Question

How do you write \[64{{w}^{2}}+169{{v}^{2}}\] in the form of \[A\left( Bx+C \right)\left( Dx+E \right)\]?

Answer 1

Hint: This question is from the topic of algebra. In solving this question, we will first understand about the term iota and also know some formulas of iota. After that, we will solve the term \[64{{w}^{2}}+169{{v}^{2}}\] and make that term in the form of \[A\left( Bx+C \right)\left( Dx+E \right)\]. After that, we will get our answer.

Complete step by step answer:
Let us solve this question.
In this question, we have asked to write the term \[64{{w}^{2}}+169{{v}^{2}}\] in the form of \[A\left( Bx+C \right)\left( Dx+E \right)\]. Or, we can say that we have to solve the term \[64{{w}^{2}}+169{{v}^{2}}\] in such a way so that the solve term comes in the form of \[A\left( Bx+C \right)\left( Dx+E \right)\].
Before solving this question, let us first know about the term iota.
The term iota is denoted by the term \[i\]. The value of the term iota is: \[i=\sqrt{-1}\]
We can write the equation \[i=\sqrt{-1}\] as
\[{{i}^{2}}=-1\]
\[{{i}^{3}}=-i\]
\[{{i}^{4}}=1\]
Now, we know that negative number multiplied by negative number is always positive. Or, we can say that \[\left( -1 \right)\times \left( -1 \right)=1\] and we know that the square of iota is negative of 1, so we can write \[\left( -1 \right)\times {{i}^{2}}=1\] or we can write this as \[-{{i}^{2}}=1\]. Now, seeing the term \[64{{w}^{2}}+169{{v}^{2}}\], we can write this term as
\[64{{w}^{2}}+169{{v}^{2}}=64{{w}^{2}}-{{i}^{2}}\times 169{{v}^{2}}\]
As we know that the square of 8 is 64 and square of 13 is 169, so we can write the above equation as
\[\Rightarrow 64{{w}^{2}}+169{{v}^{2}}={{8}^{2}}\times {{w}^{2}}-{{i}^{2}}\times {{13}^{2}}\times {{v}^{2}}\]
The above equation can also be written as
\[\Rightarrow 64{{w}^{2}}+169{{v}^{2}}={{\left( 8w \right)}^{2}}-{{\left( i13v \right)}^{2}}\]
Now, using the formula: \[{{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)\], we can write the above equation as
\[\Rightarrow 64{{w}^{2}}+169{{v}^{2}}=\left( 8w+i\times 13v \right)\left( 8w-i\times 13v \right)\]
So, we have solved this question now.
The term \[64{{w}^{2}}+169{{v}^{2}}\] in the form of \[A\left( Bx+C \right)\left( Dx+E \right)\] is \[\left( 8w+13iv \right)\left( 8w-13iv \right)\].

Note: We should have a better knowledge in the topic of algebra to solve this type of question easily. To solve this type of question easily, we should remember the formulas:
\[{{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)\]
Iota : \[i=\sqrt{-1}\]
\[{{i}^{2}}=-1\]
\[{{i}^{3}}=-i\]
\[{{i}^{4}}=1\]
Remember that, the negative number multiplied by a negative number is always a positive number.