How do you simplify $- \sqrt{40}$ $-sqrt40$ ?

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How do you simplify $- \sqrt{40}$ $-sqrt40$ ?

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Answer 1 · 2018-04-17T00:37:42

$- 2 \sqrt{10}$ $-2sqrt10$

Explanation:

$- \sqrt{40} \to$ $-sqrt40 rarr$ Look for any perfect squares

$- \sqrt{4 \cdot 10} \to$ $-sqrt(4*10) rarr$ $4$ $4$ is a perfect square

$- 2 \sqrt{10}$ $-2sqrt10$

Answer 2 · 2018-04-17T00:40:23

$- 2 \sqrt{10}$ $-2sqrt10$

Explanation:

Note here that:

$\sqrt{a b} = {(a b)}^{\frac{1}{2}} \Rightarrow a^{\frac{1}{2}} \cdot b^{\frac{1}{2}}$ $sqrt(ab)=(ab)^(1/2)=>a^(1/2)*b^(1/2)$

Factor $40$ $40$

$\Rightarrow - \sqrt{4 \cdot 10}$ $=>-sqrt(4*10)$

$\Rightarrow - {(4)}^{\frac{1}{2}} \cdot 10^{\frac{1}{2}}$ $=>-(4)^(1/2)*10^(1/2)$

$\Rightarrow - \sqrt{4} \cdot \sqrt{10}$ $=>-sqrt4*sqrt10$

$\Rightarrow - 2 \sqrt{10}$ $=>-2sqrt10$

Answer 3 · 2018-04-17T00:41:32

$- 2 \sqrt{10}$ $-2sqrt10$

Explanation:

$- \sqrt{40}$ $-sqrt40$ is the same thing as $- \sqrt{4 \cdot 10}$ $-sqrt(4*10)$ . Since 4 is a perfect square, (2*2=4) or ( $\sqrt{4} = 2$ $sqrt(4)=2$ ), you can take four out of the radical to simplify to $- 2 \sqrt{10}$ $-2sqrt10$ .

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How do you simplify $- \sqrt{40}$ $-sqrt40$ ?

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How do you simplify $- \sqrt{40}$ $-sqrt40$ ?