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Question Number 223655 by Rojarani last updated on 01/Aug/25
Commented by MathematicalUser2357 last updated on 01/Aug/25
is this like:
$${is}\:{this}\:{like}: \\ $$
Commented by MathematicalUser2357 last updated on 01/Aug/25
Commented by Rojarani last updated on 01/Aug/25
Sir yes.
$$\:{Sir}\:{yes}. \\ $$
Commented by yerlow last updated on 02/Aug/25
Let A = (√x) and B = (√y) So that x+y = A^2 +B^2 . { ((A(1+(1/(A^2 +B^2 )))=(3/2))),((B(1−(1/(A^2 +B^2 )))=(6/5))) :} Define S = A^2 +B^2 . { ((A=(3/2)∙(S/(S+1)))),(((6/5)∙(S/(S−1)))) :} We will use this to find S. Since A^2 +B^2 =S, Substitute the expressions. (((3S)/(2(S+1))))^2 +(((6S)/(5(S−1))))^2 = S ((9S^2 )/(4(S+1)^2 ))+((36S^2 )/(25(S−1)))=S Multiply both sides by the common denominator 100(S+1)^2 (S−1)^2 225S^2 (S−1)+144S^2 (S+1)^2 =100(S+1)^2 (S−1)^2 Expand both sides. For LHS:225S^2 (S^2 −2S+1)+144S^2 (S^2 + 2S+1) =S^2 (225(S^2 −2S+1)+144(S^2 +2S+1)) =S^2 (369S^2 −162S+369) For RHS:100S^5 −100S Divide both sides by S ≠ 0 S(369S^2 −162S+369)=100S^4 −100 And Bring all terms to one side. 100S^4 −369S^3 +162S^2 −369S−100=0 Solving the quarctic would be hard.. Im going to check if S = 4 works: 100(4)−369(4)^3 +162(4)^2 −369(4)−100 =100(256)−369(64)+162(16)−369(4)−100 =25600−23520+2592−1476−100=0 So S = 4 is a real Root. Im going to Factor (S−4) using polynomial division. 100S^4 −369S^3 +162S^2 −100÷(S−4) (S−4)(100S^3 +31S^2 +38S+25)=0 Now.. 100S^3 +31S^2 +38S+25=0 Unfortunately, I dont know how to solve cubic equations. Using a CAS (numerically) S_2 =−(7/8)+((3i)/2) S_3 =−(7/8)−((3i)/2) Back−Substituting Complex S, Computing A^2 = x, B^2 =y Eventually this gives: x=−(7/(16))±((3i)/2),y=((63)/(400))±((27i)/(50))✓ And the real solution: x=((36)/(25)),y=((63)/(400))✓ (Because after solving the quarctic to get the real solution, (((3/2)(4))/((4)−1))=(6/5)⇒x=A^2 =((36)/(25))✓ (((6/5)(4))/((4)−1))=((24)/(15))=(8/5)⇒y=B^2 =((64)/(25))✓
Answered by yerlow last updated on 02/Aug/25
Real Solution:x=((36)/(25)), y = ((64)/(25)) Complex Solution: x = −(7/(16))±((3i)/2),y=((63)/(400))±((27i)/(50)) Check my plain text comment to see the proof. I answered there because of a bug sir.
$$\mathrm{Real}\:\mathrm{Solution}:{x}=\frac{\mathrm{36}}{\mathrm{25}},\:{y}\:=\:\frac{\mathrm{64}}{\mathrm{25}} \\ $$$$\mathrm{Complex}\:\mathrm{Solution}:\:{x}\:=\:−\frac{\mathrm{7}}{\mathrm{16}}\pm\frac{\mathrm{3}{i}}{\mathrm{2}},{y}=\frac{\mathrm{63}}{\mathrm{400}}\pm\frac{\mathrm{27}{i}}{\mathrm{50}} \\ $$$$\mathrm{Check}\:\mathrm{my}\:\mathrm{plain}\:\mathrm{text}\:\mathrm{comment}\:\mathrm{to}\:\mathrm{see}\:\mathrm{the}\:\mathrm{proof}. \\ $$$$\mathrm{I}\:\mathrm{answered}\:\mathrm{there}\:\mathrm{because}\:\mathrm{of}\:\mathrm{a}\:\mathrm{bug}\:\mathrm{sir}. \\ $$
Answered by mr W last updated on 02/Aug/25
say a=(√x) >0, b=(√y) >0 a(1+(1/(a^2 +b^2 )))=(3/2) b(1−(1/(a^2 +b^2 )))=(6/5) say u=a^2 +b^2 a=(3/(2(1+(1/u))))=((3u)/(2(u+1))) b=(6/(5(1−(1/u))))=((6u)/(5(u−1))) >0 ⇒u>1 u=a^2 +b^2 =((9u^2 )/(4(u+1)^2 ))+((36u^2 )/(25(u−1)^2 )) 100(u^2 −1)^2 =225u(u−1)^2 +144u(u+1)^2 100u^4 −369u^3 −38u^2 −369u+100=0 (u−4)(4u−1)(25u^2 +14u+25)=0 ⇒u=4 (or (1/4)<1 or ((−7±24i)/(25)) ⇒rejected) ⇒a=(3/(2(1+(1/4))))=(6/5) ⇒b=(6/(5(1−(1/4))))=(8/5) only real solution is u=4, a=(6/5), b=(8/5) ⇒ x=a^2 =((36)/(25)), y=b^2 =((64)/(25)) ✓
$${say}\:{a}=\sqrt{{x}}\:>\mathrm{0},\:{b}=\sqrt{{y}}\:>\mathrm{0} \\ $$$${a}\left(\mathrm{1}+\frac{\mathrm{1}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\right)=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$${b}\left(\mathrm{1}−\frac{\mathrm{1}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\right)=\frac{\mathrm{6}}{\mathrm{5}} \\ $$$${say}\:{u}={a}^{\mathrm{2}} +{b}^{\mathrm{2}} \\ $$$${a}=\frac{\mathrm{3}}{\mathrm{2}\left(\mathrm{1}+\frac{\mathrm{1}}{{u}}\right)}=\frac{\mathrm{3}{u}}{\mathrm{2}\left({u}+\mathrm{1}\right)} \\ $$$${b}=\frac{\mathrm{6}}{\mathrm{5}\left(\mathrm{1}−\frac{\mathrm{1}}{{u}}\right)}=\frac{\mathrm{6}{u}}{\mathrm{5}\left({u}−\mathrm{1}\right)}\:>\mathrm{0}\:\Rightarrow{u}>\mathrm{1} \\ $$$${u}={a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\frac{\mathrm{9}{u}^{\mathrm{2}} }{\mathrm{4}\left({u}+\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{36}{u}^{\mathrm{2}} }{\mathrm{25}\left({u}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\mathrm{100}\left({u}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} =\mathrm{225}{u}\left({u}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{144}{u}\left({u}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\mathrm{100}{u}^{\mathrm{4}} −\mathrm{369}{u}^{\mathrm{3}} −\mathrm{38}{u}^{\mathrm{2}} −\mathrm{369}{u}+\mathrm{100}=\mathrm{0} \\ $$$$\left({u}−\mathrm{4}\right)\left(\mathrm{4}{u}−\mathrm{1}\right)\left(\mathrm{25}{u}^{\mathrm{2}} +\mathrm{14}{u}+\mathrm{25}\right)=\mathrm{0} \\ $$$$\Rightarrow{u}=\mathrm{4}\:\left({or}\:\frac{\mathrm{1}}{\mathrm{4}}<\mathrm{1}\:{or}\:\frac{−\mathrm{7}\pm\mathrm{24}{i}}{\mathrm{25}}\:\Rightarrow{rejected}\right) \\ $$$$\Rightarrow{a}=\frac{\mathrm{3}}{\mathrm{2}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}\right)}=\frac{\mathrm{6}}{\mathrm{5}} \\ $$$$\Rightarrow{b}=\frac{\mathrm{6}}{\mathrm{5}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}\right)}=\frac{\mathrm{8}}{\mathrm{5}} \\ $$$${only}\:{real}\:{solution}\:{is}\: \\ $$$${u}=\mathrm{4},\:{a}=\frac{\mathrm{6}}{\mathrm{5}},\:{b}=\frac{\mathrm{8}}{\mathrm{5}} \\ $$$$\Rightarrow\:{x}={a}^{\mathrm{2}} =\frac{\mathrm{36}}{\mathrm{25}},\:{y}={b}^{\mathrm{2}} =\frac{\mathrm{64}}{\mathrm{25}}\:\checkmark \\ $$
Commented by Rojarani last updated on 02/Aug/25
Sir, excellent.
$${Sir},\:{excellent}. \\ $$
Answered by Ghisom last updated on 02/Aug/25
the fastest path I found is let y=px square & transform both equations x^2 −(((9p+1))/(4(p+1)))x+(1/((p+1)^2 ))=0 x^2 −((2(43p+18))/(25p(p+1)))x+(1/((p+1)^2 ))=0 subtract ((225p^2 −319p−144)/(100p(p+1)))x=0 obviously x≠0 ⇒ p=−(9/(25))∨p=((16)/9) inserting above we get p=−(9/(25))∧x=−(7/(16))±(3/2)i ⇒ y=((63)/(400))∓((27)/(50))i p=((16)/9)∧x=(9/(100))∨x=((36)/(25)) ⇒ y=(4/(25))∨y=((64)/(25)) testing ⇒ x=(9/(100))∧y=(4/(25)) is false (due to squaring)
$$\mathrm{the}\:\mathrm{fastest}\:\mathrm{path}\:\mathrm{I}\:\mathrm{found}\:\mathrm{is} \\ $$$$\mathrm{let}\:{y}={px} \\ $$$$\mathrm{square}\:\&\:\mathrm{transform}\:\mathrm{both}\:\mathrm{equations} \\ $$$${x}^{\mathrm{2}} −\frac{\left(\mathrm{9}{p}+\mathrm{1}\right)}{\mathrm{4}\left({p}+\mathrm{1}\right)}{x}+\frac{\mathrm{1}}{\left({p}+\mathrm{1}\right)^{\mathrm{2}} }=\mathrm{0} \\ $$$${x}^{\mathrm{2}} −\frac{\mathrm{2}\left(\mathrm{43}{p}+\mathrm{18}\right)}{\mathrm{25}{p}\left({p}+\mathrm{1}\right)}{x}+\frac{\mathrm{1}}{\left({p}+\mathrm{1}\right)^{\mathrm{2}} }=\mathrm{0} \\ $$$$\mathrm{subtract} \\ $$$$\frac{\mathrm{225}{p}^{\mathrm{2}} −\mathrm{319}{p}−\mathrm{144}}{\mathrm{100}{p}\left({p}+\mathrm{1}\right)}{x}=\mathrm{0} \\ $$$$\mathrm{obviously}\:{x}\neq\mathrm{0} \\ $$$$\Rightarrow \\ $$$${p}=−\frac{\mathrm{9}}{\mathrm{25}}\vee{p}=\frac{\mathrm{16}}{\mathrm{9}} \\ $$$$\mathrm{inserting}\:\mathrm{above}\:\mathrm{we}\:\mathrm{get} \\ $$$${p}=−\frac{\mathrm{9}}{\mathrm{25}}\wedge{x}=−\frac{\mathrm{7}}{\mathrm{16}}\pm\frac{\mathrm{3}}{\mathrm{2}}\mathrm{i}\:\Rightarrow\:{y}=\frac{\mathrm{63}}{\mathrm{400}}\mp\frac{\mathrm{27}}{\mathrm{50}}\mathrm{i} \\ $$$${p}=\frac{\mathrm{16}}{\mathrm{9}}\wedge{x}=\frac{\mathrm{9}}{\mathrm{100}}\vee{x}=\frac{\mathrm{36}}{\mathrm{25}}\:\Rightarrow\:{y}=\frac{\mathrm{4}}{\mathrm{25}}\vee{y}=\frac{\mathrm{64}}{\mathrm{25}} \\ $$$$\mathrm{testing}\:\Rightarrow \\ $$$${x}=\frac{\mathrm{9}}{\mathrm{100}}\wedge{y}=\frac{\mathrm{4}}{\mathrm{25}}\:\mathrm{is}\:\mathrm{false}\:\left(\mathrm{due}\:\mathrm{to}\:\mathrm{squaring}\right) \\ $$

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