How do you simplify #sqrt7(sqrt14 + sqrt3)#?

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Answer 1 · 2017-10-08T13:12:16

See a solution process below:

First, expand the terms in parenthesis by multiplying each term within the parenthesis by the term outside the parenthesis:

#sqrt( color(red)(7) ) ( sqrt(color(blue)(14)) + sqrt(color(blue)(3)) ) =>#

#(sqrt(color(red)(7)) xx sqrt(color(blue)(14))) + (sqrt(color(red)(7)) xx sqrt(color(blue)(3)))#

Next, we can use this rule for radicals to rewrite the expression:

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

#sqrt(color(red)(7) xx color(blue)(14)) + sqrt(color(red)(7) xx color(blue)(3)) =>#

#sqrt(98) + sqrt(21)#

We can rewrite the term on the left as:

#sqrt(49 xx 2) + sqrt(21)#

We can use the reverse of the rule above to simplify the term on the left:

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

#sqrt(color(red)(49) xx color(blue)(2)) + sqrt(21) =>#

#(sqrt(color(red)(49)) xx sqrt(color(blue)(2))) + sqrt(21) =>#

#(7 xx sqrt(color(blue)(2))) + sqrt(21) =>#

#7sqrt(color(blue)(2)) + sqrt(21)#