How do you simplify # (sqrt3 - sqrt5) (sqrt5 + sqrt7)#?

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Answer 1 · 2017-08-14T09:10:08

#sqrt15+sqrt21-5-sqrt35#

Use the foil method:
#(a+b)(c+d)=ac+ad+bc+bd#
#sqrt15+sqrt21-5-sqrt35#

Answer 2 · 2017-08-14T09:11:59

#-5+sqrt(15)+sqrt(21)-sqrt(35)#

you can distribute out the brackets and simplify it
#(sqrt(3)-sqrt(5))(sqrt(5)+sqrt(7))#

#=sqrt(3)*sqrt(5)+sqrt(3)*sqrt(7)-sqrt(5)*sqrt(5)-sqrt(5)sqrt(7)#

because
#sqrt(a)*sqrt(b)=sqrt(ab)#

Our equation becomes
#=sqrt(15)+sqrt(21)-sqrt(25)-sqrt(35)#

#=-5+sqrt(15)+sqrt(21)-sqrt(35)#