Question

Simplify sqrt72 -3sqrt12 + sqrt 192

Answer 1

Answer:

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

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:

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

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

Answer 2

Answer:

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

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

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:

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

Add the like terms:

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

and we're done!