How do you simplify #sqrt88 * sqrt33#?

Question

How do you simplify #sqrt88  * sqrt33#?

Benjamin Altschul · Accepted Answer

#22sqrt6#. See below

#sqrt88·sqrt33=sqrt(11·2^3)·sqrt(11·3)=sqrt11·sqrt(2^3)·sqrt11·sqrt3# Since is a product, conmutativity allows us to reorganize it. #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#

Emery Allen · Answer

You factorize both arguments:

#sqrt88=sqrt(2xx2xx2xx11)=2sqrt(2xx11)=2sqrt2xxsqrt11# #sqrt33=sqrt(3xx11)=sqrt3xxsqrt11# Now, increase: #2sqrt2xxsqrt11xxsqrt3xxsqrt11# Reorganize: #=2sqrt2xxsqrt3xx(sqrt11xxsqrt11)# #=2sqrt2xxsqrt3xx11=(2xx11)xxsqrt(2xx3)# #=22sqrt6#

Mila Anderson · Answer

undefined To simplify $ \sqrt{88} \times \sqrt{33} $, you can use the property of square roots which states that the product of two square roots is equal to the square root of the product of their radicands.

Therefore,

$$ \sqrt{88} \times \sqrt{33} = \sqrt{88 \times 33} $$

Now, calculate the product of 88 and 33:

$$ 88 \times 33 = 2904 $$

So,

$$ \sqrt{88} \times \sqrt{33} = \sqrt{2904} $$

To simplify $ \sqrt{2904} $, find the prime factorization of 2904:

$$ 2904 = 2^3 \times 3 \times 11^2 $$

Now, take out pairs of the same numbers to simplify the square root:

$$ \sqrt{2904} = \sqrt{2^2 \times 3 \times 11^2} $$

$$ \sqrt{2904} = 2 \times 11 \times \sqrt{3} $$

$$ \sqrt{2904} = 22 \sqrt{3} $$

So,

$$ \sqrt{88} \times \sqrt{33} = 22 \sqrt{3} $$