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Question Number 124463 by mathocean1 last updated on 03/Dec/20

w e a r e i n N^{3} .

(S) : {\begin{cases} p^{2} + q^{2} = r^{2} \\ q + p + r = 24 a n d r < p + q \end{cases}

1) S h o w t h a t (p; q; r) i s s o l u t i o n o f (S)

i f o n l y r < 12 a n d p; q a r e s o l u t i o n o f

t h e e q u a t i o n : n^{2} - (24 - r) n + 24 (12 - r) = 0^{}

w h e r e n i s u n k n o w n .

Commented by mathocean1 last updated on 03/Dec/20

t h a n k s s i r

Answered by floor(10²Eta[1]) last updated on 03/Dec/20

(I) p + q = 24 - r, but r < p + q

\Rightarrow 24 - r > r \Rightarrow r < 12

(II) p^{2} + q^{2} = r^{2}

p^{2} + q^{2} + 2 pq = r^{2} + 2 pq

{(p + q)}^{2} = r^{2} + 2 pq

{(24 - r)}^{2} = r^{2} + 2 pq

24^{2} - 48 r + r^{2} = r^{2} + 2 pq

24 (24 - 2 r) = 2 pq

pq = 24 (12 - r)

and also p + q = 24 - r

let p and q be the solutions of the the

quadratic equation p (n), since we know

the sum and product of the roots ∴

p (n) = n^{2} - (24 - r) n + 24 (12 - r)

Commented by mathocean1 last updated on 03/Dec/20

t h a n k s s i r \dots

s o w e c a n s o l v e t h e n (S) ?

i t s e e m d i f f i c u l t f o r m e s o l v e i t \dots

Commented by floor(10²Eta[1]) last updated on 07/Dec/20

you {can}^{'} t solve S because you have

2 equations but you have 3 variables