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Question Number 64580 by jimful last updated on 19/Jul/19
a^2 +b^2 =a+b ,a^2 −b^2 =ab a=? b=? and what if a^2 +b^2 =a−b?
$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} ={a}+{b}\:,{a}^{\mathrm{2}} −{b}^{\mathrm{2}} ={ab} \\ $$$${a}=?\:{b}=? \\ $$$${and}\:{what}\:{if}\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} ={a}−{b}? \\ $$
Answered by meme last updated on 19/Jul/19
a^2 +b^2 =a+b a^2 −b^2 =ab 2a^2 =ab(a+b) 2a=ab+ab^2 2=b+b^2 b=((−1−3)/2)=−2 or b=((−1+3)/2)=1 for b=−2 a^2 −4=−2a a^2 +2a−4=0 a=((−2−(√(20)))/2) or a=((−2+(√(20)))/2) for b=1 a^2 −1=a a^2 −a−1=0 a=((1−(√5))/2) or a=((1+(√5))/2) a^2 +b^2 = a−b if
$$\:{a}^{\mathrm{2}} \:+{b}^{\mathrm{2}} \:={a}+{b}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$${a}^{\mathrm{2}} −{b}^{\mathrm{2}} ={ab} \\ $$$$\mathrm{2}{a}^{\mathrm{2}} ={ab}\left({a}+{b}\right) \\ $$$$\mathrm{2}{a}={ab}+{ab}^{\mathrm{2}} \\ $$$$\mathrm{2}={b}+{b}^{\mathrm{2}} \\ $$$${b}=\frac{−\mathrm{1}−\mathrm{3}}{\mathrm{2}}=−\mathrm{2}\:{or}\:{b}=\frac{−\mathrm{1}+\mathrm{3}}{\mathrm{2}}=\mathrm{1} \\ $$$${for}\:{b}=−\mathrm{2} \\ $$$${a}^{\mathrm{2}} −\mathrm{4}=−\mathrm{2}{a} \\ $$$${a}^{\mathrm{2}} +\mathrm{2}{a}−\mathrm{4}=\mathrm{0} \\ $$$${a}=\frac{−\mathrm{2}−\sqrt{\mathrm{20}}}{\mathrm{2}}\:{or}\:{a}=\frac{−\mathrm{2}+\sqrt{\mathrm{20}}}{\mathrm{2}} \\ $$$${for}\:{b}=\mathrm{1} \\ $$$${a}^{\mathrm{2}} −\mathrm{1}={a} \\ $$$${a}^{\mathrm{2}} −{a}−\mathrm{1}=\mathrm{0} \\ $$$${a}=\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\:{or}\:{a}=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\:{a}−{b}\:\:\:\:\:{if} \\ $$$$ \\ $$$$ \\ $$
Answered by MJS last updated on 19/Jul/19
(1) a^2 +b^2 =a+b (2) a^2 −b^2 =ab (1)−(2) 2b^2 =a(1−b)+b ⇒ a=−((b(2b−1))/(b−1)) (1) ((6b−5)/((b−1)^2 ))+5b^2 +4b+5=−(1/(b−1))−b−1 b^2 (b^2 −b+(1/5))=0 ⇒ b_1 =0 b_2 =(1/2)−((√5)/(10)) b_3 =(1/2)+((√5)/(10)) ⇒ a_1 =0 a_2 =(1/2)−((3(√5))/(10)) a_3 =(1/2)+((3(√5))/(10))
$$\left(\mathrm{1}\right)\:\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} ={a}+{b} \\ $$$$\left(\mathrm{2}\right)\:\:{a}^{\mathrm{2}} −{b}^{\mathrm{2}} ={ab} \\ $$$$\left(\mathrm{1}\right)−\left(\mathrm{2}\right)\:\:\mathrm{2}{b}^{\mathrm{2}} ={a}\left(\mathrm{1}−{b}\right)+{b} \\ $$$$\Rightarrow\:{a}=−\frac{{b}\left(\mathrm{2}{b}−\mathrm{1}\right)}{{b}−\mathrm{1}} \\ $$$$\left(\mathrm{1}\right)\:\:\frac{\mathrm{6}{b}−\mathrm{5}}{\left({b}−\mathrm{1}\right)^{\mathrm{2}} }+\mathrm{5}{b}^{\mathrm{2}} +\mathrm{4}{b}+\mathrm{5}=−\frac{\mathrm{1}}{{b}−\mathrm{1}}−{b}−\mathrm{1} \\ $$$${b}^{\mathrm{2}} \left({b}^{\mathrm{2}} −{b}+\frac{\mathrm{1}}{\mathrm{5}}\right)=\mathrm{0} \\ $$$$\Rightarrow\:{b}_{\mathrm{1}} =\mathrm{0}\:\:{b}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{5}}}{\mathrm{10}}\:\:{b}_{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{5}}}{\mathrm{10}} \\ $$$$\Rightarrow\:{a}_{\mathrm{1}} =\mathrm{0}\:\:{a}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{3}\sqrt{\mathrm{5}}}{\mathrm{10}}\:\:{a}_{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{3}\sqrt{\mathrm{5}}}{\mathrm{10}} \\ $$
Answered by MJS last updated on 19/Jul/19
(1) a^2 +b^2 =a−b (2) a^2 −b^2 =ab (1)−(2) 2b^2 =a(1−b)−b ⇒ a=−((b(2b+1))/(b−1)) (1) ((3(10b−7))/((b−1)^2 ))+5b^2 +12b+21=−(3/(b−1))−3b−3 b^2 (b^2 +b−(1/5))=0 ⇒ b_1 =−(1/2)−((3(√5))/(10)) b_2 =0 b_3 =−(1/2)+((3(√5))/(10)) ⇒ a_1 =(1/2)+((√5)/(10)) a_2 =0 a_3 =(1/2)−((√5)/(10))
$$\left(\mathrm{1}\right)\:\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} ={a}−{b} \\ $$$$\left(\mathrm{2}\right)\:\:{a}^{\mathrm{2}} −{b}^{\mathrm{2}} ={ab} \\ $$$$\left(\mathrm{1}\right)−\left(\mathrm{2}\right)\:\:\mathrm{2}{b}^{\mathrm{2}} ={a}\left(\mathrm{1}−{b}\right)−{b} \\ $$$$\Rightarrow\:{a}=−\frac{{b}\left(\mathrm{2}{b}+\mathrm{1}\right)}{{b}−\mathrm{1}} \\ $$$$\left(\mathrm{1}\right)\:\:\frac{\mathrm{3}\left(\mathrm{10}{b}−\mathrm{7}\right)}{\left({b}−\mathrm{1}\right)^{\mathrm{2}} }+\mathrm{5}{b}^{\mathrm{2}} +\mathrm{12}{b}+\mathrm{21}=−\frac{\mathrm{3}}{{b}−\mathrm{1}}−\mathrm{3}{b}−\mathrm{3} \\ $$$${b}^{\mathrm{2}} \left({b}^{\mathrm{2}} +{b}−\frac{\mathrm{1}}{\mathrm{5}}\right)=\mathrm{0} \\ $$$$\Rightarrow\:{b}_{\mathrm{1}} =−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{3}\sqrt{\mathrm{5}}}{\mathrm{10}}\:\:{b}_{\mathrm{2}} =\mathrm{0}\:\:{b}_{\mathrm{3}} =−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{3}\sqrt{\mathrm{5}}}{\mathrm{10}} \\ $$$$\Rightarrow\:{a}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{5}}}{\mathrm{10}}\:\:{a}_{\mathrm{2}} =\mathrm{0}\:\:{a}_{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{5}}}{\mathrm{10}} \\ $$

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