Question

If the values of \[{{a}^{2}}+{{b}^{2}}=13\] and \[ab=6\] find: \[3{{\left( a+b \right)}^{2}}-2{{(a-b)}^{2}}\].
\[\begin{align}
 & \text{A}\text{. 14} \\
 & \text{B}\text{. 89} \\
 & \text{C}\text{. 45} \\
 & \text{D}\text{. 73} \\
\end{align}\]

Answer 1

Hint: From the question, we were given that \[{{a}^{2}}+{{b}^{2}}=13\] and \[ab=6\]. Let us assume \[{{a}^{2}}+{{b}^{2}}=13\] as equation (1). In the similar manner, let us assume \[ab=6\] as equation (2). Now we should find the value of \[3{{\left( a+b \right)}^{2}}-2{{(a-b)}^{2}}\]. We know that \[{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}\] and \[{{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}\]. By using the formulae, we should find the value of \[3{{\left( a+b \right)}^{2}}-2{{(a-b)}^{2}}\] in terms of \[{{a}^{2}}+{{b}^{2}}\] and \[ab\]. Let us consider this as equation (3). Then we will substitute equation (1) and equation (2) in equation (3).

Complete step-by-step answer:
Before solving the question, we should know that \[{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}\] and\[{{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}\].
From the question, we were given that \[{{a}^{2}}+{{b}^{2}}=13\] and \[ab=6\].
Let us consider
\[\begin{align}
  & {{a}^{2}}+{{b}^{2}}=13.....(1) \\
 & ab=6.......(2) \\
\end{align}\]
Now we have to find the value of \[3{{\left( a+b \right)}^{2}}-2{{(a-b)}^{2}}\].
We know that \[{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}\] and \[{{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}\].
\[\begin{align}
  & \Rightarrow 3{{\left( a+b \right)}^{2}}-2{{(a-b)}^{2}}=3\left( {{a}^{2}}+2ab+{{b}^{2}} \right)-2\left( {{a}^{2}}-2ab+{{b}^{2}} \right) \\
 & \Rightarrow 3{{\left( a+b \right)}^{2}}-2{{(a-b)}^{2}}=\left( 3{{a}^{2}}+6ab+3{{b}^{2}} \right)-\left( 2{{a}^{2}}-4ab+2{{b}^{2}} \right) \\
 & \Rightarrow 3{{\left( a+b \right)}^{2}}-2{{(a-b)}^{2}}=3{{a}^{2}}+6ab+3{{b}^{2}}-2{{a}^{2}}+4ab-2{{b}^{2}} \\
 & \Rightarrow 3{{\left( a+b \right)}^{2}}-2{{(a-b)}^{2}}={{a}^{2}}+10ab+{{b}^{2}} \\
 & \Rightarrow 3{{\left( a+b \right)}^{2}}-2{{(a-b)}^{2}}={{a}^{2}}+{{b}^{2}}+10ab.......(3) \\
\end{align}\]
Now let us substitute equation (1) and equation (2) in equation (3), then we get
\[\begin{align}
  & \Rightarrow 3{{\left( a+b \right)}^{2}}-2{{(a-b)}^{2}}=13+10\left( 6 \right) \\
 & \Rightarrow 3{{\left( a+b \right)}^{2}}-2{{(a-b)}^{2}}=13+60 \\
 & \Rightarrow 3{{\left( a+b \right)}^{2}}-2{{(a-b)}^{2}}=73......(4) \\
\end{align}\]
From equation (4), it is clear that the value of \[3{{\left( a+b \right)}^{2}}-2{{(a-b)}^{2}}\] is equal to 73.

So, the correct answer is “Option D”.

Note: Students may have a misconception that \[{{\left( a-b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}\] and \[{{\left( a+b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}\]. If this misconception is flowed, then the solution may go wrong as shown below:
From the question, we were given that \[{{a}^{2}}+{{b}^{2}}=13\] and \[ab=6\].
Let us consider
\[\begin{align}
  & {{a}^{2}}+{{b}^{2}}=13.....(1) \\
 & ab=6.......(2) \\
\end{align}\]
Now we have to find the value of \[3{{\left( a+b \right)}^{2}}-2{{(a-b)}^{2}}\].
We know that \[{{\left( a-b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}\] and \[{{\left( a+b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}\].
\[\begin{align}
  & \Rightarrow 3{{\left( a+b \right)}^{2}}-2{{(a-b)}^{2}}=3\left( {{a}^{2}}-2ab+{{b}^{2}} \right)-2\left( {{a}^{2}}+2ab+{{b}^{2}} \right) \\
 & \Rightarrow 3{{\left( a+b \right)}^{2}}-2{{(a-b)}^{2}}=\left( 3{{a}^{2}}-6ab+3{{b}^{2}} \right)-\left( 2{{a}^{2}}+4ab+2{{b}^{2}} \right) \\
 & \Rightarrow 3{{\left( a+b \right)}^{2}}-2{{(a-b)}^{2}}=3{{a}^{2}}-6ab+3{{b}^{2}}-2{{a}^{2}}-4ab-2{{b}^{2}} \\
 & \Rightarrow 3{{\left( a+b \right)}^{2}}-2{{(a-b)}^{2}}={{a}^{2}}-10ab+{{b}^{2}} \\
 & \Rightarrow 3{{\left( a+b \right)}^{2}}-2{{(a-b)}^{2}}={{a}^{2}}+{{b}^{2}}-10ab.......(3) \\
\end{align}\]
Now let us substitute equation (1) and equation (2) in equation (3), then we get
\[\begin{align}
  & \Rightarrow 3{{\left( a+b \right)}^{2}}-2{{(a-b)}^{2}}=13-10\left( 6 \right) \\
 & \Rightarrow 3{{\left( a+b \right)}^{2}}-2{{(a-b)}^{2}}=13-60 \\
 & \Rightarrow 3{{\left( a+b \right)}^{2}}-2{{(a-b)}^{2}}=-47......(4) \\
\end{align}\]
From equation (4), it is clear that the value of \[3{{\left( a+b \right)}^{2}}-2{{(a-b)}^{2}}\] is equal to -47.
But we know that the value of \[3{{\left( a+b \right)}^{2}}-2{{(a-b)}^{2}}\] is equal to 73.
Hence, to correct answers these misconceptions should be avoided.