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Question Number 78762 by ajfour last updated on 20/Jan/20

3 {a c r}^{2} (1 - r) + 3 a p r (1 - r) (p a + q b)

+ 3 b q r (1 - r) (p a + q b)

= 3 {(1 - r)}^{2} {(p a + q b)}^{2} + r^{2} b^{2}

F i n d p, q, r s u c h t h a t t h e e q u a t i o n

i s s a t i s f i e d f o r g e n e r a l a n y

v a l u e s o f a, b, c .

Commented by ajfour last updated on 20/Jan/20

r e a l v a l u e s a r e r e q u i r e d, a n y

m e t h o d t o t h i s ?

Answered by mr W last updated on 20/Jan/20

3 r^{2} (1 - r) a c + 3 p^{2} r (1 - r) a^{2} + 6 p q r (1 - r) a b + 3 q^{2} r (1 - r) b^{2}

= 3 p^{2} {(1 - r)}^{2} a^{2} + [3 q^{2} {(1 - r)}^{2} - r^{2}] b^{2} + 6 p q {(1 - r)}^{2} a b

3 r^{2} (1 - r) = 0 \Rightarrow r = 0 o r 1

3 p^{2} r (1 - r) = 3 p^{2} {(1 - r)}^{2} \Rightarrow r = 0, p = 0 o r r = 1, p = a n y

3 q^{2} r (1 - r) = 3 q^{2} {(1 - r)}^{2} - r^{2} \Rightarrow r = 0, q = 0 o r r = 1, q = i m p o s s i b l e

6 p q r (1 - r) = 6 p q {(1 - r)}^{2} \Rightarrow r = p = q = 0

\Rightarrow o n l y p o s s i b i l i t y i s p = q = r = 0 .

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