Let a,b,c be positive roots of an cubic equation such that ab+bc+ac+abc=4, show using vieta′s relations that (a/(a+2))+(b/(b+2))+(c/(c+2))=1


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Question Number 216646 by York12 last updated on 13/Feb/25

$Let a, b, c be positive roots of an cubic equation such that$ $a b + b c + a c + a b c = 4, show using {vieta}^{'} s relations that$ $\frac{a}{a + 2} + \frac{b}{b + 2} + \frac{c}{c + 2} = 1$
	Answered by A5T last updated on 14/Feb/25

	$\frac{a (b + 2) (c + 2) + b (a + 2) (c + 2) + c (a + 2) (b + 2)}{(a + 2) (b + 2) (c + 2)} = 1$ $\Leftrightarrow 3 abc + 4 ab + 4 ac + 4 a + 4 bc + 4 b + 4 c = (a + 2) (b + 2) (c + 2)$ $\Leftrightarrow 16 - abc + 4 (a + b + c)$ $= abc + 2 ab + 2 ac + 4 a + 2 bc + 4 b + 4 c + 8$ $\Leftrightarrow 4 - abc = (ab + bc + ca) which is true .$
	Commented by York12 last updated on 14/Feb/25

	$Thank you but I need a solution based on vieta$
	Answered by Rasheed.Sindhi last updated on 14/Feb/25
	$Let the equation is: x^3 −px^2 +qx−r=0 According to Vieta′s formulas: { ((a+b+c=p)),((ab+bc+ca=q)),((abc=r)) :} Given: ab+bc+ac+abc=4⇒q+r=4 (a/(a+2))+(b/(b+2))+(c/(c+2))=1 ⇒(1−(2/(a+2)))+(1−(2/(b+2)))+(1−(2/(c+2)))=1 ⇒3−2((1/(a+2))+(1/(b+2))+(1/(c+2)))=1 ⇒(1/(a+2))+(1/(b+2))+(1/(c+2))=1 ⇒(b+2)(c+2)+(a+2)(c+2)+(a+2)(b+2) =(a+2)(b+2)(c+2) ⇒bc+2b+2c+4+ac+2a+2c+4+ab+2a+2b+4 =abc+2(ab+bc+ca)+4(a+b+c)+8 ⇒ab+bc+ca+4(a+b+c)+12 =abc+2(ab+bc+ca)+4(a+b+c)+8 ⇒q+4p+12=r+2q+4p+8 ⇒q+12=r+2q+8 ⇒q+r=4 which is according to given. Hence (a/(a+2))+(b/(b+2))+(c/(c+2))=1$
	$L e t t h e e q u a t i o n i s : x^{3} - {p x}^{2} + q x - r = 0$ $A c c o r d i n g t o V {i e t a}^{'} s f o r m u l a s :$ ${\begin{cases} a + b + c = p \\ a b + b c + c a = q \\ a b c = r \end{cases}$ $G i v e n : a b + b c + a c + a b c = 4 \Rightarrow q + r = 4$ $\frac{a}{a + 2} + \frac{b}{b + 2} + \frac{c}{c + 2} = 1$ $\Rightarrow (1 - \frac{2}{a + 2}) + (1 - \frac{2}{b + 2}) + (1 - \frac{2}{c + 2}) = 1$ $\Rightarrow 3 - 2 (\frac{1}{a + 2} + \frac{1}{b + 2} + \frac{1}{c + 2}) = 1$ $\Rightarrow \frac{1}{a + 2} + \frac{1}{b + 2} + \frac{1}{c + 2} = 1$ $\Rightarrow (b + 2) (c + 2) + (a + 2) (c + 2) + (a + 2) (b + 2)$ $= (a + 2) (b + 2) (c + 2)$ $\Rightarrow b c + 2 b + 2 c + 4 + a c + 2 a + 2 c + 4 + a b + 2 a + 2 b + 4$ $= a b c + 2 (a b + b c + c a) + 4 (a + b + c) + 8$ $\Rightarrow a b + b c + c a + 4 (a + b + c) + 12$ $= a b c + 2 (a b + b c + c a) + 4 (a + b + c) + 8$ $\Rightarrow q + 4 p + 12 = r + 2 q + 4 p + 8$ $\Rightarrow q + 12 = r + 2 q + 8$ $\Rightarrow q + r = 4$ $w h i c h i s a c c o r d i n g t o g i v e n .$ $H e n c e$ $\frac{a}{a + 2} + \frac{b}{b + 2} + \frac{c}{c + 2} = 1$
	Answered by Rasheed.Sindhi last updated on 14/Feb/25

	$\frac{a}{a + 2} + \frac{b}{b + 2} + \frac{c}{c + 2} = 1$ $\frac{a}{a + 2} - 1 + \frac{b}{b + 2} - 1 + \frac{c}{c + 2} - 1 = 1 - 3$ $\frac{- 2}{a + 2} + \frac{- 2}{b + 2} + \frac{- 2}{c + 2} = - 2$ $\frac{1}{a + 2} + \frac{1}{b + 2} + \frac{1}{c + 2} = 1$ $▸ (b + 2) (c + 2) + (a + 2) (c + 2) + (a + 2) (b + 2)$ $= (a + 2) (b + 2) (c + 2)$ $▸ a b + b c + c a + 12 + 4 (a + b + c)$ $= a b c + 2 (a b + b c + c a) + 4 (a + b + c) + 8$ $▸ a b + b c + c a + 12$ $= a b c + 2 (a b + b c + c a) + 8$ $▸ a b + b c + c a + a b c = 4 (G i v e n)$ $H e n c e$ $\frac{a}{a + 2} + \frac{b}{b + 2} + \frac{c}{c + 2} = 1$
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