QUESTION IMAGE
Question
as an imaginary nu
- √(-81)
of i.
To solve the expression \( 2 \cdot \sqrt{-81} \) (assuming that's the intended expression, as we work with imaginary numbers here), we can follow these steps:
Step 1: Recall the definition of the imaginary unit \( i \)
The imaginary unit \( i \) is defined as \( i = \sqrt{-1} \). This allows us to rewrite square roots of negative numbers in terms of \( i \).
Step 2: Simplify \( \sqrt{-81} \)
We can factor out the negative sign inside the square root:
Using the property of square roots \( \sqrt{ab} = \sqrt{a} \cdot \sqrt{b} \) (valid for real numbers \( a \geq 0, b \geq 0 \); here we extend it to include negative numbers by using \( i \)):
We know \( \sqrt{81} = 9 \) and \( \sqrt{-1} = i \), so:
Step 3: Multiply by the coefficient \( 2 \)
Now substitute \( \sqrt{-81} = 9i \) back into the original expression \( 2 \cdot \sqrt{-81} \):
Step 4: Simplify the product
Multiply \( 2 \) and \( 9 \):
So, \( 2 \cdot \sqrt{-81} = \boldsymbol{18i} \).
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To solve the expression \( 2 \cdot \sqrt{-81} \) (assuming that's the intended expression, as we work with imaginary numbers here), we can follow these steps:
Step 1: Recall the definition of the imaginary unit \( i \)
The imaginary unit \( i \) is defined as \( i = \sqrt{-1} \). This allows us to rewrite square roots of negative numbers in terms of \( i \).
Step 2: Simplify \( \sqrt{-81} \)
We can factor out the negative sign inside the square root:
Using the property of square roots \( \sqrt{ab} = \sqrt{a} \cdot \sqrt{b} \) (valid for real numbers \( a \geq 0, b \geq 0 \); here we extend it to include negative numbers by using \( i \)):
We know \( \sqrt{81} = 9 \) and \( \sqrt{-1} = i \), so:
Step 3: Multiply by the coefficient \( 2 \)
Now substitute \( \sqrt{-81} = 9i \) back into the original expression \( 2 \cdot \sqrt{-81} \):
Step 4: Simplify the product
Multiply \( 2 \) and \( 9 \):
So, \( 2 \cdot \sqrt{-81} = \boldsymbol{18i} \).