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Question Number 195301 by SajaRashki last updated on 29/Jul/23
{ ((x^2 +y=11)),((x+y^2 =7)) :}⇒ x,y=?
$$\begin{cases}{{x}^{\mathrm{2}} +{y}=\mathrm{11}}\\{{x}+{y}^{\mathrm{2}} =\mathrm{7}}\end{cases}\Rightarrow\:{x},{y}=? \\ $$
Answered by AST last updated on 30/Jul/23
y=11−x^2 ⇒(11−x^2 )^2 =7−x...(i) (i)⇒121−22x^2 +x^4 =7−x⇒x^4 −22x^2 +x+114=0 We get x=3 x+x^2 +y+y^2 =18 x=3⇒y^2 +y−6=0⇒y=−3 or 2 ⇒(x,y)=(3,2)
$${y}=\mathrm{11}−{x}^{\mathrm{2}} \Rightarrow\left(\mathrm{11}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} =\mathrm{7}−{x}…\left({i}\right) \\ $$$$\left({i}\right)\Rightarrow\mathrm{121}−\mathrm{22}{x}^{\mathrm{2}} +{x}^{\mathrm{4}} =\mathrm{7}−{x}\Rightarrow{x}^{\mathrm{4}} −\mathrm{22}{x}^{\mathrm{2}} +{x}+\mathrm{114}=\mathrm{0} \\ $$$${We}\:{get}\:{x}=\mathrm{3} \\ $$$${x}+{x}^{\mathrm{2}} +{y}+{y}^{\mathrm{2}} =\mathrm{18} \\ $$$${x}=\mathrm{3}\Rightarrow{y}^{\mathrm{2}} +{y}−\mathrm{6}=\mathrm{0}\Rightarrow{y}=−\mathrm{3}\:{or}\:\mathrm{2} \\ $$$$\Rightarrow\left({x},{y}\right)=\left(\mathrm{3},\mathrm{2}\right) \\ $$
Commented by SajaRashki last updated on 29/Jul/23
thank you sir; but this question have 4 answer; and you have reached only one answer
$${thank}\:{you}\:{sir};\:{but}\:{this}\:{question} \\ $$$${have}\:\mathrm{4}\:{answer};\:{and}\:{you}\:{have}\:{reached} \\ $$$${only}\:{one}\:{answer} \\ $$
Commented by AST last updated on 30/Jul/23
I didn′t bother with the non-integer part.
$${I}\:{didn}'{t}\:{bother}\:{with}\:{the}\:{non}-{integer}\:{part}. \\ $$
Commented by SajaRashki last updated on 29/Jul/23
it has 3 other answers which are real number; look at Frix answer
$${it}\:{has}\:\mathrm{3}\:{other}\:{answers}\:{which}\:{are} \\ $$$${real}\:{number};\:{look}\:{at}\:{Frix}\:{answer} \\ $$
Commented by Frix last updated on 30/Jul/23
Sorry Sir but you are wrong. You can see 4 real solutions in the picture you posted. You would not see complex solutions in a real plot at all. The ≈10^(−15) i only come in because of the approximation method.
$$\mathrm{Sorry}\:\mathrm{Sir}\:\mathrm{but}\:\mathrm{you}\:\mathrm{are}\:\mathrm{wrong}.\:\mathrm{You}\:\mathrm{can}\:\mathrm{see} \\ $$$$\mathrm{4}\:\mathrm{real}\:\mathrm{solutions}\:\mathrm{in}\:\mathrm{the}\:\mathrm{picture}\:\mathrm{you}\:\mathrm{posted}. \\ $$$$\mathrm{You}\:\mathrm{would}\:\mathrm{not}\:\mathrm{see}\:\mathrm{complex}\:\mathrm{solutions}\:\mathrm{in}\:\mathrm{a} \\ $$$$\mathrm{real}\:\mathrm{plot}\:\mathrm{at}\:\mathrm{all}. \\ $$$$\mathrm{The}\:\approx\mathrm{10}^{−\mathrm{15}} \mathrm{i}\:\mathrm{only}\:\mathrm{come}\:\mathrm{in}\:\mathrm{because}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{approximation}\:\mathrm{method}. \\ $$
Commented by Frix last updated on 30/Jul/23
See my solution below. The 3^(rd) degree polynomial has 3 real solutions. If I made a mistake, please correct it.
$$\mathrm{See}\:\mathrm{my}\:\mathrm{solution}\:\mathrm{below}.\:\mathrm{The}\:\mathrm{3}^{\mathrm{rd}} \:\mathrm{degree} \\ $$$$\mathrm{polynomial}\:\mathrm{has}\:\mathrm{3}\:\mathrm{real}\:\mathrm{solutions}.\:\mathrm{If}\:\mathrm{I}\:\mathrm{made} \\ $$$$\mathrm{a}\:\mathrm{mistake},\:\mathrm{please}\:\mathrm{correct}\:\mathrm{it}. \\ $$
Answered by Frix last updated on 29/Jul/23
y=11−x^2 x^4 −22x^2 +x+114=0 (x−3)(x^3 +3x^2 −13x−38)=0 x_1 =3 x^3 +3x^2 −13x−38=0 x=t−1 t^3 −16t−23=0 3 real solutions ⇒ use trigonometric method to get exact values
$${y}=\mathrm{11}−{x}^{\mathrm{2}} \\ $$$${x}^{\mathrm{4}} −\mathrm{22}{x}^{\mathrm{2}} +{x}+\mathrm{114}=\mathrm{0} \\ $$$$\left({x}−\mathrm{3}\right)\left({x}^{\mathrm{3}} +\mathrm{3}{x}^{\mathrm{2}} −\mathrm{13}{x}−\mathrm{38}\right)=\mathrm{0} \\ $$$${x}_{\mathrm{1}} =\mathrm{3} \\ $$$${x}^{\mathrm{3}} +\mathrm{3}{x}^{\mathrm{2}} −\mathrm{13}{x}−\mathrm{38}=\mathrm{0} \\ $$$${x}={t}−\mathrm{1} \\ $$$${t}^{\mathrm{3}} −\mathrm{16}{t}−\mathrm{23}=\mathrm{0} \\ $$$$\mathrm{3}\:\mathrm{real}\:\mathrm{solutions}\:\Rightarrow\:\mathrm{use}\:\mathrm{trigonometric}\:\mathrm{method} \\ $$$$\mathrm{to}\:\mathrm{get}\:\mathrm{exact}\:\mathrm{values} \\ $$
Commented by SajaRashki last updated on 29/Jul/23
thank you sir; +38 or −38?
$${thank}\:{you}\:{sir};\:+\mathrm{38}\:{or}\:−\mathrm{38}? \\ $$
Commented by Frix last updated on 29/Jul/23
Typo, −38 is correct
$$\mathrm{Typo},\:−\mathrm{38}\:\mathrm{is}\:\mathrm{correct} \\ $$
Commented by Frix last updated on 29/Jul/23
t_1 =((8(√3))/3)sin ((π+sin^(−1) ((69(√3))/(128)))/3) ⇒ x≈3.58442834∧y≈−1.84812653 t_2 =−((8(√3))/3)cos ((π+2sin^(−1) ((69(√3))/(128)))/6) ⇒ x≈−3.77931025∧y≈−3.28318599 t_3 =−((8(√3))/3)sin ((sin^(−1) ((69(√3))/(128)))/3) ⇒ x≈−2.80511809∧y≈3.13131252 and of course x=3∧y=2
$${t}_{\mathrm{1}} =\frac{\mathrm{8}\sqrt{\mathrm{3}}}{\mathrm{3}}\mathrm{sin}\:\frac{\pi+\mathrm{sin}^{−\mathrm{1}} \:\frac{\mathrm{69}\sqrt{\mathrm{3}}}{\mathrm{128}}}{\mathrm{3}}\:\Rightarrow\:{x}\approx\mathrm{3}.\mathrm{58442834}\wedge{y}\approx−\mathrm{1}.\mathrm{84812653} \\ $$$${t}_{\mathrm{2}} =−\frac{\mathrm{8}\sqrt{\mathrm{3}}}{\mathrm{3}}\mathrm{cos}\:\frac{\pi+\mathrm{2sin}^{−\mathrm{1}} \:\frac{\mathrm{69}\sqrt{\mathrm{3}}}{\mathrm{128}}}{\mathrm{6}}\:\Rightarrow\:{x}\approx−\mathrm{3}.\mathrm{77931025}\wedge{y}\approx−\mathrm{3}.\mathrm{28318599} \\ $$$${t}_{\mathrm{3}} =−\frac{\mathrm{8}\sqrt{\mathrm{3}}}{\mathrm{3}}\mathrm{sin}\:\frac{\mathrm{sin}^{−\mathrm{1}} \:\frac{\mathrm{69}\sqrt{\mathrm{3}}}{\mathrm{128}}}{\mathrm{3}}\:\Rightarrow\:{x}\approx−\mathrm{2}.\mathrm{80511809}\wedge{y}\approx\mathrm{3}.\mathrm{13131252} \\ $$$$\mathrm{and}\:\mathrm{of}\:\mathrm{course}\:{x}=\mathrm{3}\wedge{y}=\mathrm{2} \\ $$
Commented by SajaRashki last updated on 29/Jul/23
very very nice sir; and infinity thank you
$${very}\:{very}\:{nice}\:{sir};\:{and}\:{infinity}\:{thank}\:{you} \\ $$
Commented by Frix last updated on 30/Jul/23
To clarify: t^3 +pt+q=0 If (p^3 /(27))+(q^2 /4)<0 Cardano doesn′t work. This is even wrong in wolframalpha. We need the Trigonometric Method leading to t_k =((2(√(−3p)))/3)sin ((2kπ+sin^(−1) ((3(√3)q)/(2(√(−p^3 )))))/3) with k=1, 2, 3
$$\mathrm{To}\:\mathrm{clarify}: \\ $$$${t}^{\mathrm{3}} +{pt}+{q}=\mathrm{0} \\ $$$$\mathrm{If}\:\frac{{p}^{\mathrm{3}} }{\mathrm{27}}+\frac{{q}^{\mathrm{2}} }{\mathrm{4}}<\mathrm{0}\:\mathrm{Cardano}\:\mathrm{doesn}'\mathrm{t}\:\mathrm{work}.\:\mathrm{This} \\ $$$$\mathrm{is}\:\mathrm{even}\:\mathrm{wrong}\:\mathrm{in}\:\mathrm{wolframalpha}.\:\mathrm{We}\:\mathrm{need} \\ $$$$\mathrm{the}\:\mathrm{Trigonometric}\:\mathrm{Method}\:\mathrm{leading}\:\mathrm{to} \\ $$$${t}_{{k}} =\frac{\mathrm{2}\sqrt{−\mathrm{3}{p}}}{\mathrm{3}}\mathrm{sin}\:\frac{\mathrm{2}{k}\pi+\mathrm{sin}^{−\mathrm{1}} \:\frac{\mathrm{3}\sqrt{\mathrm{3}}{q}}{\mathrm{2}\sqrt{−{p}^{\mathrm{3}} }}}{\mathrm{3}}\:\mathrm{with}\:{k}=\mathrm{1},\:\mathrm{2},\:\mathrm{3} \\ $$
Commented by AST last updated on 30/Jul/23
Wolfram alpha also got real roots to the cubic equation but it seems making the original question the input changes the solution to have two complex roots.. I don′t know what algorithm was used, so I don′t know anything about its validity.
$${Wolfram}\:{alpha}\:{also}\:{got}\:{real}\:{roots}\:{to}\:{the}\:{cubic} \\ $$$${equation}\:{but}\:{it}\:{seems}\:{making}\:{the}\:{original} \\ $$$${question}\:{the}\:{input}\:{changes}\:{the}\:{solution}\:{to} \\ $$$${have}\:{two}\:{complex}\:{roots}..\:{I}\:{don}'{t}\:{know}\:{what} \\ $$$${algorithm}\:{was}\:{used},\:{so}\:{I}\:{don}'{t}\:{know}\:{anything} \\ $$$${about}\:{its}\:{validity}. \\ $$
Commented by Frix last updated on 30/Jul/23
My solution proves there are 4 real intersections. It′s obviously a problem of the approximation. An imaginary part of ≈.0000000000000001i is strange. But as I wrote before, your picture also shows 4 real solutions. Do you trust more in an algorithm?
$$\mathrm{My}\:\mathrm{solution}\:\mathrm{proves}\:\mathrm{there}\:\mathrm{are}\:\mathrm{4}\:\mathrm{real}\:\mathrm{intersections}. \\ $$$$\mathrm{It}'\mathrm{s}\:\mathrm{obviously}\:\mathrm{a}\:\mathrm{problem}\:\mathrm{of}\:\mathrm{the}\:\mathrm{approximation}. \\ $$$$\mathrm{An}\:\mathrm{imaginary}\:\mathrm{part}\:\mathrm{of}\:\approx.\mathrm{0000000000000001i} \\ $$$$\mathrm{is}\:\mathrm{strange}.\:\mathrm{But}\:\mathrm{as}\:\mathrm{I}\:\mathrm{wrote}\:\mathrm{before},\:\mathrm{your} \\ $$$$\mathrm{picture}\:\mathrm{also}\:\mathrm{shows}\:\mathrm{4}\:\mathrm{real}\:\mathrm{solutions}.\:\mathrm{Do}\:\mathrm{you} \\ $$$$\mathrm{trust}\:\mathrm{more}\:\mathrm{in}\:\mathrm{an}\:\mathrm{algorithm}? \\ $$
Commented by AST last updated on 30/Jul/23
It′s a problem of approximation.. The efficiency of the algorithm depends on the programmer,computer,etc.. The question should have 4 real x solutions based on the solution to the cubic equation.
$${It}'{s}\:{a}\:{problem}\:{of}\:{approximation}..\: \\ $$$${The}\:{efficiency}\:{of}\:{the}\:{algorithm}\:{depends}\:{on} \\ $$$${the}\:{programmer},{computer},{etc}..\:{The}\:{question} \\ $$$${should}\:{have}\:\mathrm{4}\:{real}\:{x}\:{solutions}\:{based}\:{on}\:{the} \\ $$$${solution}\:{to}\:{the}\:{cubic}\:{equation}. \\ $$

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