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Question Number 20307 by Tinkutara last updated on 25/Aug/17
Show that a(b − c)x^2 + b(c − a)xy + c(a − b)y^2 will be a perfect square if a, b, c are in H.P.
Showthata(b−c)x2+b(c−a)xy+c(a−b)y2willbeaperfectsquareifa,b,careinH.P.
Answered by ajfour last updated on 25/Aug/17
=x^2 [c(a−b)((y/x))^2 +b(c−a)((y/x))+a(b−c)] =c(a−b)x^2 {((y/x))^2 +2[((b(c−a))/(2c(a−b)))]x+ [((b(c−a))/(2c(a−b)))]^2 }−((b^2 (c−a)^2 )/(4c(a−b)))+a(b−c) =c(a−b)x^2 {(y/x)+((b(c−a))/(2c(a−b)))}^2 + ((4ac(a−b)(b−c)−b^2 (c−a)^2 )/(4c(a−b))) ⇒ expression is a perfect square if the constant term is zero, ⇒ 4ac(ab−ac−b^2 +bc) =b^2 c^2 −2b^2 ac+a^2 b^2 ⇒ 4a^2 bc−4a^2 c^2 −4b^2 ac+4abc^2 =b^2 c^2 −2b^2 ac+a^2 b^2 ⇒ 4abc(a+c)=4a^2 c^2 +b^2 (a^2 +c^2 +2ac) ⇒ 4abc(a+c)=4a^2 c^2 +b^2 (a+c)^2 ⇒ [b(a+c)−2ac]^2 =0 ⇒ b=((2ac)/(a+c)) ⇒ a, b, c need be in H.P. •
=x2[c(a−b)(yx)2+b(c−a)(yx)+a(b−c)]=c(a−b)x2{(yx)2+2[b(c−a)2c(a−b)]x+[b(c−a)2c(a−b)]2}−b2(c−a)24c(a−b)+a(b−c)=c(a−b)x2{yx+b(c−a)2c(a−b)}2+4ac(a−b)(b−c)−b2(c−a)24c(a−b)⇒expressionisaperfectsquareiftheconstanttermiszero,⇒4ac(ab−ac−b2+bc)=b2c2−2b2ac+a2b2⇒4a2bc−4a2c2−4b2ac+4abc2=b2c2−2b2ac+a2b2⇒4abc(a+c)=4a2c2+b2(a2+c2+2ac)⇒4abc(a+c)=4a2c2+b2(a+c)2⇒[b(a+c)−2ac]2=0⇒b=2aca+c⇒a,b,cneedbeinH.P.∙
Commented by Tinkutara last updated on 25/Aug/17
Thank you very much Sir!
ThankyouverymuchSir!
Answered by $@ty@m last updated on 26/Aug/17
discriminant=0 ⇒{b(c−a)y}^2 −4ac(b−c)(a−b)y^2 =0 ⇒b^2 (c−a)^2 −4ac(b−c)(a−b)=0 ⇒b^2 (c−a)^2 =4ac(b−c)(a−b) ⇒{a(b−c)+c(a−b)}^2 =4ac(b−c)(a−b) ⇒{a(b−c)−c(a−b)}^2 =0 ⇒a(b−c)=c(a−b) ⇒ab−ac=ca−cb ⇒(1/c)−(1/b)=(1/b)−(1/a) (dividing by abc) ⇒a, b, c are in H.P.
discriminant=0⇒{b(c−a)y}2−4ac(b−c)(a−b)y2=0⇒b2(c−a)2−4ac(b−c)(a−b)=0⇒b2(c−a)2=4ac(b−c)(a−b)⇒{a(b−c)+c(a−b)}2=4ac(b−c)(a−b)⇒{a(b−c)−c(a−b)}2=0⇒a(b−c)=c(a−b)⇒ab−ac=ca−cb⇒1c−1b=1b−1a(dividingbyabc)⇒a,b,careinH.P.
Commented by $@ty@m last updated on 25/Aug/17
A quadratic expression is a perfect square⇒discriminant=0
Aquadraticexpressionisaperfectsquare⇒discriminant=0
Commented by ajfour last updated on 25/Aug/17
thanks.
thanks.
Commented by Tinkutara last updated on 25/Aug/17
Thank you very much Satyam!
ThankyouverymuchSatyam!
Commented by $@ty@m last updated on 26/Aug/17
∧ simpler method
∧simplermethod

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