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.

Answer 1

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.

Complete step-by-step 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).