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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

[tex] \displaystyle \sqrt{72} - 3 \sqrt{12} + \sqrt{192} [/tex]

Recall that

[tex] \sqrt{ab} = \sqrt{a} \sqrt{b} , \forall \text{a and b such that a$\geq$0,b$\geq$0} [/tex]

Utilizing the formula yields,

[tex] \sqrt{72} \implies \sqrt{36 \cdot 2} \implies 6 \sqrt{2} [/tex]

[tex] \sqrt{12} \implies \sqrt{4\cdot 3} \implies 2 \sqrt{3} [/tex]

[tex] \sqrt{192} \implies \sqrt{64\cdot 3} \implies 8 \sqrt{3} [/tex]

So,

[tex]6 \sqrt{2}-3\cdot2 \sqrt{3}+8 \sqrt{3}[/tex]

Carry out multiplication:

[tex]\implies 6 \sqrt{2}-6 \sqrt{3}+8 \sqrt{3}[/tex]

Add the like terms:

[tex]\boxed{6 \sqrt{2}+2 \sqrt{3}}[/tex]

and we're done!
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[tex]6\sqrt{2}+2{\sqrt3}[/tex]

Step-by-step explanation:

Given expression:

[tex]\sqrt{72}-3\sqrt{12}+\sqrt{192}[/tex]

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

[tex]\implies \sqrt{36 \cdot 2}-3\sqrt{4 \cdot 3}+\sqrt{64 \cdot 3}[/tex]

Apply the radical rule [tex]\sqrt{a \cdot b}=\sqrt{a}\sqrt{b}[/tex] :

[tex]\implies \sqrt{36}\sqrt{2}-3\sqrt{4}\sqrt{3}+\sqrt{64}{\sqrt3}[/tex]

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

[tex]\implies \sqrt{6^2}\sqrt{2}-3\sqrt{2^2}\sqrt{3}+\sqrt{8^2}{\sqrt3}[/tex]

Apply the radical rule [tex]\sqrt{a^2}=a[/tex] :

[tex]\implies 6\sqrt{2}-3\cdot 2\sqrt{3}+8{\sqrt3}[/tex]

Simplify:

[tex]\implies 6\sqrt{2}-6\sqrt{3}+8{\sqrt3}[/tex]

[tex]\implies 6\sqrt{2}+2{\sqrt3}[/tex]
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