Question

If $a+b+c=12$ and $a^2+b^2+c^2=50$, find $ab+bc+ca$

Answer 1

Using the identity $(a+b+c{)}^2=a^2+b^2+c^2+2ab+2bc+2ca$.

It is given that $a+b+c=12$ and $a^2+b^2+c^2=50$, therefore,

$(a+b+c{)}^2=a^2+b^2+c^2+2ab+2bc+2ca\Rightarrow (a+b+c{)}^2=a^2+b^2+c^2+2(ab+bc+ca)\Rightarrow (12{)}^2=50+2(ab+bc+ca)\Rightarrow 144=50+2(ab+bc+ca)\Rightarrow 2(ab+bc+ca)=144-50\Rightarrow 2(ab+bc+ca)=94\Rightarrow ab+bc+ca=\frac{94}{2}=47$

Hence, $ab+bc+ca=47$.