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Question Number 51431 by Tawa1 last updated on 26/Dec/18

Find x and y x^2 + y^2 = 25 ...... (i) x^3 + y^3 = 91 ....... (ii)

$$\mathrm{Find}\:\:\mathrm{x}\:\mathrm{and}\:\mathrm{y} \\ $$$$\:\:\:\:\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{y}^{\mathrm{2}} \:\:=\:\:\mathrm{25}\:\:\:\:\:......\:\:\left(\mathrm{i}\right) \\ $$$$\:\:\:\:\mathrm{x}^{\mathrm{3}} \:+\:\mathrm{y}^{\mathrm{3}} \:\:=\:\:\mathrm{91}\:\:\:.......\:\left(\mathrm{ii}\right) \\ $$

Commented by Tawa1 last updated on 27/Dec/18

God bless you sir.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

Commented by Abdo msup. last updated on 26/Dec/18

let x+y =s and xy =p we have s^2 =x^2 +y^2 +2xy =25 +2p slso s^3 =x^3 +y^3 +3xy(x+y) =91 +3sp ⇒s^2 −2p=25 and s^3 −3sp =91 ⇒ s^3 −2ps =25 s ⇒s^3 −3ps +ps =25s ⇒ 91 +ps=25s ⇒(25−p)s=91 ⇒ (25−p)^2 s^2 =91^2 ⇒(25−p)^2 (2p+25)=91^2 ⇒ (p^2 −50p +25^2 )(2p+25)=91^2 ⇒ 2p^3 +25p^2 −100p^2 −50.25p +2.25^2 p +25^3 =91^2 ⇒ 2p^3 −75p^2 +25^3 −91^2 =0 ⇒ p^3 −((75)/2)p^2 +((25^3 −91^2 )/2) =0 we must solve the equation x^3 +ax^2 +b =0 let use the changement x =t −(a/3) (e) ⇔ t^3 −3t^2 .(a/3) +3t.(a^2 /9) −(a^3 /(27)) +a(t^2 −((2at)/3) +(a^2 /9))+b=0 ⇒t^3 −at^2 + ((a^2 t)/3) −(a^3 /(27)) +at^2 −2((a^2 t)/3) +(a^3 /9) +b =0 ⇒ t^3 −(a^2 /3)t +((2a^3 )/9) +b =0 after we use the chsngement t =u+v....be continued...

$${let}\:{x}+{y}\:={s}\:{and}\:{xy}\:={p}\:\:{we}\:{have}\:{s}^{\mathrm{2}} ={x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} +\mathrm{2}{xy} \\ $$$$=\mathrm{25}\:+\mathrm{2}{p}\:\:{slso}\:\:{s}^{\mathrm{3}} ={x}^{\mathrm{3}} \:+{y}^{\mathrm{3}} \:+\mathrm{3}{xy}\left({x}+{y}\right) \\ $$$$=\mathrm{91}\:+\mathrm{3}{sp}\:\Rightarrow{s}^{\mathrm{2}} −\mathrm{2}{p}=\mathrm{25}\:{and}\:{s}^{\mathrm{3}} −\mathrm{3}{sp}\:=\mathrm{91}\:\Rightarrow \\ $$$${s}^{\mathrm{3}} −\mathrm{2}{ps}\:=\mathrm{25}\:{s}\:\Rightarrow{s}^{\mathrm{3}} −\mathrm{3}{ps}\:+{ps}\:=\mathrm{25}{s}\:\Rightarrow \\ $$$$\mathrm{91}\:+{ps}=\mathrm{25}{s}\:\Rightarrow\left(\mathrm{25}−{p}\right){s}=\mathrm{91}\:\Rightarrow \\ $$$$\left(\mathrm{25}−{p}\right)^{\mathrm{2}} {s}^{\mathrm{2}} \:=\mathrm{91}^{\mathrm{2}} \:\Rightarrow\left(\mathrm{25}−{p}\right)^{\mathrm{2}} \left(\mathrm{2}{p}+\mathrm{25}\right)=\mathrm{91}^{\mathrm{2}} \:\Rightarrow \\ $$$$\left({p}^{\mathrm{2}} −\mathrm{50}{p}\:+\mathrm{25}^{\mathrm{2}} \right)\left(\mathrm{2}{p}+\mathrm{25}\right)=\mathrm{91}^{\mathrm{2}} \:\Rightarrow \\ $$$$\mathrm{2}{p}^{\mathrm{3}} \:+\mathrm{25}{p}^{\mathrm{2}} −\mathrm{100}{p}^{\mathrm{2}} −\mathrm{50}.\mathrm{25}{p}\:+\mathrm{2}.\mathrm{25}^{\mathrm{2}} {p}\:+\mathrm{25}^{\mathrm{3}} =\mathrm{91}^{\mathrm{2}} \:\Rightarrow \\ $$$$\mathrm{2}{p}^{\mathrm{3}} −\mathrm{75}{p}^{\mathrm{2}} \:+\mathrm{25}^{\mathrm{3}} −\mathrm{91}^{\mathrm{2}} =\mathrm{0}\:\:\Rightarrow \\ $$$${p}^{\mathrm{3}} \:−\frac{\mathrm{75}}{\mathrm{2}}{p}^{\mathrm{2}} \:+\frac{\mathrm{25}^{\mathrm{3}} −\mathrm{91}^{\mathrm{2}} }{\mathrm{2}}\:=\mathrm{0}\:\:{we}\:{must}\:{solve}\:{the} \\ $$$${equation}\:{x}^{\mathrm{3}} \:+{ax}^{\mathrm{2}} \:+{b}\:=\mathrm{0}\:\:\:\:{let}\:{use}\:{the}\:{changement} \\ $$$${x}\:={t}\:−\frac{{a}}{\mathrm{3}} \\ $$$$\left({e}\right)\:\Leftrightarrow\:{t}^{\mathrm{3}} \:−\mathrm{3}{t}^{\mathrm{2}} .\frac{{a}}{\mathrm{3}}\:+\mathrm{3}{t}.\frac{{a}^{\mathrm{2}} }{\mathrm{9}}\:−\frac{{a}^{\mathrm{3}} }{\mathrm{27}}\:+{a}\left({t}^{\mathrm{2}} −\frac{\mathrm{2}{at}}{\mathrm{3}}\:+\frac{{a}^{\mathrm{2}} }{\mathrm{9}}\right)+{b}=\mathrm{0} \\ $$$$\Rightarrow{t}^{\mathrm{3}} −{at}^{\mathrm{2}} \:+\:\frac{{a}^{\mathrm{2}} {t}}{\mathrm{3}}\:−\frac{{a}^{\mathrm{3}} }{\mathrm{27}}\:+{at}^{\mathrm{2}} \:\:−\mathrm{2}\frac{{a}^{\mathrm{2}} {t}}{\mathrm{3}}\:+\frac{{a}^{\mathrm{3}} }{\mathrm{9}}\:+{b}\:=\mathrm{0}\:\Rightarrow \\ $$$${t}^{\mathrm{3}} \:−\frac{{a}^{\mathrm{2}} }{\mathrm{3}}{t}\:\:+\frac{\mathrm{2}{a}^{\mathrm{3}} }{\mathrm{9}}\:+{b}\:=\mathrm{0}\:\:{after}\:{we}\:{use}\:{the}\:{chsngement} \\ $$$${t}\:={u}+{v}....{be}\:{continued}... \\ $$

Commented by Abdo msup. last updated on 26/Dec/18

after finding s and p xand y are roots of X^2 −sX +p =0

$${after}\:{finding}\:\:{s}\:{and}\:{p}\:\:{xand}\:{y}\:{are}\:{roots}\:{of} \\ $$$${X}^{\mathrm{2}} \:−{sX}\:+{p}\:=\mathrm{0}\: \\ $$

Answered by mr W last updated on 26/Dec/18

let x+y=a, xy=b x^3 +y^3 =(x+y)(x^2 +y^2 −xy) ⇒91=(x+y)(25−xy) ⇒91=a(25−b) ...(1) x^2 +y^2 =(x+y)^2 −2xy ⇒25=(x+y)^2 −2xy ⇒25=a^2 −2b ...(2) from (2): ⇒b=((a^2 −25)/2) 91=a(25−((a^2 −25)/2)) a^3 −75a+182=0 (a−7)(a^2 +7a−26)=0 ⇒a=7 ⇒a=((−7±(√(153)))/2) with a=7: b=((49−25)/2)=12 x,y are roots of t^2 −7t+12=0 (t−3)(t−4)=0 ⇒x,y=3,4 with a=((−7−(√(153)))/2) ⇒b=((1−7a)/2)=((51+7(√(153)))/4) x,y are roots of t^2 +((7+(√(153)))/2)t+((51+7(√(153)))/4)=0 ⇒x,y=(1/4)(−7−(√(153))±i(√(2+14(√(153))))) with a=((−7+(√(153)))/2) ⇒b=((1−7a)/2)=((51−7(√(153)))/4) x,y are roots of t^2 +((7−(√(153)))/2)t+((51−7(√(153)))/4)=0 ⇒x,y=(1/4)(−7+(√(153))±(√(−2+14(√(153))))) ⇒x,y≈−1.928, 4.613

$${let}\:{x}+{y}={a},\:{xy}={b} \\ $$$${x}^{\mathrm{3}} +{y}^{\mathrm{3}} =\left({x}+{y}\right)\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} −{xy}\right) \\ $$$$\Rightarrow\mathrm{91}=\left({x}+{y}\right)\left(\mathrm{25}−{xy}\right) \\ $$$$\Rightarrow\mathrm{91}={a}\left(\mathrm{25}−{b}\right)\:\:\:...\left(\mathrm{1}\right) \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\left({x}+{y}\right)^{\mathrm{2}} −\mathrm{2}{xy} \\ $$$$\Rightarrow\mathrm{25}=\left({x}+{y}\right)^{\mathrm{2}} −\mathrm{2}{xy} \\ $$$$\Rightarrow\mathrm{25}={a}^{\mathrm{2}} −\mathrm{2}{b}\:\:\:...\left(\mathrm{2}\right) \\ $$$${from}\:\left(\mathrm{2}\right): \\ $$$$\Rightarrow{b}=\frac{{a}^{\mathrm{2}} −\mathrm{25}}{\mathrm{2}} \\ $$$$\mathrm{91}={a}\left(\mathrm{25}−\frac{{a}^{\mathrm{2}} −\mathrm{25}}{\mathrm{2}}\right) \\ $$$${a}^{\mathrm{3}} −\mathrm{75}{a}+\mathrm{182}=\mathrm{0} \\ $$$$\left({a}−\mathrm{7}\right)\left({a}^{\mathrm{2}} +\mathrm{7}{a}−\mathrm{26}\right)=\mathrm{0} \\ $$$$\Rightarrow{a}=\mathrm{7} \\ $$$$\Rightarrow{a}=\frac{−\mathrm{7}\pm\sqrt{\mathrm{153}}}{\mathrm{2}} \\ $$$$ \\ $$$${with}\:{a}=\mathrm{7}: \\ $$$${b}=\frac{\mathrm{49}−\mathrm{25}}{\mathrm{2}}=\mathrm{12} \\ $$$${x},{y}\:{are}\:{roots}\:{of} \\ $$$${t}^{\mathrm{2}} −\mathrm{7}{t}+\mathrm{12}=\mathrm{0} \\ $$$$\left({t}−\mathrm{3}\right)\left({t}−\mathrm{4}\right)=\mathrm{0} \\ $$$$\Rightarrow{x},{y}=\mathrm{3},\mathrm{4} \\ $$$$ \\ $$$${with}\:{a}=\frac{−\mathrm{7}−\sqrt{\mathrm{153}}}{\mathrm{2}} \\ $$$$\Rightarrow{b}=\frac{\mathrm{1}−\mathrm{7}{a}}{\mathrm{2}}=\frac{\mathrm{51}+\mathrm{7}\sqrt{\mathrm{153}}}{\mathrm{4}} \\ $$$${x},{y}\:{are}\:{roots}\:{of} \\ $$$${t}^{\mathrm{2}} +\frac{\mathrm{7}+\sqrt{\mathrm{153}}}{\mathrm{2}}{t}+\frac{\mathrm{51}+\mathrm{7}\sqrt{\mathrm{153}}}{\mathrm{4}}=\mathrm{0} \\ $$$$\Rightarrow{x},{y}=\frac{\mathrm{1}}{\mathrm{4}}\left(−\mathrm{7}−\sqrt{\mathrm{153}}\pm{i}\sqrt{\mathrm{2}+\mathrm{14}\sqrt{\mathrm{153}}}\right) \\ $$$$ \\ $$$${with}\:{a}=\frac{−\mathrm{7}+\sqrt{\mathrm{153}}}{\mathrm{2}} \\ $$$$\Rightarrow{b}=\frac{\mathrm{1}−\mathrm{7}{a}}{\mathrm{2}}=\frac{\mathrm{51}−\mathrm{7}\sqrt{\mathrm{153}}}{\mathrm{4}} \\ $$$${x},{y}\:{are}\:{roots}\:{of} \\ $$$${t}^{\mathrm{2}} +\frac{\mathrm{7}−\sqrt{\mathrm{153}}}{\mathrm{2}}{t}+\frac{\mathrm{51}−\mathrm{7}\sqrt{\mathrm{153}}}{\mathrm{4}}=\mathrm{0} \\ $$$$\Rightarrow{x},{y}=\frac{\mathrm{1}}{\mathrm{4}}\left(−\mathrm{7}+\sqrt{\mathrm{153}}\pm\sqrt{−\mathrm{2}+\mathrm{14}\sqrt{\mathrm{153}}}\right) \\ $$$$\Rightarrow{x},{y}\approx−\mathrm{1}.\mathrm{928},\:\mathrm{4}.\mathrm{613} \\ $$

Commented by Tawa1 last updated on 27/Dec/18

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

Answered by peter frank last updated on 26/Dec/18

let a=x^2 b=y^2 a+b=25....(i)×x ax+by=91...(ii) ax+bx=25x ax+by=91 b(x−y)=25x−91...(iii) a+b=25....(i)×y ax+by=91...(ii) ay+by=25y ax+by=91 a(y−x)=25y−91....(iv) b(x−y)=25x−91...(iii) −a(x−y)=−25y+91....(iv) (x−y)[−a+b]=25(x−y) b−a=25......v ....

$${let} \\ $$$${a}={x}^{\mathrm{2}} \\ $$$${b}={y}^{\mathrm{2}} \\ $$$${a}+{b}=\mathrm{25}....\left({i}\right)×{x} \\ $$$${ax}+{by}=\mathrm{91}...\left({ii}\right) \\ $$$${ax}+{bx}=\mathrm{25}{x} \\ $$$${ax}+{by}=\mathrm{91} \\ $$$${b}\left({x}−{y}\right)=\mathrm{25}{x}−\mathrm{91}...\left({iii}\right) \\ $$$${a}+{b}=\mathrm{25}....\left({i}\right)×{y} \\ $$$${ax}+{by}=\mathrm{91}...\left({ii}\right) \\ $$$${ay}+{by}=\mathrm{25}{y} \\ $$$${ax}+{by}=\mathrm{91} \\ $$$${a}\left({y}−{x}\right)=\mathrm{25}{y}−\mathrm{91}....\left({iv}\right) \\ $$$${b}\left({x}−{y}\right)=\mathrm{25}{x}−\mathrm{91}...\left({iii}\right) \\ $$$$−{a}\left({x}−{y}\right)=−\mathrm{25}{y}+\mathrm{91}....\left({iv}\right) \\ $$$$\left({x}−{y}\right)\left[−{a}+{b}\right]=\mathrm{25}\left({x}−{y}\right) \\ $$$${b}−{a}=\mathrm{25}......{v} \\ $$$$ \\ $$$$.... \\ $$

Commented by Tawa1 last updated on 27/Dec/18

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

Answered by behi83417@gmail.com last updated on 27/Dec/18

x+y=p,xy=q ⇒ { ((p^2 −2q=25)),((p(p^2 −3q)=91)) :}⇒ { ((3p^3 −6pq=75p)),((2p^3 −6pq=182)) :} ⇒p^3 −75p+182=0 ⇒(p−7)(p^2 +7p−26)=0 ⇒p=7,[((−7±(√(49+4×26)))/2)=2.68,−9.68] p=7⇒49−2q=25⇒q=12 p=2.68⇒7.18−2q=25⇒q=−8.91 p=−9.68⇒93.7−2q=25⇒q=34.35 1.[p=7,q=12]⇒z^2 −7z+12=0 ⇒[z_1 =x=3,z_2 =y=4]∨[x=4,y=3] 2.[p=2.68,q=−8.91]⇒z^2 −2.68z−8.91=0 ⇒[z_(1,2) =x∨y=((2.68±(√(2.68^2 +4×8.91)))/2)] ⇒[z_(1,2) =x∨y= 4.61 ∨−1.93] 3.[p=−9.69,q=34.35]⇒z^2 +9.69z+34.35=0 ⇒[z_(1,2) =((−9.69±(√(9.69^2 −4×34.35)))/2)] ⇒[z_(1,2) =x∨y=−4.85±3.3i] .■

$${x}+{y}={p},{xy}={q} \\ $$$$\Rightarrow\begin{cases}{{p}^{\mathrm{2}} −\mathrm{2}{q}=\mathrm{25}}\\{{p}\left({p}^{\mathrm{2}} −\mathrm{3}{q}\right)=\mathrm{91}}\end{cases}\Rightarrow\begin{cases}{\mathrm{3}{p}^{\mathrm{3}} −\mathrm{6}{pq}=\mathrm{75}{p}}\\{\mathrm{2}{p}^{\mathrm{3}} −\mathrm{6}{pq}=\mathrm{182}}\end{cases} \\ $$$$\Rightarrow{p}^{\mathrm{3}} −\mathrm{75}{p}+\mathrm{182}=\mathrm{0} \\ $$$$\Rightarrow\left({p}−\mathrm{7}\right)\left({p}^{\mathrm{2}} +\mathrm{7}{p}−\mathrm{26}\right)=\mathrm{0} \\ $$$$\Rightarrow{p}=\mathrm{7},\left[\frac{−\mathrm{7}\pm\sqrt{\mathrm{49}+\mathrm{4}×\mathrm{26}}}{\mathrm{2}}=\mathrm{2}.\mathrm{68},−\mathrm{9}.\mathrm{68}\right] \\ $$$${p}=\mathrm{7}\Rightarrow\mathrm{49}−\mathrm{2}{q}=\mathrm{25}\Rightarrow{q}=\mathrm{12} \\ $$$${p}=\mathrm{2}.\mathrm{68}\Rightarrow\mathrm{7}.\mathrm{18}−\mathrm{2}{q}=\mathrm{25}\Rightarrow{q}=−\mathrm{8}.\mathrm{91} \\ $$$${p}=−\mathrm{9}.\mathrm{68}\Rightarrow\mathrm{93}.\mathrm{7}−\mathrm{2}{q}=\mathrm{25}\Rightarrow{q}=\mathrm{34}.\mathrm{35} \\ $$$$\mathrm{1}.\left[{p}=\mathrm{7},{q}=\mathrm{12}\right]\Rightarrow{z}^{\mathrm{2}} −\mathrm{7}{z}+\mathrm{12}=\mathrm{0} \\ $$$$\Rightarrow\left[{z}_{\mathrm{1}} ={x}=\mathrm{3},{z}_{\mathrm{2}} ={y}=\mathrm{4}\right]\vee\left[{x}=\mathrm{4},{y}=\mathrm{3}\right] \\ $$$$\mathrm{2}.\left[{p}=\mathrm{2}.\mathrm{68},{q}=−\mathrm{8}.\mathrm{91}\right]\Rightarrow{z}^{\mathrm{2}} −\mathrm{2}.\mathrm{68}{z}−\mathrm{8}.\mathrm{91}=\mathrm{0} \\ $$$$\Rightarrow\left[{z}_{\mathrm{1},\mathrm{2}} ={x}\vee{y}=\frac{\mathrm{2}.\mathrm{68}\pm\sqrt{\mathrm{2}.\mathrm{68}^{\mathrm{2}} +\mathrm{4}×\mathrm{8}.\mathrm{91}}}{\mathrm{2}}\right] \\ $$$$\Rightarrow\left[{z}_{\mathrm{1},\mathrm{2}} ={x}\vee{y}=\:\mathrm{4}.\mathrm{61}\:\vee−\mathrm{1}.\mathrm{93}\right] \\ $$$$\mathrm{3}.\left[{p}=−\mathrm{9}.\mathrm{69},{q}=\mathrm{34}.\mathrm{35}\right]\Rightarrow{z}^{\mathrm{2}} +\mathrm{9}.\mathrm{69}{z}+\mathrm{34}.\mathrm{35}=\mathrm{0} \\ $$$$\Rightarrow\left[{z}_{\mathrm{1},\mathrm{2}} =\frac{−\mathrm{9}.\mathrm{69}\pm\sqrt{\mathrm{9}.\mathrm{69}^{\mathrm{2}} −\mathrm{4}×\mathrm{34}.\mathrm{35}}}{\mathrm{2}}\right] \\ $$$$\Rightarrow\left[{z}_{\mathrm{1},\mathrm{2}} ={x}\vee{y}=−\mathrm{4}.\mathrm{85}\pm\mathrm{3}.\mathrm{3}\boldsymbol{\mathrm{i}}\right]\:\:\:.\blacksquare \\ $$

Commented by Tawa1 last updated on 27/Dec/18

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

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