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Question Number 88245 by jagoll last updated on 09/Apr/20

	Answered by john santu last updated on 09/Apr/20

	$\Rightarrow \frac{Δ}{4} = 0, [Δ = d i s c r i m i n a n t]$ $(a b + b c) - (a^{2} + b^{2}) (b^{2} + c^{2}) = 0$ $(a^{2} b^{2} + 2 {a b}^{2} c + b^{2} c^{2}) - (a^{2} b^{2} + a^{2} c + b^{4} + b^{2} c^{2}) = 0$ $2 {a b}^{2} c - a^{2} c^{2} - b^{4} = 0$ ${(b^{2} - a c)}^{2} = 0 \Rightarrow b^{2} = a c$ $r = \frac{b}{a} = \frac{c}{b}$ $t h a t p r o v e t h e s e a r e i n G P$ $r = x . l e t r = \frac{b}{a} = \frac{c}{b}$ $(a^{2} + b^{2}) r^{2} - 2 b (a + c) r + b^{2} + c^{2}$ $= (a^{2} + b^{2}) {(\frac{c}{b})}^{2} - 2 b (a + c) (\frac{c}{b}) +$ $(b^{2} + c^{2})$ $= \frac{{(a c)}^{2}}{b^{2}} + c^{2} - 2 a c - c^{2} + b^{2}$ $w e k n o w a c = b^{2}, l e t s u b s t i t u t e$ $= \frac{b^{4}}{b^{2}} + c^{2} - 2 b^{2} - c^{2} + b^{2}$ $= b^{2} - b^{2} = 0 ⋮$
	Commented by jagoll last updated on 09/Apr/20

	$thank you$
	Answered by $@ty@m123 last updated on 09/Apr/20
	${2b(c+a)}^2 =4(a^2 +b^2 )(b^2 +c^2 ) b^2 (c^2 +a^2 +2ca)=a^2 b^2 +a^2 c^2 +b^4 +b^2 c^2 2b^2 ac=b^4 +a^2 c^2 b^2 +a^2 c^2 −2b^2 ac=0 (b^2 −ac)^2 =0 b^2 =ac ...(1) Now x=−((−2b(c+a))/(2(a^2 +b^2 ))) x=((b(c+a))/(a^2 +ac)) x=((b(c+a))/(a(c+a))) x=(b/a) x=((bc)/b^2 ) { from(1) x=(c/b) ... (2) From (1) and (2) we get the result.$
	${2 b (c + a)}^{2} = 4 (a^{2} + b^{2}) (b^{2} + c^{2})$ $b^{2} (c^{2} + a^{2} + 2 c a) = a^{2} b^{2} + a^{2} c^{2} + b^{4} + b^{2} c^{2}$ $2 b^{2} a c = b^{4} + a^{2} c^{2}$ $b^{2} + a^{2} c^{2} - 2 b^{2} a c = 0$ ${(b^{2} - a c)}^{2} = 0$ $b^{2} = a c . . . (1)$ $N o w$ $x = - \frac{- 2 b (c + a)}{2 (a^{2} + b^{2})}$ $x = \frac{b (c + a)}{a^{2} + a c}$ $x = \frac{b (c + a)}{a (c + a)}$ $x = \frac{b}{a}$ $x = \frac{b c}{b^{2}} {f r o m (1)$ $x = \frac{c}{b} . . . (2)$ $F r o m (1) a n d (2)$ $w e g e t t h e r e s u l t .$
	Commented by jagoll last updated on 09/Apr/20

	$thank you$
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