Question

if a+b+c=5 and a^2+b^2+c^2=29, then the value of 3ab+3bc+3ca is

Answer 1

a^2+b^2+c^2=29......(1)
a+b+c=5
Squaring both sides
(a+b+c)^2=25
a^2+b^2+c^2+2ab+2bc+2ca=25
putting value of a^2+b^2+c^2 from eq. 1
29+2ab+2bc+2bc=25
2(ab+bc+ca)=25-29=-4
(ab+bc+ca)=(-2)
3(ab+bc+ca)=3×(-2)
3ab+3bc+3ca=(-6)

Answer 2

a+b+c=5
squaring both sides,we get,
(a+b+c)^2=25
a^2+b^2+c^2+2 (ab+bc+ca)=25
29+2 (ab+bc+ca)=25 (given a^2+b^2+c^2=29) now,2 (ab+bc+ca)=25-29
2 (ab+bc+ca)=-4
ab+bc+ca=-4/2=-2

Hence,3(ab+bc+ca)=3×(-2)= -6
3ab+3bc+3ca= -6 Ans.