Solve-a-2-c-2-196-i-b-2-c-a-2-169-ii-c-2-b-c-2-225-iii-

Question Number 104377 by I want to learn more last updated on 21/Jul/20

Solve : a^{2} + c^{2} = 196 \dots (i)

b^{2} + {(c - a)}^{2} = 169 \dots (ii)

c^{2} + {(b - c)}^{2} = 225 \dots . (iii)

Commented by Rasheed.Sindhi last updated on 21/Jul/20

I s t h e q u e s t i o n p e r f e c t l y r i g h t ?

Commented by 1549442205PVT last updated on 21/Jul/20

Three equatios for three unknowns

I think that it is perfect

Commented by I want to learn more last updated on 21/Jul/20

It is correct sir, but i cannot solve it .

Commented by Rasheed.Sindhi last updated on 21/Jul/20

T h e q u e s t i o n m a y b e a s :

a^{2} + c^{2} = 196 \dots (i)

b^{2} + {(c - a)}^{2} = 169 \dots (ii)

c^{2} + {(b - a)}^{2} = 225 \dots . (iii)

O r a s

a^{2} + {(b - c)}^{2} = 196 \dots (i)

b^{2} + {(c - a)}^{2} = 169 \dots (ii)

c^{2} + {(b - a)}^{2} = 225 \dots . (iii)

? ? ? ? ?

Commented by Rasheed.Sindhi last updated on 21/Jul/20

T h a n k s S i r!

Commented by I want to learn more last updated on 22/Jul/20

Not {(b - a)}^{2} sir . {(b - c)}^{2} .

Answered by 1549442205PVT last updated on 23/Jul/20

From (i) and (ii) we get

b^{2} + c^{2} + a^{2} - 2 ac = 169 \Rightarrow b^{2} - 2 ac = - 27

\Rightarrow a = \frac{b^{2} + 27}{2 c} (1) \Rightarrow a^{2} = \frac{b^{4} + {54 b}^{2} + 729}{{4 c}^{2}} . Replace into

(i) we get b^{4} + {54 b}^{2} + {4 c}^{4} - {784 c}^{2} = 0 (2)

From (iii) we get b = c \pm \sqrt{225 - c^{2}} (3)

\Rightarrow b^{2} = 225 \pm 2 c \sqrt{225 - c^{2}} (4)

b^{4} = 50626 \pm 900 c \sqrt{225 - c^{2}} + {900 c}^{2} - {4 c}^{4} (5)

Replace (4) (5) into (2) we get

50625 \pm 900 c \sqrt{225 - c^{2}} + {900 c}^{2} - {4 c}^{4} +

12150 \pm 108 c \sqrt{225 - c^{2}} + 729 + {4 c}^{4} - {784 c}^{2} + {900 c}^{2} - {4 c}^{4} = 0

\Leftrightarrow {116 c}^{2} + 63504 \pm 1008 c \sqrt{225 - c^{2}} = 0

\Rightarrow {116 c}^{2} + 63504 = \pm 1008 c \sqrt{225 - c^{2}} .

Squaring two sides of the above eqs . we get

{13456 c}^{4} + 14 732 {928 c}^{2} + 4 032 758 016 = 1 016 {064 c}^{2} (225 - c^{2})

\Leftrightarrow 1 029 {520 c}^{4} - 213 881 {472 c}^{2} + 4 032 758 016 = 0 (6)

Solve (6) by caculator we get

c^{2} \in {\frac{15876}{85}; \frac{15876}{757}} \Rightarrow c \in {\pm 13.66661884, \pm 4.57954789}

a) For c = 13.66661884 from (3) we get

b \in {19.84913689, 7.484100795} . Replace

into (1) we get a \in {15.4020625, 3.03702641}

Check directly we see only the triple

(a, b, c) = (3.0370; 7.484; 13.6666} i s accepted

b) For c = 4.579547894, similarly we get

(a, b, c) = (13.22980; - 9.7042800; 4.5795)

c) For c = - 13.66661884 from (3) we get

b \in {- 7.484100795, - 19.84913689}

a \in {- 3.03702641, - 15.4020625}

(a, b, c) = (- 3.037, - 7.4841, - 13.6666)

d) For c = - 4.57954789 we get

b \in {9.704280069, - 18.86337585}

a \in {- 13.22980503, - 41.79746098}

(a, b, c) = (- 13.22980, 9.70428, - 4.5795)

Thus, we get four triples (a, b, c) satisfy

the given system of the equtions

Commented by I want to learn more last updated on 24/Jul/20

Wow, thanks sir . I appreciate .