How do you simplify #4(sqrt2-sqrt7)#?

1 Answer
George C.
Jul 3, 2017

#4(sqrt(2)-sqrt(7)) = 4sqrt(2)-4sqrt(7)#

Explanation:

The two square roots #sqrt(2)# and #sqrt(7)# are essentially unrelated to one another. So about all we can do to simplify the given expression is multiply it out to get:

#4(sqrt(2)-sqrt(7)) = 4sqrt(2)-4sqrt(7)#

#color(white)()#
Bonus

If you encountered #4(sqrt(2)-sqrt(7))# as the denominator of a rational expression, then you could rationalise the denominator by multiplying by #sqrt(2)+sqrt(7)#. For example:

#1/(4(sqrt(2)-sqrt(7))) = (sqrt(2)+sqrt(7))/(4(sqrt(2)-sqrt(7))(sqrt(2)+sqrt(7)))#

#color(white)(1/(4(sqrt(2)-sqrt(7)))) = (sqrt(2)+sqrt(7))/(4((sqrt(2))^2-(sqrt(7))^2))#

#color(white)(1/(4(sqrt(2)-sqrt(7)))) = (sqrt(2)+sqrt(7))/(4(2-7))#

#color(white)(1/(4(sqrt(2)-sqrt(7)))) = (sqrt(2)+sqrt(7))/(-20)#

#color(white)(1/(4(sqrt(2)-sqrt(7)))) = -sqrt(2)/20-sqrt(7)/20#

Related questions
See all questions in Multiplication and Division of Radicals
Impact of this question
999 views around the world
You can reuse this answer
Creative Commons License