How do you simplify #(3+ sqrt 5) times (5- sqrt5)#?

1 Answer
Jun 4, 2016

Answer:

#10+2sqrt5#

Explanation:

By expanding the brackets using FOIL or similar method.

#rArr(3+sqrt5)(5-sqrt5)=15-3sqrt5+5sqrt5-(sqrt5)^2#
#"---------------------------------------------------------------"#

now in general #sqrtaxxsqrta=a#

#rArr(sqrt5)^2=sqrt5xxsqrt5=5#

radicals may be collected together in a similar manner to 'collecting like terms' algebraically.

That is #4sqrt3+5sqrt3=(4+5)sqrt3=9sqrt3#

#rArr-3sqrt5+5sqrt5=(-3+5)sqrt5=2sqrt5#
#"---------------------------------------------------------------"#

#rArr(3+sqrt5)(5-sqrt5)=15-3sqrt5+5sqrt5-(sqrt5)^2#

#=15+2sqrt5-5=10+2sqrt5#