
Find the value of \[\sqrt 2 \sqrt { - 3} \].
A. \[\sqrt 6 i\]
B. - \[\sqrt 6 i\]
C. \[\sqrt 3 i\]
D. \[\sqrt 6 \]
Answer
164.1k+ views
Hint: In this question, we have to determine the value of \[\sqrt 2 \sqrt { - 3} \]. For this, we have to use the concept of imaginary unit. That is “iota”. The square root of a negative number \[\sqrt { - 3} \] is the primary square root of the positive radicand multiplied by i. After simplification, we get the desired result.
Formula used: Consider the following useful identity of iota for solving this example..
\[{i^2} = - 1\]
\[i = \sqrt { - 1} \]
Here, i is the imaginary unit called ‘iota’.
Complete step-by-step answer:
Consider an expression \[\sqrt 2 \sqrt { - 3} \]
Now, let us simplify the above expression.
\[\sqrt 2 \sqrt { - 3} = \sqrt 2 \sqrt {3 \times \left( { - 1} \right)} \]
\[\sqrt 2 \sqrt { - 3} = \sqrt 2 \sqrt 3 \sqrt { - 1} \]
But we know that \[{i^2} = - 1 \Rightarrow i = \sqrt { - 1} \]
\[\sqrt 2 \sqrt { - 3} = \sqrt 2 \sqrt {3 \times {i^2}} \]
By simplifying further, we get
\[\sqrt 2 \sqrt { - 3} = \left( {\sqrt {2 \times 3} } \right)i\]
\[\sqrt 2 \sqrt { - 3} = \left( i{\sqrt 6 } \right)\]
Hence, the value of an expression \[\sqrt 2 \sqrt { - 3} \] is \[\left( i{\sqrt 6 } \right)\] .
Therefore, the correct option is \[\left( i{\sqrt 6 } \right)\].
Additional Information: Iota is a Greek letter used in mathematics to represent the imaginary component of a complex number. The all-encompassing set of Complex Numbers was produced once the imaginary I was linked with the collection of Real Numbers. Since there is no real number whose square is a negative integer, imaginary numbers cannot be expressed by a real number. To solve this difficulty, the letter i is invented to signify \[\sqrt { - 1} \].
Note: Many students make mistakes in the simplification part. Many times students write iota, that is i inside the square root sign while finding the square root of a negative number instead of writing it outside the square root sign. This is the only way through which we can solve this example in an easy manner.
Formula used: Consider the following useful identity of iota for solving this example..
\[{i^2} = - 1\]
\[i = \sqrt { - 1} \]
Here, i is the imaginary unit called ‘iota’.
Complete step-by-step answer:
Consider an expression \[\sqrt 2 \sqrt { - 3} \]
Now, let us simplify the above expression.
\[\sqrt 2 \sqrt { - 3} = \sqrt 2 \sqrt {3 \times \left( { - 1} \right)} \]
\[\sqrt 2 \sqrt { - 3} = \sqrt 2 \sqrt 3 \sqrt { - 1} \]
But we know that \[{i^2} = - 1 \Rightarrow i = \sqrt { - 1} \]
\[\sqrt 2 \sqrt { - 3} = \sqrt 2 \sqrt {3 \times {i^2}} \]
By simplifying further, we get
\[\sqrt 2 \sqrt { - 3} = \left( {\sqrt {2 \times 3} } \right)i\]
\[\sqrt 2 \sqrt { - 3} = \left( i{\sqrt 6 } \right)\]
Hence, the value of an expression \[\sqrt 2 \sqrt { - 3} \] is \[\left( i{\sqrt 6 } \right)\] .
Therefore, the correct option is \[\left( i{\sqrt 6 } \right)\].
Additional Information: Iota is a Greek letter used in mathematics to represent the imaginary component of a complex number. The all-encompassing set of Complex Numbers was produced once the imaginary I was linked with the collection of Real Numbers. Since there is no real number whose square is a negative integer, imaginary numbers cannot be expressed by a real number. To solve this difficulty, the letter i is invented to signify \[\sqrt { - 1} \].
Note: Many students make mistakes in the simplification part. Many times students write iota, that is i inside the square root sign while finding the square root of a negative number instead of writing it outside the square root sign. This is the only way through which we can solve this example in an easy manner.
Recently Updated Pages
Geometry of Complex Numbers – Topics, Reception, Audience and Related Readings

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

Atomic Structure - Electrons, Protons, Neutrons and Atomic Models

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Degree of Dissociation and Its Formula With Solved Example for JEE

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

NCERT Solutions for Class 11 Maths Chapter 4 Complex Numbers and Quadratic Equations

Instantaneous Velocity - Formula based Examples for JEE

NCERT Solutions for Class 11 Maths Chapter 6 Permutations and Combinations

NCERT Solutions for Class 11 Maths Chapter 8 Sequences and Series
