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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?
a2+b2=a+b,a2−b2=aba=?b=?andwhatifa2+b2=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
a2+b2=a+ba2−b2=ab2a2=ab(a+b)2a=ab+ab22=b+b2b=−1−32=−2orb=−1+32=1forb=−2a2−4=−2aa2+2a−4=0a=−2−202ora=−2+202forb=1a2−1=aa2−a−1=0a=1−52ora=1+52a2+b2=a−bif
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))
(1)a2+b2=a+b(2)a2−b2=ab(1)−(2)2b2=a(1−b)+b⇒a=−b(2b−1)b−1(1)6b−5(b−1)2+5b2+4b+5=−1b−1−b−1b2(b2−b+15)=0⇒b1=0b2=12−510b3=12+510⇒a1=0a2=12−3510a3=12+3510
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))
(1)a2+b2=a−b(2)a2−b2=ab(1)−(2)2b2=a(1−b)−b⇒a=−b(2b+1)b−1(1)3(10b−7)(b−1)2+5b2+12b+21=−3b−1−3b−3b2(b2+b−15)=0⇒b1=−12−3510b2=0b3=−12+3510⇒a1=12+510a2=0a3=12−510

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