if-x-2-2cx-5d-0-has-root-a-and-b-also-x-2-2ax-5b-0-has-root-c-and-d-then-a-b-c-and-d-are-different-real-number-What-the-value-of-a-b-c-d-

Question Number 2859 by Syaka last updated on 29/Nov/15

i f x^{2} - 2 c x - 5 d = 0 h a s r o o t a a n d b,

a l s o x^{2} - 2 a x - 5 b = 0 h a s r o o t c a n d d

t h e n a, b, c a n d d a r e d i f f e r e n t r e a l n u m b e r .

W h a t t h e v a l u e o f a + b + c + d = ?

Commented by Syaka last updated on 29/Nov/15

N i c e O n e S i r \dots . T h a n k s t o S i r R a s h e e d a n d S i r P r a k a s h f o r S o l u t i o n

Answered by Rasheed Soomro last updated on 29/Nov/15

x^{2} - 2 c x - 5 d = 0 a, b a r e r o o t s

x^{2} - 2 a x - 5 b = 0 c, d a r e r o o s

a + b + c + d = ?

I f {a x}^{2} + b x + c = 0 h a s α a n d β r o o t s

α + β = \frac{- b}{a} a n d α β = \frac{c}{a}

S u m

a + b = \frac{- (- 2 c)}{1} = 2 c \dots \dots \dots . (i)

c + d = \frac{- (- 2 a)}{1} = 2 a \dots \dots \dots . (i i

A d d i n g (i) a n d (i i)

a + b + c + d = 2 c + 2 a = 2 (a + c) \dots \dots \dots \dots \dots \dots \dots (1)

a + c = b + d

P r o d u c t

a b = - 5 d a n d c d = - 5 b \Rightarrow a b c d = 25 b d \Rightarrow a c = 25 \Rightarrow c = \frac{25}{a} \dots \dots . (2)

F r o m (1) a n d (2)

a + b + c + d = 2 (a + \frac{25}{a})

a = \frac{- (- 2 c) + \sqrt{{(- 2 c)}^{2} - 4 (1) (- 5 d)}}{2 (1)} = \frac{2 c + 2 \sqrt{c^{2} + 5 d}}{2}

a = c + \sqrt{c^{2} + 5 d}

b = c - \sqrt{c^{2} + 5 d} \Rightarrow a - b = 2 \sqrt{c^{2} + 5 d} \Rightarrow a^{2} + b^{2} - 2 a b = 4 c^{2} + 20 d

\Rightarrow a^{2} + b^{2} - 4 c^{2} = 2 a b + 20 d \dots \dots \dots \dots \dots \dots \dots \dots . . (A)

c = \frac{- (- 2 a) + \sqrt{{(- 2 a)}^{2} - 4 (1) (- 5 b)}}{2 (1)} = \frac{2 a + 2 \sqrt{a^{2} + 5 b}}{2}

c = a + \sqrt{a^{2} + 5 b}

d = a - \sqrt{a^{2} + 5 b} \Rightarrow c - d = 2 \sqrt{a^{2} + 5 b} \Rightarrow c^{2} + d^{2} - 2 c d = 4 a^{2} + 20 b

\Rightarrow - 4 a^{2} + c^{2} + d^{2} = 2 c d + 20 b \dots \dots \dots \dots \dots \dots \dots \dots . (B)

F o r c o m p l e t e s o l u t i o n (i n d e p e n d a n t o f a, b, c a n d d i - e c o n s t a n t)

s e e t h e c o m m e n t o f S i r p r a k a s h j a i n b e l o w

Commented by prakash jain last updated on 29/Nov/15

I think a + b + c + d should evaluate to a constant

value since you have 4 variables and 4 equation

that you derived .

Commented by Syaka last updated on 29/Nov/15

a n d a + b + c + d I n e e d e x a c t v a l u e, S i r

Commented by RasheedAhmad last updated on 29/Nov/15

N i c e S i r!

Commented by prakash jain last updated on 29/Nov/15

Four equation given by Rasheed

a + b = 2 c

c + d = 2 a

a b = - 5 d

c d = - 5 b

c = \frac{25}{a}

a + b = 2 c \Rightarrow b = \frac{50 - a^{2}}{a}

d = \frac{- 5 b}{c} = \frac{- 5 (50 - a^{2})}{a \frac{25}{a}} = - \frac{50 - a^{2}}{5}

c + d = 2 a

a = b = c = d = 0 is trival solution \dots (A)

\frac{25}{a} - \frac{50 - a^{2}}{5} = 2 a

125 - 50 a + a^{3} = 10 a^{2}

a^{3} - 10 a^{2} - 50 a + 125 = 0

a^{3} + 5 a^{2} - 15 a^{2} - 75 a + 25 a + 125 = 0

(a + 5) (a^{2} - 15 a + 25) = 0

a = - 5 o r a = \frac{15 \pm \sqrt{225 - 100}}{2} = \frac{15 \pm \sqrt{5}}{2}

a = - 5, c = - 5, b = - 5, c = - 5 \dots . (B)

a = \frac{15 - \sqrt{5}}{2}, c = \frac{25}{a} = \frac{50}{15 - \sqrt{5}} = \frac{50 (15 + \sqrt{5})}{100} = \frac{15 + \sqrt{5}}{2}

b = 2 c - a = \frac{15 + 15 \sqrt{5}}{2}, d = \frac{15 - 15 \sqrt{5}}{2} \dots (C)

s i m i l a r y o u c a n g e t t h e l a s t s o l u t i o n f o r a = \frac{15 + \sqrt{5}}{2}

a = \frac{15 + \sqrt{5}}{2}, c = \frac{15 - \sqrt{5}}{2}, b = \frac{15 - 15 \sqrt{5}}{2}, d = \frac{15 + 15 \sqrt{5}}{2} . . (D)

C ans D are only valid solutions for

a, b, c, d are different .

in both cases

a + b + c + d = 30