How do you evaluate 2√3 3√5 - 4√5?

1 Answer
IDKwhatName
Jun 28, 2017

#6sqrt(15)-4sqrt(5)#

Explanation:

The equation can be rewritten as #(2sqrt(3)*3sqrt(5))-4sqrt(5)#.

#asqrt(b)*csqrt(d) -= (a*c)sqrt(b*d)#, so #2sqrt(3)*3sqrt(5) = (2*3)sqrt(3*5) = 6sqrt(15)#.

You now have #6sqrt(15)-4sqrt(5)#, as 15 has no factors which are perfect squares, #6sqrt(15)# cannot be simplified further, leaving you with #6sqrt(15)-4sqrt(5)#.

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