac{\mathrm{9}}{\mathrm{4}}{p}+\frac{\mathrm{5}}{\mathrm{4}}=\mathrm{0}\:\Rightarrow \\ $$$$\:\Rightarrow\:\left({p}−\mathrm{1}\right)\left({p}^{\mathrm{2}} +{p}−\frac{\mathrm{5}}{\mathrm{4}}\right)=\mathrm{0} \\ $$$${p}_{\mathrm{1}} =\mathrm{1}\:\Rightarrow\:{q}_{\mathrm{1}} =\pm\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$${p}_{\mathrm{2}} =−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{6}}}{\mathrm{2}}\:\Rightarrow\:{q}_{\mathrm{2}} =\pm\frac{\sqrt{−\mathrm{17}+\mathrm{12}\sqrt{\mathrm{6}}}}{\mathrm{5}}\mathrm{i} \\ $$$${p}_{\mathrm{3}} =−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{6}}}{\mathrm{2}}\:\Rightarrow\:{q}_{\mathrm{3}} =\pm\frac{\sqrt{\mathrm{17}+\mathrm{12}\sqrt{\mathrm{6}}}}{\mathrm{5}} \\ $$$$ \\ $$$${x}_{\mathrm{1}} =\mathrm{1}\pm\frac{\sqrt{\mathrm{2}}}{\mathrm{2}};\:{y}_{\mathrm{1}} =\mathrm{1}\mp\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$${x}_{\mathrm{2}} =−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{6}}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{6}}}}{\mathrm{2}}\mathrm{i}; \\ $$$$\:\:\:\:\:{y}_{\mathrm{2}} =−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{6}}}{\mathrm{2}}\mp\frac{\sqrt{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{6}}}}{\mathrm{2}}\mathrm{i} \\ $$$${x}_{\mathrm{3}} =−\frac{\mathrm{1}}{\mathrm{2}}+\sqrt{\mathrm{6}}\pm\frac{\sqrt{−\mathrm{1}+\mathrm{2}\sqrt{\mathrm{6}}}}{\mathrm{2}}; \\ $$$$\:\:\:\:\:{y}_{\mathrm{3}} =−\frac{\mathrm{1}}{\mathrm{2}}+\sqrt{\mathrm{6}}\mp\frac{\sqrt{−\mathrm{1}+\mathrm{2}\sqrt{\mathrm{6}}}}{\mathrm{2}} \\ $$$$ \\ $$

Commented by math2018 last updated on 04/Jun/18

Thanks for your work.

$${Thanks}\:{for}\:{your}\:{work}. \\ $$

Commented by MJS last updated on 04/Jun/18

you′re welcome

$$\mathrm{you}'\mathrm{re}\:\mathrm{welcome} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com