How do you simplify $sqrt88 * sqrt33$ ?

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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:

$=2sqrt2xxsqrt3xx(sqrt11xxsqrt11)$

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

$=22sqrt6$

How do you simplify sqrt88 * sqrt33sqrt88 * sqrt33?