Question

The sum of the squares of the three numbers is 138 while the sum of the products take two at a time 131 the sum is

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