Let the polynomial p(x)=5x^3 +3x^2 −10 have roots a,b and c. What is the value of (a/(b+c))+(b/(c+a))+(c/(a+b))?


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Question Number 199092 by necx122 last updated on 27/Oct/23

$L e t t h e p o l y n o m i a l p (x) = 5 x^{3} + 3 x^{2} - 10$ $h a v e r o o t s a, b a n d c . W h a t i s t h e v a l u e$ $o f \frac{a}{b + c} + \frac{b}{c + a} + \frac{c}{a + b} ?$
	Answered by Rasheed.Sindhi last updated on 28/Oct/23

	$p (x) = 5 x^{3} + 3 x^{2} - 10$ $a + b + c = - \frac{3}{5}, a b + b c + c a = 0, a b c = 2$ $a^{2} + b^{2} + c^{2} = {(a + b + c)}^{2} - 2 (a b + b c + c a)$ $a^{2} + b^{2} + c^{2} = {(- \frac{3}{5})}^{2} - 2 (0) = \frac{9}{25}$ $\frac{a}{b + c} + 1 + \frac{b}{c + a} + 1 + \frac{c}{a + b} + 1 - 3$ $= \frac{a + b + c}{b + c} + \frac{b + c + a}{c + a} + \frac{c + a + b}{a + b} - 3$ $= (a + b + c) (\frac{1}{b + c} + \frac{1}{c + a} + \frac{1}{a + b}) - 3$ $= - \frac{3}{5} (\frac{1}{- \frac{3}{5} - a} + \frac{1}{- \frac{3}{5} - b} + \frac{1}{- \frac{3}{5} - c}) - 3$ $= - \frac{3}{5} (\frac{5}{- 3 - 5 a} + \frac{5}{- 3 - 5 b} + \frac{5}{- 3 - 5 c}) - 3$ $= 3 (\frac{1}{3 + 5 a} + \frac{1}{3 + 5 b} + \frac{1}{3 + 5 c}) - 3$ $= 3 (\frac{(3 + 5 a) (3 + 5 b) + (3 + 5 b) (3 + 5 c) + (3 + 5 a) (3 + 5 c)}{(3 + 5 a) (3 + 5 b) (3 + 5 c)}) - 3$ $= 3 (\frac{9 + 15 a + 15 b + 25 a b + 9 + 15 b + 15 c + 25 b c + 9 + 15 c + 15 a + 25 c a}{27 + 45 (a + b + c) + 75 (a b + b c + c a) + 125 a b c}) - 3$ $= 3 (\frac{27 + 30 (a + b + c) + 25 (a b + b c + c a)}{27 + 45 (a + b + c) + 75 (a b + b c + c a) + 125 a b c}) - 3$ $= 3 (\frac{27 + 30 (- \frac{3}{5}) + 25 (0)}{27 + 45 (- \frac{3}{5}) + 75 (0) + 125 (2)}) - 3$ $= 3 (\frac{27 - 18}{27 - 27 + 250}) - 3 = \frac{27 - 750}{250} = - \frac{723}{250}$
	Commented by deleteduser1 last updated on 28/Oct/23
	$This method may not always work when b,c ∈C\R$
	$T h i s m e t h o d m a y n o t a l w a y s w o r k w h e n b, c \in C ∖ R$
	Commented by Rasheed.Sindhi last updated on 28/Oct/23

	${Y o u}^{'} r e v e r y r i g h t A S T s i r!$ $T h a n X!$
	Commented by mr W last updated on 28/Oct/23

	$R a s h e e d s i r :$ $i t h i n k y o u r p a t h i s c o r r e c t . b u t t a k e$ $c a r e i n l a s t s t e p : a b c = 2 \neq 10, t h e n$ $y o u a l s o g e t - \frac{723}{250} .$
	Commented by Rasheed.Sindhi last updated on 28/Oct/23

	$A l o t o f T h α n k s s i r!$
	Answered by deleteduser1 last updated on 28/Oct/23

	$L e t a = a; b = x + y i, c = e + f i$ $a + b + c = \frac{- 3}{5} \Rightarrow f = - y (o t h e r w i s e, s u m {w o n}^{'} t \in R)$ $a b c = 2 \Rightarrow a (x + y i) (e - y i) = 2 \Rightarrow e = x$ $\Rightarrow c = x - y i \Rightarrow a b c = a (x^{2} + y^{2}) = 2; a + 2 x = \frac{- 3}{5};$ $a b + b c + c a = 0 \Rightarrow a (x + y i) + x^{2} + y^{2} + a (x - y i) = 0$ $\Rightarrow 2 a x + x^{2} + y^{2} = 0 \Rightarrow a (- 2 a x) = 2 \Rightarrow a^{2} x = - 1$ $\frac{a}{b + c} + \frac{b}{c + a} + \frac{c}{a + b} = (a + b + c) (\frac{1}{b + c} + \frac{1}{c + a} + \frac{1}{a + b}) - 3$ $= (a + 2 x) (\frac{1}{2 x} + \frac{2 (x + a)}{x^{2} + a^{2} + 2 a x + y^{2}}) - 3 = (\frac{- 3}{5}) (\frac{1}{2 x} + \frac{2 (x + a)}{a^{2}}) - 3$ $= (\frac{- 3}{5}) (\frac{a^{2} + 4 x^{2} + 4 a x}{2 a^{2} x}) - 3 = (\frac{- 3}{5}) \frac{{(a + 2 x)}^{2}}{2 a^{2} x} - 3$ $= (\frac{- 3}{5}) \frac{(\frac{9}{25})}{\frac{- 2}{1}} - 3 = (\frac{9}{- 50} \times \frac{- 3}{5}) - 3 = \frac{27}{250} - 3 = - \frac{723}{250}$
	Answered by mr W last updated on 28/Oct/23

	$\underset{―}{M e t h o d I}$ $l e t s = a + b + c = - \frac{3}{5}$ $a b + b c + c a = 0$ $a b c = \frac{10}{5} = 2$ $\frac{a}{b + c} + \frac{b}{c + a} + \frac{c}{a + b}$ $= \frac{a}{s - a} + \frac{b}{s - b} + \frac{c}{s - c}$ $= \frac{(a + b + c) s^{2} - 2 (a b + b c + c a) s + 3 a b c}{(s - a) (s - b) (s - c)}$ $= \frac{(a + b + c) s^{2} - 2 (a b + b c + c a) s + 3 a b c}{s^{3} - (a + b + c) s^{2} + (a b + b c + c a) s - a b c}$ $= \frac{s^{3} - 2 (a b + b c + c a) s + 3 a b c}{(a b + b c + c a) s - a b c}$ $= \frac{s^{3} + 3 a b c}{- a b c}$ $= - \frac{s^{3}}{a b c} - 3$ $= - {(- \frac{3}{5})}^{3} \times \frac{1}{2} - 3 = - \frac{723}{250} ✓$
	Commented by mr W last updated on 28/Oct/23

	$\underset{―}{M e t h o d I I}$ $l e t s = a + b + c = - \frac{3}{5}$ $l e t u = \frac{x}{s - x}, i . e . x = \frac{s u}{u + 1}$ $5 {(\frac{s u}{u + 1})}^{3} + 3 {(\frac{s u}{u + 1})}^{2} - 10 = 0$ $5 {(s u)}^{3} + 3 {(s u)}^{2} (u + 1) - 10 {(u + 1)}^{3} = 0$ $(5 s^{3} + 3 s^{2} - 10) u^{3} + 3 (s^{2} - 10) u^{2} - 30 u - 10 = 0$ $\frac{a}{b + c} + \frac{b}{c + a} + \frac{c}{a + b} = \frac{a}{s - a} + \frac{b}{s - b} + \frac{c}{s - c}$ $= \frac{x_{1}}{s - x_{1}} + \frac{x_{2}}{s - x_{2}} + \frac{x_{3}}{s - x_{3}} = u_{1} + u_{2} + u_{3}$ $= - \frac{3 (s^{2} - 10)}{5 s^{3} + 3 s^{2} - 10} = - \frac{723}{250} ✓$
	Commented by mr W last updated on 28/Oct/23

	$M e t h o d I I I$ $s = a + b + c = - \frac{3}{5}$ $a b + b c + c a = 0$ $a b c = \frac{10}{5} = 2$ $\frac{a}{b + c} + \frac{b}{c + a} + \frac{c}{a + b}$ $= \frac{a + b + c}{b + c} + \frac{a + b + c}{c + a} + \frac{a + b + c}{a + b} - 3$ $= s (\frac{1}{s - a} + \frac{1}{s - b} + \frac{1}{s - c}) - 3$ $= s \times \frac{3 s^{2} - 2 (a + b + c) s + a b + b c + c a}{(s - a) (s - b) (s - c)} - 3$ $= s \times \frac{3 s^{2} - 2 s^{2} + a b + b c + c a}{s^{3} - (a + b + c) s^{2} + (a b + b c + c a) s - a b c} - 3$ $= s \times \frac{s^{2} + a b + b c + c a}{(a b + b c + c a) s - a b c} - 3$ $= \frac{s^{3}}{- a b c} - 3$ $= - \frac{1}{2} {(- \frac{3}{5})}^{3} - 3 = - \frac{723}{250} ✓$
	Commented by Rasheed.Sindhi last updated on 28/Oct/23

	$G reat sir!$
	Commented by necx122 last updated on 28/Oct/23
	This is one thing I love about this platform, the seriousness in giving us answers that are accurate and the time spent in offering this untamed service to us. Personally, I'm grateful sirs for this several methods; I understand them. @Rashes.Sindhi and @Mr. W
	Answered by ajfour last updated on 28/Oct/23
	$a+b+c=−(3/5)=s ab+bc+ca=0 abc=2 b+c=−((bc)/a)=−(2/a^2 ) Q=Σ((s−(b+c))/(b+c)) (Q+3)(−(5/3))=Σ(1/(b+c)) ((5Q)/3)+5=(1/2)Σa^2 ((5Q)/3)+5=(1/2){s^2 −2(ab+bc+ca)} 5((Q/3)+1)=(1/2)(s^2 −0)=(9/(50)) Q=3((9/(250))−1)=−((723)/(250))$
	$a + b + c = - \frac{3}{5} = s$ $a b + b c + c a = 0$ $a b c = 2$ $b + c = - \frac{b c}{a} = - \frac{2}{a^{2}}$ $Q = Σ \frac{s - (b + c)}{b + c}$ $(Q + 3) (- \frac{5}{3}) = Σ \frac{1}{b + c}$ $\frac{5 Q}{3} + 5 = \frac{1}{2} Σ a^{2}$ $\frac{5 Q}{3} + 5 = \frac{1}{2} {s^{2} - 2 (a b + b c + c a)}$ $5 (\frac{Q}{3} + 1) = \frac{1}{2} (s^{2} - 0) = \frac{9}{50}$ $Q = 3 (\frac{9}{250} - 1) = - \frac{723}{250}$
	Commented by necx122 last updated on 28/Oct/23
	just when I was in awe of the amount of steps used in solving the question here is Mr. Ajfour with another mind blowing approach. I'm grateful sir. Thank you.
	Answered by Frix last updated on 28/Oct/23

	$x^{3} + {p x}^{2} + q x + r = 0$ $\frac{x_{1}}{x_{2} + x_{3}} + \frac{x_{2}}{x_{1} + x_{3}} + \frac{x_{3}}{x_{1} + x_{2}} = - 2 + \frac{p^{3} + r}{p q - r}$
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