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How do you simplify #-sqrt5*sqrt20#?

Answer 1

Eva Abramson

#-10#

#"using the "color(blue)"laws of radicals"#

#•color(white)(x)sqrtaxxsqrtbhArrsqrt(ab)#

#•color(white)(x)sqrtaxxsqrta=a#

#-sqrt5xxsqrt20#

#=-5xxsqrt(4xx5)#

#=-5xx2sqrt5#

#=-sqrt5xxsqrt5xx2=-5xx2=-10#

#color(red)"OR"#

#-sqrt5xxsqrt20=-sqrt(5xx20)=-sqrt100=-10#

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Answer 2

Eva Adamson

We can use a "shortcut" method, and multiply the numbers inside the radicals to get

#-sqrt100#

This just simplifies to

#-10#

A more systematic approach would be to see if we can simplify #sqrt20#. #20# is the same as #4*5#, so we can rewrite #color(blue)(sqrt20)# as

#-sqrt5*color(blue)(sqrt4*sqrt5)#

Since we are just multiplying, we can rewrite this as

#-sqrt5*sqrt5*sqrt4#

This simplifies to

#-5sqrt4#

#-5*2=-10#

Either way, we get #-10#.

Hope this helps!

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Answer 3

Hazel Anglin

To simplify -sqrt5sqrt20, you can multiply the numbers under the square roots together. This gives you -sqrt(520), which simplifies to -sqrt100. The square root of 100 is 10, so the final simplified answer is -10.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.