Solve: ${{a}^{2}}-{{\left( 2a+3b \right)}^{2}}$.
Answer
543.3k+ views
Hint: In this problem, we have to perform two of the four fundamental mathematical operations, that is, subtraction and multiplication. Here, we have the two given terms in our expression as variables and not certain numbers that have constant values, but still the basis of difference and product remains the same. We will apply the BODMAS rule to solve this problem.
Complete step-by-step answer:
The expression given to us in the problem is equal to: ${{a}^{2}}-{{\left( 2a+3b \right)}^{2}}$. This can be re-written by multiplying the second term with itself. It is done as follows:
$={{a}^{2}}-\left( 2a+3b \right)\times \left( 2a+3b \right)$
In the above expression, we can see that there are two operators, namely subtraction and multiplication. The BODMAS rule governs us in solving such problems. It states the order in which the operators should be operated. And according to this rule, multiplication should be carried out before subtraction.
Thus, proceeding according to the rule, we get:
$\begin{align}
& ={{a}^{2}}-\left( 2a+3b \right)\times \left( 2a+3b \right) \\
& ={{a}^{2}}-\left( 4{{a}^{2}}+6ab+6ab+9{{b}^{2}} \right) \\
& ={{a}^{2}}-\left( 4{{a}^{2}}+12ab+9{{b}^{2}} \right) \\
\end{align}$
Now, we can simplify our expression by releasing each term inside the bracket with a negative sign. This can be done as follows:
$={{a}^{2}}-\left( 4{{a}^{2}} \right)-\left( 12ab \right)-\left( 9{{b}^{2}} \right)$
Now, we will simplify similar terms and all the other dissimilar terms will just be part of our final answer.
$\begin{align}
& ={{a}^{2}}-4{{a}^{2}}-12ab-9{{b}^{2}} \\
& =-3{{a}^{2}}-12ab-9{{b}^{2}} \\
\end{align}$
Hence, on solving ${{a}^{2}}-{{\left( 2a+3b \right)}^{2}}$, we get the final solution as $\left( -3{{a}^{2}}-12ab-9{{b}^{2}} \right)$.
Note: There is one another method by which we can solve the above problem. It requires the use of formula for difference of two squares which is given by: ${{x}^{2}}-{{y}^{2}}=\left( x+y \right)\left( x-y \right)$. The expression given in our problem is also a difference of squares, so one can apply the above formula also, to get the same solution.
Complete step-by-step answer:
The expression given to us in the problem is equal to: ${{a}^{2}}-{{\left( 2a+3b \right)}^{2}}$. This can be re-written by multiplying the second term with itself. It is done as follows:
$={{a}^{2}}-\left( 2a+3b \right)\times \left( 2a+3b \right)$
In the above expression, we can see that there are two operators, namely subtraction and multiplication. The BODMAS rule governs us in solving such problems. It states the order in which the operators should be operated. And according to this rule, multiplication should be carried out before subtraction.
Thus, proceeding according to the rule, we get:
$\begin{align}
& ={{a}^{2}}-\left( 2a+3b \right)\times \left( 2a+3b \right) \\
& ={{a}^{2}}-\left( 4{{a}^{2}}+6ab+6ab+9{{b}^{2}} \right) \\
& ={{a}^{2}}-\left( 4{{a}^{2}}+12ab+9{{b}^{2}} \right) \\
\end{align}$
Now, we can simplify our expression by releasing each term inside the bracket with a negative sign. This can be done as follows:
$={{a}^{2}}-\left( 4{{a}^{2}} \right)-\left( 12ab \right)-\left( 9{{b}^{2}} \right)$
Now, we will simplify similar terms and all the other dissimilar terms will just be part of our final answer.
$\begin{align}
& ={{a}^{2}}-4{{a}^{2}}-12ab-9{{b}^{2}} \\
& =-3{{a}^{2}}-12ab-9{{b}^{2}} \\
\end{align}$
Hence, on solving ${{a}^{2}}-{{\left( 2a+3b \right)}^{2}}$, we get the final solution as $\left( -3{{a}^{2}}-12ab-9{{b}^{2}} \right)$.
Note: There is one another method by which we can solve the above problem. It requires the use of formula for difference of two squares which is given by: ${{x}^{2}}-{{y}^{2}}=\left( x+y \right)\left( x-y \right)$. The expression given in our problem is also a difference of squares, so one can apply the above formula also, to get the same solution.
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