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.


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Question Number 20307 by Tinkutara last updated on 25/Aug/17

$Show that a (b - c) x^{2} + b (c - a) x y +$ $c (a - b) y^{2} will be a perfect square if a,$ $b, c are in H . P .$
	Answered by ajfour last updated on 25/Aug/17
	$=x^2 [c(a−b)((y/x))^2 +b(c−a)((y/x))+a(b−c)] =c(a−b)x^2 {((y/x))^2 +2[((b(c−a))/(2c(a−b)))]x+ [((b(c−a))/(2c(a−b)))]^2 }−((b^2 (c−a)^2 )/(4c(a−b)))+a(b−c) =c(a−b)x^2 {(y/x)+((b(c−a))/(2c(a−b)))}^2 + ((4ac(a−b)(b−c)−b^2 (c−a)^2 )/(4c(a−b))) ⇒ expression is a perfect square if the constant term is zero, ⇒ 4ac(ab−ac−b^2 +bc) =b^2 c^2 −2b^2 ac+a^2 b^2 ⇒ 4a^2 bc−4a^2 c^2 −4b^2 ac+4abc^2 =b^2 c^2 −2b^2 ac+a^2 b^2 ⇒ 4abc(a+c)=4a^2 c^2 +b^2 (a^2 +c^2 +2ac) ⇒ 4abc(a+c)=4a^2 c^2 +b^2 (a+c)^2 ⇒ [b(a+c)−2ac]^2 =0 ⇒ b=((2ac)/(a+c)) ⇒ a, b, c need be in H.P. •$
	$= x^{2} [c (a - b) {(\frac{y}{x})}^{2} + b (c - a) (\frac{y}{x}) + a (b - c)]$ $= c (a - b) x^{2} {{(\frac{y}{x})}^{2} + 2 [\frac{b (c - a)}{2 c (a - b)}] x +$ ${[\frac{b (c - a)}{2 c (a - b)}]}^{2}} - \frac{b^{2} {(c - a)}^{2}}{4 c (a - b)} + a (b - c)$ $= c (a - b) x^{2} {\frac{y}{x} + \frac{b (c - a)}{2 c (a - b)}}^{2} +$ $\frac{4 a c (a - b) (b - c) - b^{2} {(c - a)}^{2}}{4 c (a - b)}$ $\Rightarrow e x p r e s s i o n i s a p e r f e c t s q u a r e$ $i f t h e c o n s t a n t t e r m i s z e r o,$ $\Rightarrow 4 a c (a b - a c - b^{2} + b c)$ $= b^{2} c^{2} - 2 b^{2} a c + a^{2} b^{2}$ $\Rightarrow 4 a^{2} b c - 4 a^{2} c^{2} - 4 b^{2} a c + 4 {a b c}^{2}$ $= b^{2} c^{2} - 2 b^{2} a c + a^{2} b^{2}$ $\Rightarrow 4 a b c (a + c) = 4 a^{2} c^{2} + b^{2} (a^{2} + c^{2} + 2 a c)$ $\Rightarrow 4 a b c (a + c) = 4 a^{2} c^{2} + b^{2} {(a + c)}^{2}$ $\Rightarrow {[b (a + c) - 2 a c]}^{2} = 0$ $\Rightarrow b = \frac{2 a c}{a + c}$ $\Rightarrow a, b, c n e e d b e i n H . P . ∙$
	Commented by Tinkutara last updated on 25/Aug/17

	$Thank you very much Sir!$
	Answered by $@ty@m last updated on 26/Aug/17
	$discriminant=0 ⇒{b(c−a)y}^2 −4ac(b−c)(a−b)y^2 =0 ⇒b^2 (c−a)^2 −4ac(b−c)(a−b)=0 ⇒b^2 (c−a)^2 =4ac(b−c)(a−b) ⇒{a(b−c)+c(a−b)}^2 =4ac(b−c)(a−b) ⇒{a(b−c)−c(a−b)}^2 =0 ⇒a(b−c)=c(a−b) ⇒ab−ac=ca−cb ⇒(1/c)−(1/b)=(1/b)−(1/a) (dividing by abc) ⇒a, b, c are in H.P.$
	$d i s c r i m i n a n t = 0$ $\Rightarrow {b (c - a) y}^{2} - 4 a c (b - c) (a - b) y^{2} = 0$ $\Rightarrow b^{2} {(c - a)}^{2} - 4 a c (b - c) (a - b) = 0$ $\Rightarrow b^{2} {(c - a)}^{2} = 4 a c (b - c) (a - b)$ $\Rightarrow {a (b - c) + c (a - b)}^{2} = 4 a c (b - c) (a - b)$ $\Rightarrow {a (b - c) - c (a - b)}^{2} = 0$ $\Rightarrow a (b - c) = c (a - b)$ $\Rightarrow a b - a c = c a - c b$ $\Rightarrow \frac{1}{c} - \frac{1}{b} = \frac{1}{b} - \frac{1}{a} (d i v i d i n g b y a b c)$ $\Rightarrow a, b, c a r e i n H . P .$
	Commented by $@ty@m last updated on 25/Aug/17

	$A q u a d r a t i c e x p r e s s i o n i s a p e r f e c t$ $s q u a r e \Rightarrow d i s c r i m i n a n t = 0$
	Commented by ajfour last updated on 25/Aug/17

	$t h a n k s .$
	Commented by Tinkutara last updated on 25/Aug/17

	$Thank you very much Satyam!$
	Commented by $@ty@m last updated on 26/Aug/17

	$\land s i m p l e r m e t h o d$
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