
The imaginary part of \[{\left( {3 + 2\sqrt { - 54} } \right)^{1/2}} - {\left( {3 - 2\sqrt { - 54} } \right)^{1/2}}\;\] can be:
A. \[\sqrt 6 \]
B. \[ 2\sqrt 6 \]
C. \[6\]
D. \[ - \sqrt 6 \]
Answer
161.4k+ views
Hint: In this question, we need to find the imaginary part of \[{\left( {3 + 2\sqrt { - 54} } \right)^{1/2}} - {\left( {3 - 2\sqrt { - 54} } \right)^{1/2}}\;\]. For this, we need to simplify the given expression using “i” to get the required result. Here, ‘i’ is the iota sign used to indicate the imaginary number.
Formula used: The “i” is the symbol that is used for expressing imaginary number.
\[{i^2} = - 1\]
That means \[i = \sqrt { - 1} \]
Also, the algebraic identity is used to solve the given question.
\[{\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}\]
Here, a and b are integers.
Complete step-by-step answer:
We know that the given expression is \[{\left( {3 + 2\sqrt { - 54} } \right)^{1/2}} - {\left( {3 - 2\sqrt { - 54} } \right)^{1/2}}\;\]
Let us simplify this expression first.
Let us assume that \[E = {\left( {3 + 2\sqrt { - 54} } \right)^{1/2}} - {\left( {3 - 2\sqrt { - 54} } \right)^{1/2}}\;\;\]
\[E = {\left( {3 + \left( {2\sqrt {54 \times \left( { - 1} \right)} } \right)} \right)^{1/2}} - {\left( {3 - 2\sqrt {54 \times \left( { - 1} \right)} } \right)^{1/2}}\;\;\]
\[E = {\left( {3 + \left( {2\sqrt {54} \left( {\sqrt { - 1} } \right)} \right)} \right)^{1/2}} - {\left( {3 - 2\sqrt {54} \left( {\sqrt { - 1} } \right)} \right)^{1/2}}\;\;\]
Put \[i = \sqrt { - 1} \] in the above expression.
So, we get
\[E = {\left( {3 + \left( {\left( i \right)2\sqrt {54} } \right)} \right)^{1/2}} - {\left( {3 - \left( i \right)2\sqrt {54} } \right)^{1/2}}\;\;\]
\[E = {\left( {3 + \left( {\left( i \right)2\sqrt {9 \times 6} } \right)} \right)^{1/2}} - {\left( {3 - \left( i \right)2\sqrt {9 \times 6} } \right)^{1/2}}\;\;\]
But we know that \[\sqrt 9 = 3\]
Thus, we get
\[E = {\left( {3 + \left( {2 \times \left( i \right)3\sqrt 6 } \right)} \right)^{1/2}} - {\left( {3 - 2 \times \left( i \right)3\sqrt 6 } \right)^{1/2}}\;\;\]
But \[{\left( {3 + \left( i \right)\sqrt 6 } \right)^2} = 3 + 2 \times \left( i \right)3\sqrt 6 \] and \[{\left( {3 - \left( i \right)\sqrt 6 } \right)^2} = 3 - 2 \times \left( i \right)3\sqrt 6 \]
So, we get
\[E = {\left( {{{\left( {3 + \left( i \right)\sqrt 6 } \right)}^2}} \right)^{1/2}} - {\left( {{{\left( {3 - \left( i \right)\sqrt 6 } \right)}^2}} \right)^{1/2}}\;\;\]
\[E = \left( {3 + \left( i \right)\sqrt 6 } \right) - \left( {3 - \left( i \right)\sqrt 6 } \right)\;\;\]
E =\[ \left( {i2\sqrt 6 } \right)\]
Hence, the imaginary part of \[{\left( {3 + 2\sqrt { - 54} } \right)^{1/2}} - {\left( {3 - 2\sqrt { - 54} } \right)^{1/2}}\;\] is \[ \left( {2\sqrt 6 } \right)\].
Therefore, the correct option is (B).
Additional Information: Imaginary numbers are numbers that, once squared, produce a negative outcome. In simple words, imaginary numbers are defined as the square root of negative numbers with no fixed value. This is mostly expressed as real numbers multiplied by an imaginary unit termed "i". The letter ‘I’ serves as the basis for any imaginary number. A complex number is a solution written with the help of this imaginary number in the form x+yi. So, we can say that a complex number, in other words, is one that contains both real and imaginary numbers.
Note: Here, students generally make mistakes in simplifying an expression. They may confuse it with the iota sign. The main trick to solve this problem is we need to consider \[3 + 2 \times \left( i \right)3\sqrt 6 \] as \[{\left( {3 + \left( i \right)\sqrt 6 } \right)^2}\] and \[3 - 2 \times \left( i \right)3\sqrt 6 \] as \[{\left( {3 - \left( i \right)\sqrt 6 } \right)^2}\]. This simplifies the given complex expression and it is easy to find the imaginary part.
Formula used: The “i” is the symbol that is used for expressing imaginary number.
\[{i^2} = - 1\]
That means \[i = \sqrt { - 1} \]
Also, the algebraic identity is used to solve the given question.
\[{\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}\]
Here, a and b are integers.
Complete step-by-step answer:
We know that the given expression is \[{\left( {3 + 2\sqrt { - 54} } \right)^{1/2}} - {\left( {3 - 2\sqrt { - 54} } \right)^{1/2}}\;\]
Let us simplify this expression first.
Let us assume that \[E = {\left( {3 + 2\sqrt { - 54} } \right)^{1/2}} - {\left( {3 - 2\sqrt { - 54} } \right)^{1/2}}\;\;\]
\[E = {\left( {3 + \left( {2\sqrt {54 \times \left( { - 1} \right)} } \right)} \right)^{1/2}} - {\left( {3 - 2\sqrt {54 \times \left( { - 1} \right)} } \right)^{1/2}}\;\;\]
\[E = {\left( {3 + \left( {2\sqrt {54} \left( {\sqrt { - 1} } \right)} \right)} \right)^{1/2}} - {\left( {3 - 2\sqrt {54} \left( {\sqrt { - 1} } \right)} \right)^{1/2}}\;\;\]
Put \[i = \sqrt { - 1} \] in the above expression.
So, we get
\[E = {\left( {3 + \left( {\left( i \right)2\sqrt {54} } \right)} \right)^{1/2}} - {\left( {3 - \left( i \right)2\sqrt {54} } \right)^{1/2}}\;\;\]
\[E = {\left( {3 + \left( {\left( i \right)2\sqrt {9 \times 6} } \right)} \right)^{1/2}} - {\left( {3 - \left( i \right)2\sqrt {9 \times 6} } \right)^{1/2}}\;\;\]
But we know that \[\sqrt 9 = 3\]
Thus, we get
\[E = {\left( {3 + \left( {2 \times \left( i \right)3\sqrt 6 } \right)} \right)^{1/2}} - {\left( {3 - 2 \times \left( i \right)3\sqrt 6 } \right)^{1/2}}\;\;\]
But \[{\left( {3 + \left( i \right)\sqrt 6 } \right)^2} = 3 + 2 \times \left( i \right)3\sqrt 6 \] and \[{\left( {3 - \left( i \right)\sqrt 6 } \right)^2} = 3 - 2 \times \left( i \right)3\sqrt 6 \]
So, we get
\[E = {\left( {{{\left( {3 + \left( i \right)\sqrt 6 } \right)}^2}} \right)^{1/2}} - {\left( {{{\left( {3 - \left( i \right)\sqrt 6 } \right)}^2}} \right)^{1/2}}\;\;\]
\[E = \left( {3 + \left( i \right)\sqrt 6 } \right) - \left( {3 - \left( i \right)\sqrt 6 } \right)\;\;\]
E =\[ \left( {i2\sqrt 6 } \right)\]
Hence, the imaginary part of \[{\left( {3 + 2\sqrt { - 54} } \right)^{1/2}} - {\left( {3 - 2\sqrt { - 54} } \right)^{1/2}}\;\] is \[ \left( {2\sqrt 6 } \right)\].
Therefore, the correct option is (B).
Additional Information: Imaginary numbers are numbers that, once squared, produce a negative outcome. In simple words, imaginary numbers are defined as the square root of negative numbers with no fixed value. This is mostly expressed as real numbers multiplied by an imaginary unit termed "i". The letter ‘I’ serves as the basis for any imaginary number. A complex number is a solution written with the help of this imaginary number in the form x+yi. So, we can say that a complex number, in other words, is one that contains both real and imaginary numbers.
Note: Here, students generally make mistakes in simplifying an expression. They may confuse it with the iota sign. The main trick to solve this problem is we need to consider \[3 + 2 \times \left( i \right)3\sqrt 6 \] as \[{\left( {3 + \left( i \right)\sqrt 6 } \right)^2}\] and \[3 - 2 \times \left( i \right)3\sqrt 6 \] as \[{\left( {3 - \left( i \right)\sqrt 6 } \right)^2}\]. This simplifies the given complex expression and it is easy to find the imaginary part.
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