Question

Sum of te squares of three is 138 sum of their products taken two at time is 131

Answer 1

Answer:

given condition

a$^{2}$+b$^{2}$+c$^{2}$=138

ab+bc+ca=131

now we know that (a+b+c)$^{2}$

=>  a$^{2}$+b$^{2}$+c$^{2}$+2(ab+bc+ca)

on taking values we get (a+b+c)$^{2}$

=>  138+2(131) = 400

or

a+b+c = 20

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@GauravSaxena01