Simplify #sqrt300-sqrt48#?

Question

2974 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-04T09:05:05

#sqrt300-sqrt48=6sqrt3#

#sqrt300#

= #sqrt(2xx2xx3xx5xx5)#

= #sqrt(ul(2xx2)xx3xxul(5xx5))#

= #2xx5xxsqrt3#

= #10sqrt3#

Similarly #sqrt48#

= #sqrt(2xx2xx2xx2xx3)#

= #sqrt(ul(2xx2)xxul(2xx2)xx3)#

= #2xx2xxsqrt3#

= #4sqrt3#

Hence #sqrt300-sqrt48#

= #10sqrt3-4sqrt3#

= #(10-4)sqrt3#

= #6sqrt3#