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What is the square root of -26 times the square root of -13?

1 Answer

Oct 8, 2015

$\sqrt{- 26} \cdot \sqrt{- 13} = - 13 \sqrt{2}$ $sqrt(-26)*sqrt(-13) = -13sqrt(2)$

Explanation:

If $a, b \geq 0$ $a, b >= 0$ then $\sqrt{a} \sqrt{b} = \sqrt{a b}$ $sqrt(a)sqrt(b) = sqrt(ab)$

If $a < 0$ $a < 0$ , then $\sqrt{a} = i \sqrt{- a}$ $sqrt(a) = i sqrt(-a)$ , where $i$ $i$ is the imaginary unit.

So:

$\sqrt{- 26} \cdot \sqrt{- 13} = i \sqrt{26} \cdot i \sqrt{13}$ $sqrt(-26) * sqrt(-13) = i sqrt(26) * i sqrt(13)$

$= i^{2} \cdot \sqrt{26} \sqrt{13}$ $= i^2 * sqrt(26)sqrt(13)$

$= - 1 \cdot \sqrt{26 \cdot 13}$ $= -1 * sqrt(26*13)$

$= - \sqrt{13^{2} \cdot 2}$ $= - sqrt(13^2 * 2)$

$= - \sqrt{13^{2}} \sqrt{2}$ $= - sqrt(13^2)sqrt(2)$

$= - 13 \sqrt{2}$ $= -13sqrt(2)$

Note that you have to be careful with square roots of negative numbers. For example:

$1 = \sqrt{1} = \sqrt{- 1 \cdot - 1} \neq \sqrt{- 1} \cdot \sqrt{- 1} = i^{2} = - 1$ $1 = sqrt(1) = sqrt(-1 * -1) != sqrt(-1)*sqrt(-1) = i^2 = -1$

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