Question

The sum of the squares of three numbers is 138, while the sum of their products taken two at a time is 131. Their sum is

Answer 1

hi friend,

let a,b,c be the three numbers

→given a²+b²+c²=138

→ab+bc+ca=131

→let a+b+c=x

squaring on both sides, we get

→(a+b+c)²=a²+b²+c²+2(ab+bc+ca)

=138+2(131)

=138+262=400

→x²=400


x=±20

so a+b+c=±20

I hope this will help u ;)

Answer 2

let no = a,b,c
1st case ab +bc+ca =131
a^2+b^2+c^2=138

(a+b+c)^2= a^2+b^2+c^2 +2(ab+bc+ca)
(a+b+c)^2= 138+2×131
=138+262
=400
a+b+c = √400
20