Question

simplify root 2 (root 6 - root 18)+root3 (root 27-root6)+3 root 2

Answer 1

I hope it is not wrong and helps you

Answer 2

Answer:

$2\sqrt{3}+3$

Step-by-step explanation:

As per the given question,

We have to simplify the given equation,

$\sqrt{2}(\sqrt{6}-\sqrt{18})+\sqrt{3}(\sqrt{27}-\sqrt{6})+3\sqrt{2}$

We have to find the value of this in the most simplified form,

On taking the common from the terms, we get,

$=\sqrt{2}(\sqrt{6}-\sqrt{18})+\sqrt{3}(\sqrt{27}-\sqrt{6})+3\sqrt{2}\\=\sqrt{2}\times \sqrt{2}(\sqrt{3}-\sqrt{9})+\sqrt{3}\times \sqrt{3}(\sqrt{9}-\sqrt{2})+3\sqrt{2}\\=2(\sqrt{3}-3)+3(3-\sqrt{2})+3\sqrt{2}\\=2(\sqrt{3}-3)+9\\=2\sqrt{3}-6+9\\=2\sqrt{3}+3$

Now, the final simplified product can be stated as,

$2\sqrt{3}+3=\sqrt{3}(2+\sqrt{3})$

So,

Now the equation is in its most simplified form.

Therefore, the simplified or required value of the given equation is given as,

$2\sqrt{3}+3=\sqrt{3}(2+\sqrt{3})$