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remark-to-question-195301-and-similar-ones-x-2-y-a-x-y-2-b-a-b-gt-0-how-many-solutions-depending-on-a-b-

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Question Number 195356 by MJS_new last updated on 31/Jul/23
remark to question 195301 and similar ones x^2 +y=a x+y^2 =b a, b >0 how many solutions depending on a, b?
$$\mathrm{remark}\:\mathrm{to}\:\mathrm{question}\:\mathrm{195301}\:\mathrm{and}\:\mathrm{similar}\:\mathrm{ones} \\ $$$${x}^{\mathrm{2}} +{y}={a} \\ $$$${x}+{y}^{\mathrm{2}} ={b} \\ $$$${a},\:{b}\:>\mathrm{0} \\ $$$$\mathrm{how}\:\mathrm{many}\:\mathrm{solutions}\:\mathrm{depending}\:\mathrm{on}\:{a},\:{b}? \\ $$
Answered by MJS_new last updated on 31/Jul/23
y=a−x^2 inserting in the 2^(nd) equation gives x^4 −2ax+x+a^2 −b=0 possible cases: I. 4 distinct real solutions a^3 −a^2 b^2 −(9/8)ab+b^3 +((27)/(256))>0 II. 2 distinct and 1 double real solutions b≠(4/3)a^2 ∧a^3 −a^2 b^2 −(9/8)ab+b^3 +((27)/(256))=0 III. 1 single and 1 triple real solutions b=(4/3)a^2 ∧a^3 −a^2 b^2 −(9/8)ab+b^3 +((27)/(256))=0 ⇔ a=b=(3/4) IV. 2 distinct real and 1 pair of conjugated complex solutions a^3 −a^2 b^2 −(9/8)ab+b^3 +((27)/(256))<0
$$\:\:\:\:\:{y}={a}−{x}^{\mathrm{2}} \\ $$$$\mathrm{inserting}\:\mathrm{in}\:\mathrm{the}\:\mathrm{2}^{\mathrm{nd}} \:\mathrm{equation}\:\mathrm{gives} \\ $$$$\:\:\:\:\:{x}^{\mathrm{4}} −\mathrm{2}{ax}+{x}+{a}^{\mathrm{2}} −{b}=\mathrm{0} \\ $$$$ \\ $$$$\mathrm{possible}\:\mathrm{cases}: \\ $$$$ \\ $$$$\mathrm{I}.\:\mathrm{4}\:\mathrm{distinct}\:\mathrm{real}\:\mathrm{solutions} \\ $$$$\:\:\:\:\:{a}^{\mathrm{3}} −{a}^{\mathrm{2}} {b}^{\mathrm{2}} −\frac{\mathrm{9}}{\mathrm{8}}{ab}+{b}^{\mathrm{3}} +\frac{\mathrm{27}}{\mathrm{256}}>\mathrm{0} \\ $$$$\mathrm{II}.\:\mathrm{2}\:\mathrm{distinct}\:\mathrm{and}\:\mathrm{1}\:\mathrm{double}\:\mathrm{real}\:\mathrm{solutions} \\ $$$$\:\:\:\:\:{b}\neq\frac{\mathrm{4}}{\mathrm{3}}{a}^{\mathrm{2}} \wedge{a}^{\mathrm{3}} −{a}^{\mathrm{2}} {b}^{\mathrm{2}} −\frac{\mathrm{9}}{\mathrm{8}}{ab}+{b}^{\mathrm{3}} +\frac{\mathrm{27}}{\mathrm{256}}=\mathrm{0} \\ $$$$\mathrm{III}.\:\mathrm{1}\:\mathrm{single}\:\mathrm{and}\:\mathrm{1}\:\mathrm{triple}\:\mathrm{real}\:\mathrm{solutions} \\ $$$$\:\:\:\:\:{b}=\frac{\mathrm{4}}{\mathrm{3}}{a}^{\mathrm{2}} \wedge{a}^{\mathrm{3}} −{a}^{\mathrm{2}} {b}^{\mathrm{2}} −\frac{\mathrm{9}}{\mathrm{8}}{ab}+{b}^{\mathrm{3}} +\frac{\mathrm{27}}{\mathrm{256}}=\mathrm{0} \\ $$$$\:\:\:\:\:\Leftrightarrow\:{a}={b}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\mathrm{IV}.\:\mathrm{2}\:\mathrm{distinct}\:\mathrm{real}\:\mathrm{and}\:\mathrm{1}\:\mathrm{pair}\:\mathrm{of}\:\mathrm{conjugated} \\ $$$$\:\:\:\:\:\:\mathrm{complex}\:\mathrm{solutions} \\ $$$$\:\:\:\:\:{a}^{\mathrm{3}} −{a}^{\mathrm{2}} {b}^{\mathrm{2}} −\frac{\mathrm{9}}{\mathrm{8}}{ab}+{b}^{\mathrm{3}} +\frac{\mathrm{27}}{\mathrm{256}}<\mathrm{0} \\ $$
Commented by MJS_new last updated on 31/Jul/23
with a=11∧b=7 we get case I
$$\mathrm{with}\:{a}=\mathrm{11}\wedge{b}=\mathrm{7}\:\mathrm{we}\:\mathrm{get}\:\mathrm{case}\:\mathrm{I} \\ $$
Commented by Frix last updated on 31/Jul/23
And without a,b >0?
$$\mathrm{And}\:\mathrm{without}\:{a},{b}\:>\mathrm{0}? \\ $$
Answered by MJS_new last updated on 31/Jul/23
without the restriction a, b >0: a, b ∈R I. 4 distinct real solutions a>0∧b>0∧a^3 −a^2 b^2 −(9/8)ab+b^3 +((27)/(256))<0 II. 2 distinct and 1 double real solutions a>0∧b>0∧b≠(4/3)a^2 ∧a^3 −a^2 b^2 −(9/8)ab+b^3 +((27)/(256))=0 III. 1 single and 1 triple real solutions a=b=(3/4) IV. 2 distinct real and 1 pair of conjugated complex solutions a^3 −a^2 b^2 −(9/8)ab+b^3 +((27)/(256))>0 V. 1 double real and 1 pair of conjugated complex solutions a^3 −a^2 b^2 −(9/8)ab+b^3 +((27)/(256))=0∧(a<0∨b<0) VI. 2 pairs of conjugated complex solutions a^3 −a^2 b^2 −(9/8)ab+b^3 +((27)/(256))<0∧(a<0∨b<0)
$$\mathrm{without}\:\mathrm{the}\:\mathrm{restriction}\:{a},\:{b}\:>\mathrm{0}:\:{a},\:{b}\:\in\mathbb{R} \\ $$$$\mathrm{I}.\:\mathrm{4}\:\mathrm{distinct}\:\mathrm{real}\:\mathrm{solutions} \\ $$$$\:\:\:\:\:{a}>\mathrm{0}\wedge{b}>\mathrm{0}\wedge{a}^{\mathrm{3}} −{a}^{\mathrm{2}} {b}^{\mathrm{2}} −\frac{\mathrm{9}}{\mathrm{8}}{ab}+{b}^{\mathrm{3}} +\frac{\mathrm{27}}{\mathrm{256}}<\mathrm{0} \\ $$$$\mathrm{II}.\:\mathrm{2}\:\mathrm{distinct}\:\mathrm{and}\:\mathrm{1}\:\mathrm{double}\:\mathrm{real}\:\mathrm{solutions} \\ $$$$\:\:\:\:\:{a}>\mathrm{0}\wedge{b}>\mathrm{0}\wedge{b}\neq\frac{\mathrm{4}}{\mathrm{3}}{a}^{\mathrm{2}} \wedge{a}^{\mathrm{3}} −{a}^{\mathrm{2}} {b}^{\mathrm{2}} −\frac{\mathrm{9}}{\mathrm{8}}{ab}+{b}^{\mathrm{3}} +\frac{\mathrm{27}}{\mathrm{256}}=\mathrm{0} \\ $$$$\mathrm{III}.\:\mathrm{1}\:\mathrm{single}\:\mathrm{and}\:\mathrm{1}\:\mathrm{triple}\:\mathrm{real}\:\mathrm{solutions} \\ $$$$\:\:\:\:\:{a}={b}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\mathrm{IV}.\:\mathrm{2}\:\mathrm{distinct}\:\mathrm{real}\:\mathrm{and}\:\mathrm{1}\:\mathrm{pair}\:\mathrm{of}\:\mathrm{conjugated} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{complex}\:\mathrm{solutions} \\ $$$$\:\:\:\:\:{a}^{\mathrm{3}} −{a}^{\mathrm{2}} {b}^{\mathrm{2}} −\frac{\mathrm{9}}{\mathrm{8}}{ab}+{b}^{\mathrm{3}} +\frac{\mathrm{27}}{\mathrm{256}}>\mathrm{0} \\ $$$$\mathrm{V}.\:\mathrm{1}\:\mathrm{double}\:\mathrm{real}\:\mathrm{and}\:\mathrm{1}\:\mathrm{pair}\:\mathrm{of}\:\mathrm{conjugated} \\ $$$$\:\:\:\:\:\:\mathrm{complex}\:\mathrm{solutions} \\ $$$$\:\:\:\:\:{a}^{\mathrm{3}} −{a}^{\mathrm{2}} {b}^{\mathrm{2}} −\frac{\mathrm{9}}{\mathrm{8}}{ab}+{b}^{\mathrm{3}} +\frac{\mathrm{27}}{\mathrm{256}}=\mathrm{0}\wedge\left({a}<\mathrm{0}\vee{b}<\mathrm{0}\right) \\ $$$$\mathrm{VI}.\:\mathrm{2}\:\mathrm{pairs}\:\mathrm{of}\:\mathrm{conjugated}\:\mathrm{complex}\:\mathrm{solutions} \\ $$$$\:\:\:\:\:{a}^{\mathrm{3}} −{a}^{\mathrm{2}} {b}^{\mathrm{2}} −\frac{\mathrm{9}}{\mathrm{8}}{ab}+{b}^{\mathrm{3}} +\frac{\mathrm{27}}{\mathrm{256}}<\mathrm{0}\wedge\left({a}<\mathrm{0}\vee{b}<\mathrm{0}\right) \\ $$

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