
How do you simplify the square root of negative \[6\] and root times square root of negative \[18\]?
Answer
464.1k+ views
Hint: In this question, we have to find out the required value from the given particulars.
We need to first find out the square root of negative \[6\] . Then we need to find out the square root of negative \[18\] , then we will multiply the two values. After doing the multiplication we can find out the required solution.
Formula used: We know the formula for imaginary number,
\[{i^2} = - 1\]
i.e., \[i = \sqrt { - 1} \]
Where i is the imaginary number.
Complete step-by-step solution:
We need to simplify the square root of negative \[6\] and root times square root of negative \[18\] .
First, we need to find out the square root of negative \[6\] .
If we use, \[i = \sqrt { - 1} \] then we get,
\[\sqrt { - 6} = \sqrt { - 1 \times 6} = \sqrt { - 1} .\sqrt 6 = i\sqrt 6 \].
Now, we need to find out the square root of negative \[18\] .
\[\sqrt { - 18} = \sqrt { - 1 \times 18} = \sqrt { - 1} .\sqrt {18} = i\sqrt {18} \]
Here, the square root of negative \[6\] end root times square root of negative \[18\]
\[ = \sqrt { - 6} \times \sqrt { - 18} \]
\[ = i\sqrt 6 \times i\sqrt {18} \]
\[ = {i^2}\sqrt 6 \sqrt {18} \]
\[ = - \sqrt {108} \] [Since for any real numbers, \[\sqrt A .\sqrt B = \sqrt {A.B} \] ]
\[ = - \sqrt {2 \times 2 \times 3 \times 3 \times 3} \]
\[ = - \left( { \pm 2 \times 3\sqrt 3 } \right)\]
\[ = \mp 6\sqrt 3 \]
Hence, simplifying the square root of negative \[6\] and root times square root of negative \[18\] are either \[6\sqrt 3 \] or, \[ - 6\sqrt 3 \].
Note: For this problem, we need to know what it is i.
A complex number is a number that can be expressed in the form a+bi where a and b are real numbers and i represents the imaginary unit, satisfying the equation \[{i^2} = - 1\] . Since no real number satisfies this equation, i is called an imaginary number.
Square root:
In mathematics, a square root of a number x is a number y such that, \[{y^2} = x\] . In other words, a number y whose square is x.
For example, \[4, - 4\] are square roots of \[16\] , because \[{4^2} = {\left( { - 4} \right)^2} = 16\] .
We need to first find out the square root of negative \[6\] . Then we need to find out the square root of negative \[18\] , then we will multiply the two values. After doing the multiplication we can find out the required solution.
Formula used: We know the formula for imaginary number,
\[{i^2} = - 1\]
i.e., \[i = \sqrt { - 1} \]
Where i is the imaginary number.
Complete step-by-step solution:
We need to simplify the square root of negative \[6\] and root times square root of negative \[18\] .
First, we need to find out the square root of negative \[6\] .
If we use, \[i = \sqrt { - 1} \] then we get,
\[\sqrt { - 6} = \sqrt { - 1 \times 6} = \sqrt { - 1} .\sqrt 6 = i\sqrt 6 \].
Now, we need to find out the square root of negative \[18\] .
\[\sqrt { - 18} = \sqrt { - 1 \times 18} = \sqrt { - 1} .\sqrt {18} = i\sqrt {18} \]
Here, the square root of negative \[6\] end root times square root of negative \[18\]
\[ = \sqrt { - 6} \times \sqrt { - 18} \]
\[ = i\sqrt 6 \times i\sqrt {18} \]
\[ = {i^2}\sqrt 6 \sqrt {18} \]
\[ = - \sqrt {108} \] [Since for any real numbers, \[\sqrt A .\sqrt B = \sqrt {A.B} \] ]
\[ = - \sqrt {2 \times 2 \times 3 \times 3 \times 3} \]
\[ = - \left( { \pm 2 \times 3\sqrt 3 } \right)\]
\[ = \mp 6\sqrt 3 \]
Hence, simplifying the square root of negative \[6\] and root times square root of negative \[18\] are either \[6\sqrt 3 \] or, \[ - 6\sqrt 3 \].
Note: For this problem, we need to know what it is i.
A complex number is a number that can be expressed in the form a+bi where a and b are real numbers and i represents the imaginary unit, satisfying the equation \[{i^2} = - 1\] . Since no real number satisfies this equation, i is called an imaginary number.
Square root:
In mathematics, a square root of a number x is a number y such that, \[{y^2} = x\] . In other words, a number y whose square is x.
For example, \[4, - 4\] are square roots of \[16\] , because \[{4^2} = {\left( { - 4} \right)^2} = 16\] .
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