How do you simplify #sqrt88 * sqrt33#?

Question

2400 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-06T08:56:08

#22sqrt6#. See below

#sqrt88·sqrt33=sqrt(11·2^3)·sqrt(11·3)=sqrt11·sqrt(2^3)·sqrt11·sqrt3#

Because is a product we can reorganize by conmutativity

#sqrt11·sqrt(2^3)·sqrt11·sqrt3=sqrt11·sqrt11·sqrt(2^3)·sqrt3#

But #sqrt(2^3)=sqrt(2·2^2)=sqrt(2^2)·sqrt2=2sqrt2# and

#sqrt11·sqrt11=11#. And so:

#sqrt11·sqrt11·sqrt(2^3)·sqrt3=11·2·sqrt2·sqrt3=22sqrt(2·3)=22sqrt6#

Answer 2 · 2018-03-06T09:04:53

You factorize both arguments:

#sqrt88=sqrt(2xx2xx2xx11)=2sqrt(2xx11)=2sqrt2xxsqrt11#

#sqrt33=sqrt(3xx11)=sqrt3xxsqrt11#

Now multiply:

#2sqrt2xxsqrt11xxsqrt3xxsqrt11#

Rearrange:

#=2sqrt2xxsqrt3xx(sqrt11xxsqrt11)#

#=2sqrt2xxsqrt3xx11=(2xx11)xxsqrt(2xx3)#

#=22sqrt6#