Find-the-polynomial-with-roots-that-exceed-the-roots-of-f-x-3x-3-14x-2-x-62-0-by-3-Hence-determine-the-value-of-1-a-3-1-b-3-1-c-3-where-a-b-and-c-are-roots-

Question Number 198968 by necx122 last updated on 26/Oct/23

F i n d t h e p o l y n o m i a l w i t h r o o t s t h a t

e x c e e d t h e r o o t s o f

f (x) = 3 x^{3} - 14 x^{2} + x + 62 = 0 b y 3 . H e n c e

d e t e r m i n e t h e v a l u e o f \frac{1}{a + 3} + \frac{1}{b + 3} + \frac{1}{c + 3},

w h e r e a, b a n d c a r e r o o t s .

Answered by Rasheed.Sindhi last updated on 26/Oct/23

L e t a, b, c a r e r o o t s o f 3 x^{3} - 14 x^{2} + x + 62 = 0

a + b + c = \frac{14}{3}, a b + b c + c a = \frac{1}{3}, a b c = - \frac{62}{3}

T h e r e q u i r e d e q u a t i o n h a v e r o o t s a + 3,

b + 3, c + 3

∙ (a + 3) + (b + 3) + (c + 3) = a + b + c + 9 = \frac{14}{3} + 9 = \frac{41}{3}

∙ (a + 3) (b + 3) + (b + 3) (c + 3) + (c + 3) (a + 3)

= a b + 3 (a + b) + 9 + b c + 3 (b + c) + 9 + c a + 3 (c + a) + 9

= a b + b c + c a + 6 (a + b + c) + 27

= \frac{1}{3} + 6 (\frac{14}{3}) + 27 = \frac{1 + 84 + 81}{3} = \frac{166}{3}

∙ (a + 3) (b + 3) (c + 3)

= a b c + 3 (a b + b c + c a) + 9 (a + b + c) + 27

= - \frac{62}{3} + 3 (\frac{1}{3}) + 9 (\frac{14}{3}) = \frac{- 62 + 3 + 126}{3} = \frac{67}{3}

H e n c e t h e r e q u i r e d e q u a t i o n w i l l b e :

x^{3} + \frac{41}{3} x^{2} + \frac{166}{3} x + \frac{67}{3} = 0

\Rightarrow 3 x^{3} + 41 x^{2} + 166 x + 67 = 0

\frac{1}{a + 3} + \frac{1}{b + 3} + \frac{1}{c + 3}

= \frac{(a + 3) (b + 3) + (b + 3) (c + 3) + (a + 3) (c + 3)}{(a + 3) (b + 3) (c + 3)}

= \frac{\frac{166}{3}}{\frac{67}{3}} = \frac{166}{67}

Commented by necx122 last updated on 26/Oct/23

This is great and I'm grateful. well understood. Thank you sir.

Commented by mr W last updated on 26/Oct/23

i t h i n k t h e q u e s t i o n s u g g e s t s o n e t o

s h i f t t h e g i v e n p o l y n o m i a l f (x) t o t h e

r i g h t d i r e c t i o n b y 3 t o g e t t h e n e w

p o l y n o m i a l a t f i r s t . t h i s i s t h e e a s i e s t

w a y t o d e t e r m i n e t h e n e w p o l y n o m i a l .

p l e a s e r e c h e c k y o u r c a l c u l a t i o n s i r!

i t h i n k s o m e t h i n g i s w r o n g i n i t,

s e e f o l l o w i n g d i a g r a m w i t h

(1) = o r i g i n a l p o l y n o m i a l f (x)

(2) = w h a t y o u g o t

(3) = w h a t i g o t (s e e b e l o w)

Commented by mr W last updated on 26/Oct/23

Commented by AST last updated on 26/Oct/23

T h e c o r r e c t e q u a t i o n s h o u l d b e :

3 x^{2} - 41 x^{2} + 166 x - 67 = 0

Commented by mr W last updated on 26/Oct/23

i t h i n k t h e c o r r e c t e q u a t i o n i s

k (3 x^{2} - 41 x^{2} + 166 x - 148) = 0 w i t h k \in R, k \neq 0

Commented by mr W last updated on 26/Oct/23

Commented by AST last updated on 26/Oct/23

I^{'} m r e f e r r i n g t o t h e s o l u t i o n a b o v e .

T h e s o l u t i o n w o u l d h a v e b e e n c o r r e c t i f a, b a n d

c w e r e a l l r e a l n u m b e r s .

(a + 3) (b + 3) (c + 3)

= a b c + 3 (a b + b c + c a) + 9 (a + b + c) + 27 i s o n l y t r u e

w h e n a, b, c \in R

W h e n t w o (i n t h e c a s e o f a c u b i c e q u a t i o n) c o m e

f r o m C ∖ R, t h e n w e a r e n o l o n g e r d e a l i n g w i t h

v a r i a b l e s (a, b, c) b u t (a, \bar{b}, \bar{c}) w h e r e \bar{b} = x + y i .

\Rightarrow \bar{b} + 3 = (x + 3) + y i; s o w e a r e o n l y i n c r e a s i n g

t h e r e a l p a r t b y 3

Commented by AST last updated on 27/Oct/23

L e t a = a, \bar{b} = x + y i, \bar{c} = z - y i

a + \bar{b} + \bar{c} = \frac{14}{3} \Rightarrow \bar{c} = z - y i \Rightarrow a + \bar{b} + \bar{c} = a + x + z = \frac{14}{3}

a \bar{b} + \bar{b} \bar{c} + \bar{c a} = \frac{1}{3} \Rightarrow a (x + y i) + (x + y i) (z - y i)

+ a (z - y i) \Rightarrow a x + a z + x z + y^{2} + i (y z - x y) = \frac{1}{3}

\Rightarrow y z - x y = 0 \Rightarrow z = x \Rightarrow 2 a x + x^{2} + y^{2} = \frac{1}{3}

a \bar{b} \bar{c} = \frac{- 62}{3} \Rightarrow a (x^{2} + y^{2}) = \frac{- 62}{3}

a + \bar{b} + \bar{c} = \frac{14}{3} \Rightarrow a + 2 x = \frac{14}{3}

* (a + 3) + (\bar{b} + 3) + (\bar{c} + 3) = a + x + z + 9 = \frac{41}{3}

* (a + 3) (\bar{b} + 3) (\bar{c} + 3) = (a + 3) (x + 3 + y i) (x + 3 - y i)

= (a + 3) [{(x + 3)}^{2} - {(y i)}^{2}] = (a + 3) [{(x + 3)}^{2} + y^{2}]

= (a + 3) (x^{2} + y^{2} + 6 x + 9] = a (x^{2} + y^{2}) + 3 (2 a x + x^{2} + y^{2})

+ 9 (a + 2 x) + 27 = - \frac{62}{3} + 3 (\frac{1}{3}) + 9 (\frac{14}{3}) + \frac{81}{3} = \frac{148}{3}

* (a + 3) (\bar{b} + 3) + (\bar{b} + 3) (\bar{c} + 3) + (\bar{c} + 3) (a + 3)

= (a + 3) (x + y i + 3) + (x + 3 + y i) (x + 3 - y i)

+ (x - y i + 3) (a + 3)

= (a + 3) (2 x + 6) + {(x + 3)}^{2} + y^{2} = 2 a x + 6 a + 12 x + 27

+ x^{2} + y^{2} = (2 a x + x^{2} + y^{2}) + 6 (a + 2 x) + 27

= \frac{1}{3} + 6 (\frac{14}{3}) + \frac{81}{3} = \frac{166}{3}

x^{3} - \frac{41}{3} x^{2} + \frac{166}{3} x - \frac{148}{3} = 0

\Rightarrow 3 x^{3} - 41 x^{2} + 166 x - 148 = 0 \Rightarrow \frac{1}{a + 3} + \frac{1}{\bar{b} + 3} + \frac{1}{\bar{c} + 3}

= \frac{166}{148} = \frac{83}{74}

Answered by mr W last updated on 26/Oct/23

s a y t h e n e w p o l y n o m i a l i s p (x) .

r o o t s o f p (x) e x c e e d t h e r o o t s o f f (x)

b y 3, t h a t m e a n s p (x) = f (x - 3) .

p (x) = 3 {(x - 3)}^{3} - 14 {(x - 3)}^{2} + (x - 3) + 62

p (x) = 3 x^{3} - 41 x^{2} + 166 x - 148

s a y r o o t s o f p (x) = 0 a r e A, B, C,

t h e n A = a + 3, B = b + 3, C = c + 3

\frac{1}{a + 3} + \frac{1}{b + 3} + \frac{1}{c + 3} = \frac{1}{A} + \frac{1}{B} + \frac{1}{C}

= \frac{A B + B C + C A}{A B C} = \frac{\frac{166}{3}}{\frac{148}{3}} = \frac{166}{148} = \frac{83}{74} ✓

Commented by necx122 last updated on 26/Oct/23

Wow!! Having to exclaim is all i can do at this point. Funnily, both answers appear in the options.

Commented by Rasheed.Sindhi last updated on 27/Oct/23

Most efficient way!