How do you simplify $\sqrt{- 50 x^{2} y^{2}}$ $sqrt(-50x^2y^2)$ ?

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How do you simplify $\sqrt{- 50 x^{2} y^{2}}$ $sqrt(-50x^2y^2)$ ?

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Answer 1 · 2016-11-11T03:34:55

$\sqrt{- 50 x^{2} y^{2}} = \sqrt{- 1} \sqrt{50} \sqrt{x^{2}} \sqrt{y^{2}} = 5 x y i \sqrt{2}$ $sqrt(-50x^2y^2)=sqrt(-1)sqrt50sqrt(x^2)sqrt(y^2)=5xyisqrt2$

Explanation:

I'm going to rewrite this to make it easier to work with:

$\sqrt{- 50 x^{2} y^{2}} = \sqrt{- 1} \sqrt{50} \sqrt{x^{2}} \sqrt{y^{2}}$ $sqrt(-50x^2y^2)=sqrt(-1)sqrt50sqrt(x^2)sqrt(y^2)$

We have 4 square roots and I'll take them in turn:

First is $\sqrt{- 1}$ $sqrt(-1)$ . This is the definition of the term $i$ $i$ , so we have

$\sqrt{- 1} = i$ $sqrt(-1)=i$

Next up is $\sqrt{50}$ $sqrt50$ . We can break this down as follows:

$\sqrt{50} = \sqrt{25 \times 2} = \sqrt{25} \sqrt{2} = 5 \sqrt{2}$ $sqrt50=sqrt(25xx2)=sqrt25sqrt2=5sqrt2$

And lastly we have two variables, both are squared, so we can handle them this way:

$\sqrt{x^{2}} = x$ $sqrt(x^2)=x$ and $\sqrt{y^{2}} = y$ $sqrt(y^2)=y$

So we end up with:

$\sqrt{- 50 x^{2} y^{2}} = \sqrt{- 1} \sqrt{50} \sqrt{x^{2}} \sqrt{y^{2}} = 5 x y i \sqrt{2}$ $sqrt(-50x^2y^2)=sqrt(-1)sqrt50sqrt(x^2)sqrt(y^2)=5xyisqrt2$

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How do you simplify $\sqrt{- 50 x^{2} y^{2}}$ $sqrt(-50x^2y^2)$ ?

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