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 byRasheed.Sindhi last updated on 21/Jul/20

Is the question perfectly right?

Commented by1549442205PVT last updated on 21/Jul/20

Three equatios for three unknowns  I think that it is perfect

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

It is correct sir, but i cannot solve it.

Commented byRasheed.Sindhi last updated on 21/Jul/20

The question may be as:                    a^2 + c^2   =  196        ... (i)                    b^2 +(c−a)^2   =  169   ... (ii)                    c^2 +(b−a)^2   =  225     .... (iii)  Or as                    a^2 +(b−c)^2  =  196        ... (i)                    b^2 +(c−a)^2   =  169   ... (ii)                    c^2 +(b−a)^2   =  225     .... (iii)       ?????

Commented byRasheed.Sindhi last updated on 21/Jul/20

Thanks Sir!

Commented byI 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

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

Wow, thanks sir. I appreciate.