How do you simplify $\sqrt{88} \cdot \sqrt{33}$ $sqrt88 * sqrt33$ ?

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How do you simplify $\sqrt{88} \cdot \sqrt{33}$ $sqrt88 * sqrt33$ ?

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Answer 1 · 2018-03-06T08:56:08

$22 \sqrt{6}$ $22sqrt6$ . See below

Explanation:

$\sqrt{88} \cdot \sqrt{33} = \sqrt{11 \cdot 2^{3}} \cdot \sqrt{11 \cdot 3} = \sqrt{11} \cdot \sqrt{2^{3}} \cdot \sqrt{11} \cdot \sqrt{3}$ $sqrt88·sqrt33=sqrt(11·2^3)·sqrt(11·3)=sqrt11·sqrt(2^3)·sqrt11·sqrt3$

Because is a product we can reorganize by conmutativity

$\sqrt{11} \cdot \sqrt{2^{3}} \cdot \sqrt{11} \cdot \sqrt{3} = \sqrt{11} \cdot \sqrt{11} \cdot \sqrt{2^{3}} \cdot \sqrt{3}$ $sqrt11·sqrt(2^3)·sqrt11·sqrt3=sqrt11·sqrt11·sqrt(2^3)·sqrt3$

But $\sqrt{2^{3}} = \sqrt{2 \cdot 2^{2}} = \sqrt{2^{2}} \cdot \sqrt{2} = 2 \sqrt{2}$ $sqrt(2^3)=sqrt(2·2^2)=sqrt(2^2)·sqrt2=2sqrt2$ and

$\sqrt{11} \cdot \sqrt{11} = 11$ $sqrt11·sqrt11=11$ . And so:

$\sqrt{11} \cdot \sqrt{11} \cdot \sqrt{2^{3}} \cdot \sqrt{3} = 11 \cdot 2 \cdot \sqrt{2} \cdot \sqrt{3} = 22 \sqrt{2 \cdot 3} = 22 \sqrt{6}$ $sqrt11·sqrt11·sqrt(2^3)·sqrt3=11·2·sqrt2·sqrt3=22sqrt(2·3)=22sqrt6$

Answer 2 · 2018-03-06T09:04:53

You factorize both arguments:

Explanation:

$\sqrt{88} = \sqrt{2 \times 2 \times 2 \times 11} = 2 \sqrt{2 \times 11} = 2 \sqrt{2} \times \sqrt{11}$ $sqrt88=sqrt(2xx2xx2xx11)=2sqrt(2xx11)=2sqrt2xxsqrt11$

$\sqrt{33} = \sqrt{3 \times 11} = \sqrt{3} \times \sqrt{11}$ $sqrt33=sqrt(3xx11)=sqrt3xxsqrt11$

Now multiply:

$2 \sqrt{2} \times \sqrt{11} \times \sqrt{3} \times \sqrt{11}$ $2sqrt2xxsqrt11xxsqrt3xxsqrt11$

Rearrange:

$= 2 \sqrt{2} \times \sqrt{3} \times (\sqrt{11} \times \sqrt{11})$ $=2sqrt2xxsqrt3xx(sqrt11xxsqrt11)$

$= 2 \sqrt{2} \times \sqrt{3} \times 11 = (2 \times 11) \times \sqrt{2 \times 3}$ $=2sqrt2xxsqrt3xx11=(2xx11)xxsqrt(2xx3)$

$= 22 \sqrt{6}$ $=22sqrt6$

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How do you simplify $\sqrt{88} \cdot \sqrt{33}$ $sqrt88 * sqrt33$ ?

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