How do you simplify #(3+ sqrt 5) times (5- sqrt5)#?
1 Answer
Jun 4, 2016
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#