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 129394 by liberty last updated on 15/Jan/21

 { ((x^3 −3x^2 +5x+17=0)),((y^3 −3y^2 +5y+11=0)) :}   find x+y .

$$\begin{cases}{\mathrm{x}^{\mathrm{3}} −\mathrm{3x}^{\mathrm{2}} +\mathrm{5x}+\mathrm{17}=\mathrm{0}}\\{\mathrm{y}^{\mathrm{3}} −\mathrm{3y}^{\mathrm{2}} +\mathrm{5y}+\mathrm{11}=\mathrm{0}}\end{cases} \\ $$$$\:\mathrm{find}\:\mathrm{x}+\mathrm{y}\:. \\ $$

Answered by bemath last updated on 15/Jan/21

  { ((x^3 +5x+17=3x^2 )),((y^3 +5y+11=3y^2 )) :}  (1)+(2)⇒(x^3 +y^3 )+5(x+y)+28=3(x^2 +y^2 )  let x+y = u and xy = v  ⇔ u^3 −3uv+5u+28=3u^2 −6v  ⇒ u^3 −3u^2 −3uv+5u+6v+28 = 0  ⇒

$$\:\begin{cases}{\mathrm{x}^{\mathrm{3}} +\mathrm{5x}+\mathrm{17}=\mathrm{3x}^{\mathrm{2}} }\\{\mathrm{y}^{\mathrm{3}} +\mathrm{5y}+\mathrm{11}=\mathrm{3y}^{\mathrm{2}} }\end{cases} \\ $$$$\left(\mathrm{1}\right)+\left(\mathrm{2}\right)\Rightarrow\left(\mathrm{x}^{\mathrm{3}} +\mathrm{y}^{\mathrm{3}} \right)+\mathrm{5}\left(\mathrm{x}+\mathrm{y}\right)+\mathrm{28}=\mathrm{3}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \right) \\ $$$$\mathrm{let}\:\mathrm{x}+\mathrm{y}\:=\:\mathrm{u}\:\mathrm{and}\:\mathrm{xy}\:=\:\mathrm{v} \\ $$$$\Leftrightarrow\:\mathrm{u}^{\mathrm{3}} −\mathrm{3uv}+\mathrm{5u}+\mathrm{28}=\mathrm{3u}^{\mathrm{2}} −\mathrm{6v} \\ $$$$\Rightarrow\:\mathrm{u}^{\mathrm{3}} −\mathrm{3u}^{\mathrm{2}} −\mathrm{3uv}+\mathrm{5u}+\mathrm{6v}+\mathrm{28}\:=\:\mathrm{0} \\ $$$$\Rightarrow \\ $$$$ \\ $$

Answered by MJS_new last updated on 16/Jan/21

x^3 −3x^2 +5x+17=0  x=u+1  u^3 +2u+20=0  u_1 =((−10+((2(√(2031)))/9)))^(1/3) −((10+((2(√(2031)))/9)))^(1/3)   u_2 =ω((−10+((2(√(2031)))/9)))^(1/3) −ω^2 ((10+((2(√(2031)))/9)))^(1/3)   u_3 =w^2 ((−10+((2(√(2031)))/9)))^(1/3) −ω((10+((2(√(2031)))/9)))^(1/3)   with w=−(1/2)+((√3)/2)i    y^3 −3y^2 +5y+11=0  y=v+1  v^3 +2v+14=0  v_1 =((−7+((11(√(33)))/9)))^(1/3) −((7+((11(√(33)))/9)))^(1/3)   v_2 =ω((−7+((11(√(33)))/9)))^(1/3) −ω^2 ((7+((11(√(33)))/9)))^(1/3)   v_3 =ω^2 ((−7+((11(√(33)))/9)))^(1/3) −ω((7+((11(√(33)))/9)))^(1/3)     I don′t think adding x and y makes any sense,  but x_1 +y_1 =  =2+((−10+((2(√(2031)))/9)))^(1/3) −((10+((2(√(2031)))/9)))^(1/3) +((−7+((11(√(33)))/9)))^(1/3) −((7+((11(√(33)))/9)))^(1/3)   with no chance to simplify

$${x}^{\mathrm{3}} −\mathrm{3}{x}^{\mathrm{2}} +\mathrm{5}{x}+\mathrm{17}=\mathrm{0} \\ $$$${x}={u}+\mathrm{1} \\ $$$${u}^{\mathrm{3}} +\mathrm{2}{u}+\mathrm{20}=\mathrm{0} \\ $$$${u}_{\mathrm{1}} =\sqrt[{\mathrm{3}}]{−\mathrm{10}+\frac{\mathrm{2}\sqrt{\mathrm{2031}}}{\mathrm{9}}}−\sqrt[{\mathrm{3}}]{\mathrm{10}+\frac{\mathrm{2}\sqrt{\mathrm{2031}}}{\mathrm{9}}} \\ $$$${u}_{\mathrm{2}} =\omega\sqrt[{\mathrm{3}}]{−\mathrm{10}+\frac{\mathrm{2}\sqrt{\mathrm{2031}}}{\mathrm{9}}}−\omega^{\mathrm{2}} \sqrt[{\mathrm{3}}]{\mathrm{10}+\frac{\mathrm{2}\sqrt{\mathrm{2031}}}{\mathrm{9}}} \\ $$$${u}_{\mathrm{3}} ={w}^{\mathrm{2}} \sqrt[{\mathrm{3}}]{−\mathrm{10}+\frac{\mathrm{2}\sqrt{\mathrm{2031}}}{\mathrm{9}}}−\omega\sqrt[{\mathrm{3}}]{\mathrm{10}+\frac{\mathrm{2}\sqrt{\mathrm{2031}}}{\mathrm{9}}} \\ $$$$\mathrm{with}\:{w}=−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i} \\ $$$$ \\ $$$${y}^{\mathrm{3}} −\mathrm{3}{y}^{\mathrm{2}} +\mathrm{5}{y}+\mathrm{11}=\mathrm{0} \\ $$$${y}={v}+\mathrm{1} \\ $$$${v}^{\mathrm{3}} +\mathrm{2}{v}+\mathrm{14}=\mathrm{0} \\ $$$${v}_{\mathrm{1}} =\sqrt[{\mathrm{3}}]{−\mathrm{7}+\frac{\mathrm{11}\sqrt{\mathrm{33}}}{\mathrm{9}}}−\sqrt[{\mathrm{3}}]{\mathrm{7}+\frac{\mathrm{11}\sqrt{\mathrm{33}}}{\mathrm{9}}} \\ $$$${v}_{\mathrm{2}} =\omega\sqrt[{\mathrm{3}}]{−\mathrm{7}+\frac{\mathrm{11}\sqrt{\mathrm{33}}}{\mathrm{9}}}−\omega^{\mathrm{2}} \sqrt[{\mathrm{3}}]{\mathrm{7}+\frac{\mathrm{11}\sqrt{\mathrm{33}}}{\mathrm{9}}} \\ $$$${v}_{\mathrm{3}} =\omega^{\mathrm{2}} \sqrt[{\mathrm{3}}]{−\mathrm{7}+\frac{\mathrm{11}\sqrt{\mathrm{33}}}{\mathrm{9}}}−\omega\sqrt[{\mathrm{3}}]{\mathrm{7}+\frac{\mathrm{11}\sqrt{\mathrm{33}}}{\mathrm{9}}} \\ $$$$ \\ $$$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{think}\:\mathrm{adding}\:{x}\:\mathrm{and}\:{y}\:\mathrm{makes}\:\mathrm{any}\:\mathrm{sense}, \\ $$$$\mathrm{but}\:{x}_{\mathrm{1}} +{y}_{\mathrm{1}} = \\ $$$$=\mathrm{2}+\sqrt[{\mathrm{3}}]{−\mathrm{10}+\frac{\mathrm{2}\sqrt{\mathrm{2031}}}{\mathrm{9}}}−\sqrt[{\mathrm{3}}]{\mathrm{10}+\frac{\mathrm{2}\sqrt{\mathrm{2031}}}{\mathrm{9}}}+\sqrt[{\mathrm{3}}]{−\mathrm{7}+\frac{\mathrm{11}\sqrt{\mathrm{33}}}{\mathrm{9}}}−\sqrt[{\mathrm{3}}]{\mathrm{7}+\frac{\mathrm{11}\sqrt{\mathrm{33}}}{\mathrm{9}}} \\ $$$$\mathrm{with}\:\mathrm{no}\:\mathrm{chance}\:\mathrm{to}\:\mathrm{simplify} \\ $$

Answered by MJS_new last updated on 15/Jan/21

just solve both and then add?!

$$\mathrm{just}\:\mathrm{solve}\:\mathrm{both}\:\mathrm{and}\:\mathrm{then}\:\mathrm{add}?! \\ $$

Commented by MJS_new last updated on 16/Jan/21

x^3 +ax^2 +bx+c=0  x=t−(a/3)  t^3 −((a^2 −3b)/3)t+((2a^3 −9ab+27c)/(27))=0  −((a^2 −3b)/3)=p∧((2a^3 −9ab+27c)/(27))=q  t^3 +px+q=0

$${x}^{\mathrm{3}} +{ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0} \\ $$$${x}={t}−\frac{{a}}{\mathrm{3}} \\ $$$${t}^{\mathrm{3}} −\frac{{a}^{\mathrm{2}} −\mathrm{3}{b}}{\mathrm{3}}{t}+\frac{\mathrm{2}{a}^{\mathrm{3}} −\mathrm{9}{ab}+\mathrm{27}{c}}{\mathrm{27}}=\mathrm{0} \\ $$$$−\frac{{a}^{\mathrm{2}} −\mathrm{3}{b}}{\mathrm{3}}={p}\wedge\frac{\mathrm{2}{a}^{\mathrm{3}} −\mathrm{9}{ab}+\mathrm{27}{c}}{\mathrm{27}}={q} \\ $$$${t}^{\mathrm{3}} +{px}+{q}=\mathrm{0} \\ $$

Commented by bemath last updated on 15/Jan/21

by Cardano?

$$\mathrm{by}\:\mathrm{Cardano}? \\ $$

Commented by MJS_new last updated on 15/Jan/21

yes. we′ve got 2 independent equations for  2 independent unknowns... it′s like adding  your height to your weight...

$$\mathrm{yes}.\:\mathrm{we}'\mathrm{ve}\:\mathrm{got}\:\mathrm{2}\:\mathrm{independent}\:\mathrm{equations}\:\mathrm{for} \\ $$$$\mathrm{2}\:\mathrm{independent}\:\mathrm{unknowns}...\:\mathrm{it}'\mathrm{s}\:\mathrm{like}\:\mathrm{adding} \\ $$$$\mathrm{your}\:\mathrm{height}\:\mathrm{to}\:\mathrm{your}\:\mathrm{weight}... \\ $$

Commented by bemath last updated on 15/Jan/21

how to transform this both equation   to x^3 +px+q = 0?

$$\mathrm{how}\:\mathrm{to}\:\mathrm{transform}\:\mathrm{this}\:\mathrm{both}\:\mathrm{equation}\: \\ $$$$\mathrm{to}\:\mathrm{x}^{\mathrm{3}} +\mathrm{px}+\mathrm{q}\:=\:\mathrm{0}? \\ $$

Commented by liberty last updated on 16/Jan/21

thank you. i guess the question any mistake

$$\mathrm{thank}\:\mathrm{you}.\:\mathrm{i}\:\mathrm{guess}\:\mathrm{the}\:\mathrm{question}\:\mathrm{any}\:\mathrm{mistake} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com