How do you multiply the square root of 3 times the square root of 5?

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How do you multiply the square root of 3 times the square root of 5?

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Answer 1 · 2015-10-13T14:08:03

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

$\sqrt{3} \sqrt{5} = \sqrt{3 \cdot 5} = \sqrt{15}$ $sqrt(3)sqrt(5) = sqrt(3*5) = sqrt(15)$

Explanation:

${(\sqrt{a} \sqrt{b})}^{2} = \sqrt{a} \sqrt{b} \sqrt{a} \sqrt{b} = \sqrt{a} \sqrt{a} \sqrt{b} \sqrt{b} = a b$ $(sqrt(a)sqrt(b))^2 = sqrt(a)sqrt(b)sqrt(a)sqrt(b) = sqrt(a)sqrt(a)sqrt(b)sqrt(b) = ab$

So $\sqrt{a} \sqrt{b}$ $sqrt(a)sqrt(b)$ is always a square root of $a b$ $ab$ , but is it $\sqrt{a b}$ $sqrt(ab)$ or $- \sqrt{a b}$ $-sqrt(ab)$ ?

If $a, b \geq 0$ $a, b >= 0$ then $\sqrt{a}$ $sqrt(a)$ , $\sqrt{b}$ $sqrt(b)$ and $\sqrt{a b}$ $sqrt(ab)$ are all Real and non-negative, so $\sqrt{a} \sqrt{b} = \sqrt{a b}$ $sqrt(a)sqrt(b) = sqrt(ab)$

But if $a, b < 0$ $a, b < 0$ then $\sqrt{a} \sqrt{b} \neq \sqrt{a b}$ $sqrt(a)sqrt(b) != sqrt(ab)$ as we can see:

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

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How do you multiply the square root of 3 times the square root of 5?

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