Solve: a^2 + c^2 = 196 ... (i) b^2 + (c − a)^2 = 169 ... (ii) c^2 + (b − c)^2 = 225 .... (iii)


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

$Solve : a^{2} + c^{2} = 196 . . . (i)$ $b^{2} + {(c - a)}^{2} = 169 . . . (ii)$ $c^{2} + {(b - c)}^{2} = 225 . . . . (iii)$
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

$T h e q u e s t i o n m a y b e a s :$ $a^{2} + c^{2} = 196 . . . (i)$ $b^{2} + {(c - a)}^{2} = 169 . . . (ii)$ $c^{2} + {(b - a)}^{2} = 225 . . . . (iii)$ $O r a s$ $a^{2} + {(b - c)}^{2} = 196 . . . (i)$ $b^{2} + {(c - a)}^{2} = 169 . . . (ii)$ $c^{2} + {(b - a)}^{2} = 225 . . . . (iii)$ $? ? ? ? ?$
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
	$From (i) and (ii)we get b^2 +c^2 +a^2 −2ac=169⇒b^2 −2ac=−27 ⇒a=((b^2 +27)/(2c))(1)⇒a^2 =((b^4 +54b^2 +729)/(4c^2 )).Replace into (i)we get b^4 +54b^2 +4c^4 −784c^2 =0(2) From(iii)we get b=c±(√(225−c^2 )) (3) ⇒b^2 =225±2c(√(225−c^2 )) (4) b^4 =50626±900c(√(225−c^2 )) +900c^2 −4c^4 (5) Replace(4)(5) into (2) we get 50625±900c(√(225−c^2 )) +900c^2 −4c^4 + 12150±108c(√(225−c^2 ))+729+4c^4 −784c^2 +900c^2 −4c^4 =0 ⇔116c^2 +63504±1008c(√(225−c^2 )) =0 ⇒116c^2 +63504=±1008c(√(225−c^2 )) . Squaring two sides of the above eqs.we get 13456c^4 +14 732 928c^2 +4 032 758 016=1 016 064c^2 (225−c^2 ) ⇔1 029 520c^4 −213 881 472c^2 +4 032 758 016=0(6) Solve (6) by caculator we get c^2 ∈{((15876)/(85));((15876)/(757))}⇒c∈{±13.66661884,±4.57954789} a)For c=13.66661884 from (3)we get b∈{19.84913689,7.484100795}.Replace into(1)we get a∈{15.4020625,3.03702641} Check directly we see only the triple (a,b,c)=(3.0370;7.484;13.6666}is accepted b)For c=4.579547894,similarly we get (a,b,c)=(13.22980;−9.7042800;4.5795) c)For c=−13.66661884 from (3)we get b∈{−7.484100795,−19.84913689} a∈{−3.03702641,−15.4020625} (a,b,c)=(−3.037,−7.4841,−13.6666) d)For c=−4.57954789 we get b∈{9.704280069,−18.86337585} a∈{−13.22980503,−41.79746098} (a,b,c)=(−13.22980,9.70428,−4.5795) Thus,we get four triples (a,b,c) satisfy the given system of the equtions$
	$From (i) and (ii) we get$ $b^{2} + c^{2} + a^{2} - 2 ac = 169 \Rightarrow b^{2} - 2 ac = - 27$ $\Rightarrow a = \frac{b^{2} + 27}{2 c} (1) \Rightarrow a^{2} = \frac{b^{4} + {54 b}^{2} + 729}{{4 c}^{2}} . Replace into$ $(i) we get b^{4} + {54 b}^{2} + {4 c}^{4} - {784 c}^{2} = 0 (2)$ $From (iii) we get b = c \pm \sqrt{225 - c^{2}} (3)$ $\Rightarrow b^{2} = 225 \pm 2 c \sqrt{225 - c^{2}} (4)$ $b^{4} = 50626 \pm 900 c \sqrt{225 - c^{2}} + {900 c}^{2} - {4 c}^{4} (5)$ $Replace (4) (5) into (2) we get$ $50625 \pm 900 c \sqrt{225 - c^{2}} + {900 c}^{2} - {4 c}^{4} +$ $12150 \pm 108 c \sqrt{225 - c^{2}} + 729 + {4 c}^{4} - {784 c}^{2} + {900 c}^{2} - {4 c}^{4} = 0$ $\Leftrightarrow {116 c}^{2} + 63504 \pm 1008 c \sqrt{225 - c^{2}} = 0$ $\Rightarrow {116 c}^{2} + 63504 = \pm 1008 c \sqrt{225 - c^{2}} .$ $Squaring two sides of the above eqs . we get$ ${13456 c}^{4} + 14 732 {928 c}^{2} + 4 032 758 016 = 1 016 {064 c}^{2} (225 - c^{2})$ $\Leftrightarrow 1 029 {520 c}^{4} - 213 881 {472 c}^{2} + 4 032 758 016 = 0 (6)$ $Solve (6) by caculator we get$ $c^{2} \in {\frac{15876}{85}; \frac{15876}{757}} \Rightarrow c \in {\pm 13.66661884, \pm 4.57954789}$ $a) For c = 13.66661884 from (3) we get$ $b \in {19.84913689, 7.484100795} . Replace$ $into (1) we get a \in {15.4020625, 3.03702641}$ $Check directly we see only the triple$ $(a, b, c) = (3.0370; 7.484; 13.6666} i s accepted$ $b) For c = 4.579547894, similarly we get$ $(a, b, c) = (13.22980; - 9.7042800; 4.5795)$ $c) For c = - 13.66661884 from (3) we get$ $b \in {- 7.484100795, - 19.84913689}$ $a \in {- 3.03702641, - 15.4020625}$ $(a, b, c) = (- 3.037, - 7.4841, - 13.6666)$ $d) For c = - 4.57954789 we get$ $b \in {9.704280069, - 18.86337585}$ $a \in {- 13.22980503, - 41.79746098}$ $(a, b, c) = (- 13.22980, 9.70428, - 4.5795)$ $Thus, we get four triples (a, b, c) satisfy$ $the given system of the equtions$
	Commented by I want to learn more last updated on 24/Jul/20

	$Wow, thanks sir . I appreciate .$
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