What is the square root of negative 8?

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Answer 1 · 2018-03-18T21:50:20

See a solution process below:

$\sqrt{8}$ $sqrt(8)$ can be rewritten as:

$\sqrt{4 \cdot 2 \cdot - 1}$ $sqrt(4 * 2 * -1)$

We can use this rule for radicals to simplify the expression:

$\sqrt{a \cdot b} = \sqrt{a} \cdot \sqrt{b}$ $sqrt(color(red)(a) * color(blue)(b)) = sqrt(color(red)(a)) * sqrt(color(blue)(b))$

$\sqrt{4 \cdot 2 \cdot - 1} \Rightarrow$ $sqrt(color(red)(4) * color(blue)(2) * color(green)(-1)) =>$

$\sqrt{4} \cdot \sqrt{2} \cdot \sqrt{- 1} \Rightarrow$ $sqrt(color(red)(4)) * sqrt(color(blue)(2)) * sqrt(color(green)(-1)) =>$

$2 \sqrt{2} \cdot \sqrt{- 1}$ $2sqrt(color(blue)(2)) * sqrt(color(green)(-1))$

The symbol $i$ $i$ which is an imaginary number is another way to write: $\sqrt{- 1}$ $sqrt(-1)$ so we can rewrite the expression as:

$2 \sqrt{2} \cdot i \Rightarrow$ $2sqrt(color(blue)(2)) * i =>$

$2 i \sqrt{2}$ $2isqrt(color(blue)(2))$

If necessary we can approximate $\sqrt{2}$ $sqrt(2)$ as $1.414$ $1.414$ and get:

$2 \cdot 1.414 i \Rightarrow$ $2 * 1.414i =>$

$2.828 i$ $2.828i$