How do you simplify #sqrt3/-sqrt21#?

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Answer 1 · 2015-10-12T23:59:37

#-sqrt(7)/7#

The firs thing to notice here is that you can write #21# as

#21 = 7 * 3#

This means that the denominator of the fraction will be equivalent to

#sqrt(21) = sqrt(3 * 7) = sqrt(3) * sqrt(7)#

SInce #sqrt(3)# is present in both the numerator, and the denominator of the fraction, it will cancel out to give

#sqrt(3)/(-sqrt(21)) = -color(red)(cancel(color(black)(sqrt(3))))/(color(red)(cancel(color(black)(sqrt(3)))) * sqrt(7)) = - 1/sqrt(7)#

Next, rationalize the denominator by multiplying the fraction by #1 = sqrt(7)/sqrt(7)#

#-1/sqrt(7) * sqrt(7)/sqrt(7) = -sqrt(7)/(sqrt(7) * sqrt(7)) = -sqrt(7)/sqrt(7 * 7) = color(green)(-sqrt(7)/7)#