How would you simplify #sqrt48 + sqrt3#?
We can split up
The original expression can be rewritten as
Short story, try watching the result of division of the two numbers and substitute it in the bigger one.
Always try reducing the high number to something of which you know the root. Here someone should notice that:
The root of 16 is known, while the root of 3 can be factored. Therefore:
radicals in simplified form are
where a is a rational number.
to begin with , simplify
# sqrt48 #
by considering the factors of 48 , particularly 'squares'
the factors required here are 16 (square) and 3.
using the following :
# sqrta xx sqrtb hArr sqrtab #
#sqrt48 = sqrt16 xx sqrt3 = 4sqrt3 #
#sqrt48 + sqrt3 = 4sqrt3 + sqrt3 = 5sqrt3 #
To simplify radicals, we must find first its largest "perfect squares" that evenly divides to simplify.
we can simplify
using theorem from above,
since the rule of radical signs is just like variables we can combine like terms, with a slight difference.