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simplify sqrt72 -3sqrt12 + sqrt 192

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6√2+2√3

Step-by-step explanation:

We want to simplify the following radical expression

$$\displaystyle \sqrt{72} - 3 \sqrt{12} + \sqrt{192}$$

Recall that

$$\sqrt{ab} = \sqrt{a} \sqrt{b} , \forall \text{a and b such that a\geq0,b\geq0}$$

Utilizing the formula yields,

$$\sqrt{72} \implies \sqrt{36 \cdot 2} \implies 6 \sqrt{2}$$

$$\sqrt{12} \implies \sqrt{4\cdot 3} \implies 2 \sqrt{3}$$

$$\sqrt{192} \implies \sqrt{64\cdot 3} \implies 8 \sqrt{3}$$

So,

$$6 \sqrt{2}-3\cdot2 \sqrt{3}+8 \sqrt{3}$$

Carry out multiplication:

$$\implies 6 \sqrt{2}-6 \sqrt{3}+8 \sqrt{3}$$

$$\boxed{6 \sqrt{2}+2 \sqrt{3}}$$

and we're done!
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$$6\sqrt{2}+2{\sqrt3}$$

Step-by-step explanation:

Given expression:

$$\sqrt{72}-3\sqrt{12}+\sqrt{192}$$

Rewrite 72 as (36 · 2), 12 as (4 · 3), and 192 as (64 · 3):

$$\implies \sqrt{36 \cdot 2}-3\sqrt{4 \cdot 3}+\sqrt{64 \cdot 3}$$

Apply the radical rule $$\sqrt{a \cdot b}=\sqrt{a}\sqrt{b}$$ :

$$\implies \sqrt{36}\sqrt{2}-3\sqrt{4}\sqrt{3}+\sqrt{64}{\sqrt3}$$

Rewrite 36 as 6², 4 as 2², and 64 as 8²:

$$\implies \sqrt{6^2}\sqrt{2}-3\sqrt{2^2}\sqrt{3}+\sqrt{8^2}{\sqrt3}$$

Apply the radical rule $$\sqrt{a^2}=a$$ :

$$\implies 6\sqrt{2}-3\cdot 2\sqrt{3}+8{\sqrt3}$$

Simplify:

$$\implies 6\sqrt{2}-6\sqrt{3}+8{\sqrt3}$$

$$\implies 6\sqrt{2}+2{\sqrt3}$$
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