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:

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Answer 1

Answer:

Let a,b and c be 3 numbers according to the question we have,

a² + b² + c² = 138 and ab+bc+ca=131

(a+b+c)² = a² + b² + c² + 2(ab+bc+ca)

=> (a+b+c)² = 138 + 131

=> a+b+c = √400 = 20

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Answer 2

Dear Mate, here's your answer :

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ab+c+ca=131

a^2+b^2+c^2=138

We know that

(a+b+c)^2=a^2+b^2+c^2+2*(a^2+b^2+c^2)

Putting all the values in RHS we get

(a+b+c)^2= 138+2*131

=138+262=400=20^2

So a+b+c =20

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