How do you simplify #sqrt52-sqrt1300#?

1 Answer
Jun 6, 2017

Answer:

See a solution process below:

Explanation:

We can rewrite the terms within the radicals as:

#sqrt(4 * 13) - sqrt(4 * 325)#

Using this rule for multiplication of radicals we can rewrite each radical as:

#sqrt(color(red)(a) * color(blue)(b)) = sqrt(color(red)(a)) * sqrt(color(blue)(b))#

#sqrt(color(red)(4) * color(blue)(13)) - sqrt(color(red)(4) * color(blue)(325)) = (sqrt(color(red)(4)) * sqrt(color(blue)(13))) - (sqrt(color(red)(4)) * sqrt(color(blue)(325))) =#

#sqrt(color(red)(4))(sqrt(color(blue)(13)) - sqrt(color(blue)(325))) = 2(sqrt(13) - sqrt(325)#

We can now simplify the radical on the right as:

#2(sqrt(13) - sqrt(color(red)(25) * color(blue)(13))) = 2(sqrt(13) - (sqrt(color(red)(25)) * sqrt(color(blue)(13)))) =#

#2(sqrt(13) - 5sqrt(color(blue)(13))) = 2sqrt(13)(1 - 5) = 2sqrt(13) * -4 =#

#-8sqrt(13)#

Or

#-28.8# rounded to the nearest 10th