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Question Number 82067 by john santu last updated on 18/Feb/20
{ ((∣x∣ −((y+3 ))^(1/(3 )) = 1)),(((−x(√(−x)))^2 = y +10)) :} find solution
$$\begin{cases}{\mid{x}\mid\:−\sqrt[{\mathrm{3}\:}]{{y}+\mathrm{3}\:}\:=\:\mathrm{1}}\\{\left(−{x}\sqrt{−{x}}\right)^{\mathrm{2}} \:=\:{y}\:+\mathrm{10}}\end{cases} \\ $$$${find}\:{solution} \\ $$
Commented by john santu last updated on 18/Feb/20
Commented by MJS last updated on 18/Feb/20
there′s also a complex solution: x=1±i(√3) y=−2
$$\mathrm{there}'\mathrm{s}\:\mathrm{also}\:\mathrm{a}\:\mathrm{complex}\:\mathrm{solution}: \\ $$$${x}=\mathrm{1}\pm\mathrm{i}\sqrt{\mathrm{3}} \\ $$$${y}=−\mathrm{2} \\ $$
Commented by john santu last updated on 18/Feb/20
where did you get it? isn′t the condition x must be ≤ 0?
$${where}\:{did}\:{you}\:{get}\:{it}?\:{isn}'{t}\:{the}\:{condition} \\ $$$${x}\:{must}\:{be}\:\leqslant\:\mathrm{0}? \\ $$
Commented by MJS last updated on 18/Feb/20
x≤0 if x∈R if x∈C ⇒ (√(−x)) always exists
$${x}\leqslant\mathrm{0}\:\mathrm{if}\:{x}\in\mathbb{R} \\ $$$$\mathrm{if}\:{x}\in\mathbb{C}\:\Rightarrow\:\sqrt{−{x}}\:\mathrm{always}\:\mathrm{exists} \\ $$
Commented by john santu last updated on 18/Feb/20
yes sir. your right
$${yes}\:{sir}.\:{your}\:{right} \\ $$
Answered by MJS last updated on 18/Feb/20
{ ((y=∣x∣^3 −3∣x∣^2 +3∣x∣−4 ⇒ y∈R∀x∈C)),((y=−x^3 −10)) :} ⇒ −x^3 −10∈R ∣x∣^3 −3∣x∣^2 +3∣x∣−4=−x^3 −10 let x=a+bi (√(a^2 +b^2 ))(a^2 +b^2 +3)+a^3 −3(a^2 +ab^2 +b^2 −2)+i(3a^2 −b^2 )b=0 { (((√(a^2 +b^2 ))(a^2 +b^2 +3)+a^3 −3(a^2 +ab^2 +b^2 −2)=0)),(((3a^2 −b^2 )b=0 ⇒ b=0∨b=±a(√3))) :} (1) b=0 ∣a∣^3 +a^3 −3a^2 +3∣a∣+6=0 (1.1) a>0 2a^3 −3a^2 +3a+6=0 ⇒ a≈−.854 no solution (1.2) a<0 −3a^2 −3a+6=0 ⇒ a=−2∨a=1 ⇒ a=−2 x_1 =−2∧y_1 =−2 (2) b=±a(√3) 8∣a∣^3 −8a^3 −12a^2 +6∣a∣+6=0 (2.1) a<0 −16a^3 −12a^2 −6a+6=0 ⇒ a≈.427 no solution (2.2) a>0 −12a^2 +6a+6=0 ⇒ −(1/2)∨a=1 ⇒ a=1 x_(2, 3) =1±i(√3)∧y_(2, 3) =−2
$$\begin{cases}{{y}=\mid{x}\mid^{\mathrm{3}} −\mathrm{3}\mid{x}\mid^{\mathrm{2}} +\mathrm{3}\mid{x}\mid−\mathrm{4}\:\Rightarrow\:{y}\in\mathbb{R}\forall{x}\in\mathbb{C}}\\{{y}=−{x}^{\mathrm{3}} −\mathrm{10}}\end{cases} \\ $$$$\Rightarrow\:−{x}^{\mathrm{3}} −\mathrm{10}\in\mathbb{R} \\ $$$$ \\ $$$$\mid{x}\mid^{\mathrm{3}} −\mathrm{3}\mid{x}\mid^{\mathrm{2}} +\mathrm{3}\mid{x}\mid−\mathrm{4}=−{x}^{\mathrm{3}} −\mathrm{10} \\ $$$$\mathrm{let}\:{x}={a}+{b}\mathrm{i} \\ $$$$\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{3}\right)+{a}^{\mathrm{3}} −\mathrm{3}\left({a}^{\mathrm{2}} +{ab}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}\right)+\mathrm{i}\left(\mathrm{3}{a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right){b}=\mathrm{0} \\ $$$$\begin{cases}{\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{3}\right)+{a}^{\mathrm{3}} −\mathrm{3}\left({a}^{\mathrm{2}} +{ab}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}\right)=\mathrm{0}}\\{\left(\mathrm{3}{a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right){b}=\mathrm{0}\:\Rightarrow\:{b}=\mathrm{0}\vee{b}=\pm{a}\sqrt{\mathrm{3}}}\end{cases} \\ $$$$\left(\mathrm{1}\right)\:{b}=\mathrm{0} \\ $$$$\mid{a}\mid^{\mathrm{3}} +{a}^{\mathrm{3}} −\mathrm{3}{a}^{\mathrm{2}} +\mathrm{3}\mid{a}\mid+\mathrm{6}=\mathrm{0} \\ $$$$\:\:\:\:\:\left(\mathrm{1}.\mathrm{1}\right)\:{a}>\mathrm{0} \\ $$$$\:\:\:\:\:\mathrm{2}{a}^{\mathrm{3}} −\mathrm{3}{a}^{\mathrm{2}} +\mathrm{3}{a}+\mathrm{6}=\mathrm{0}\:\Rightarrow\:{a}\approx−.\mathrm{854}\:\mathrm{no}\:\mathrm{solution} \\ $$$$\:\:\:\:\:\left(\mathrm{1}.\mathrm{2}\right)\:{a}<\mathrm{0} \\ $$$$\:\:\:\:\:−\mathrm{3}{a}^{\mathrm{2}} −\mathrm{3}{a}+\mathrm{6}=\mathrm{0}\:\Rightarrow\:{a}=−\mathrm{2}\vee{a}=\mathrm{1}\:\Rightarrow\:{a}=−\mathrm{2} \\ $$$${x}_{\mathrm{1}} =−\mathrm{2}\wedge{y}_{\mathrm{1}} =−\mathrm{2} \\ $$$$ \\ $$$$\left(\mathrm{2}\right)\:{b}=\pm{a}\sqrt{\mathrm{3}} \\ $$$$\mathrm{8}\mid{a}\mid^{\mathrm{3}} −\mathrm{8}{a}^{\mathrm{3}} −\mathrm{12}{a}^{\mathrm{2}} +\mathrm{6}\mid{a}\mid+\mathrm{6}=\mathrm{0} \\ $$$$\:\:\:\:\:\left(\mathrm{2}.\mathrm{1}\right)\:{a}<\mathrm{0} \\ $$$$\:\:\:\:\:−\mathrm{16}{a}^{\mathrm{3}} −\mathrm{12}{a}^{\mathrm{2}} −\mathrm{6}{a}+\mathrm{6}=\mathrm{0}\:\Rightarrow\:{a}\approx.\mathrm{427}\:\mathrm{no}\:\mathrm{solution} \\ $$$$\:\:\:\:\:\left(\mathrm{2}.\mathrm{2}\right)\:{a}>\mathrm{0} \\ $$$$\:\:\:\:\:−\mathrm{12}{a}^{\mathrm{2}} +\mathrm{6}{a}+\mathrm{6}=\mathrm{0}\:\Rightarrow\:−\frac{\mathrm{1}}{\mathrm{2}}\vee{a}=\mathrm{1}\:\Rightarrow\:{a}=\mathrm{1} \\ $$$${x}_{\mathrm{2},\:\mathrm{3}} =\mathrm{1}\pm\mathrm{i}\sqrt{\mathrm{3}}\wedge{y}_{\mathrm{2},\:\mathrm{3}} =−\mathrm{2} \\ $$
Commented by john santu last updated on 18/Feb/20
okay sir. i agree your way
$${okay}\:{sir}.\:{i}\:{agree}\:{your}\:{way}\: \\ $$
Commented by MJS last updated on 18/Feb/20
just for fun, I always try to find complex solutions too
$$\mathrm{just}\:\mathrm{for}\:\mathrm{fun},\:\mathrm{I}\:\mathrm{always}\:\mathrm{try}\:\mathrm{to}\:\mathrm{find}\:\mathrm{complex} \\ $$$$\mathrm{solutions}\:\mathrm{too} \\ $$

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