QUESTION

# If the value of a+b+c=12 and ${{a}^{2}}+{{b}^{2}}+{{c}^{2}}=50$ then find the value of ab+bc+ca.

Hint: We have an algebraic identity for square of sum of three numbers a, b and c which is given as, ${{(a+b+c)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab+2bc+2ca$. We will use this identity and substitute given values to obtain the answer.

In the question given that the value of $a+b+c=12$ and ${{a}^{2}}+{{b}^{2}}+{{c}^{2}}=50$ then we have to find the value of $ab+bc+ca$.
To do so we will use the basic algebraic identity for square of sum of three numbers a, b and c which is given as, ${{(a+b+c)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab+2bc+2ca$.
Now, as we are given the value of $a+b+c=12$ and ${{a}^{2}}+{{b}^{2}}+{{c}^{2}}=50$, we will use above stated identity to solve further.
Let ${{(a+b+c)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab+2bc+2ca$ be equation(i).
Given $a+b+c=12$,
Squaring both sides of the above expression we have,
${{(a+b+c)}^{2}}={{(12)}^{2}}$
Using the equation (i) to open square on the left-hand side of the above expression we get,
\begin{align} & {{(a+b+c)}^{2}}={{(12)}^{2}} \\ & \Rightarrow {{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab+2bc+2ca=144 \\ \end{align}
Substituting the value of ${{a}^{2}}+{{b}^{2}}+{{c}^{2}}=50$ and taking 2 common in the above obtained expression we get,
$50+2(ab+bc+ca)=144$
Now making necessary shifting in above we get,
\begin{align} & 50+2(ab+bc+ca)=144 \\ & \Rightarrow 2(ab+bc+ca)=144-50 \\ & \Rightarrow 2(ab+bc+ca)=94 \\ & \Rightarrow (ab+bc+ca)=\dfrac{94}{2} \\ & \Rightarrow (ab+bc+ca)=47 \\ \end{align}
Therefore, we get the required result as $(ab+bc+ca)=47$.
Hence, we obtain our result as $(ab+bc+ca)=47$.

Note: In this type of question students may find the value of 2(ab+bc+ca) as a solution using identity but we need to remember that we have to find the value of (ab+bc+ca).