Question

Root 12 subtract root 18 subtract 3 root 2 sum 4 root 3

Answer 1

Answer:

The answer is -

$6\sqrt{3}$ - $6\sqrt{2}$ or 6($\sqrt{3}$ - $\sqrt{2}$)

Step-by-step explanation:

= $\sqrt{12}$ - $\sqrt{18}$ - $3\sqrt{2}$ + $4\sqrt{3}$

$\sqrt{12}$ = $\sqrt{4}$ X $\sqrt{3}$

And,

$\sqrt{4}$  = 2

$\sqrt{18}$ =  $\sqrt{9}$ X $\sqrt{2}$

And,

$\sqrt{9}$ = 3

Now,

= $\sqrt{12}$ - $\sqrt{18}$ - $3\sqrt{2}$ + $4\sqrt{3}$

=  $\sqrt{4}$ X $\sqrt{3}$ [$\sqrt{12}$ ] - $\sqrt{9}$ X $\sqrt{2}$ [$\sqrt{18}$] - $3\sqrt{2}$ + $4\sqrt{3}$

=  2[As $\sqrt{4}$ = 2] X $\sqrt{3}$ - 3[As $\sqrt{9}$ = 3] X $\sqrt{2}$ - $3\sqrt{2}$ + $4\sqrt{3}$

= 2 X $\sqrt{3}$ - 3 X $\sqrt{2}$ - $3\sqrt{2}$ + $4\sqrt{3}$

= $2\sqrt{3}$ - $3\sqrt{2}$ - $3\sqrt{2}$ + $4\sqrt{3}$

Take common by making groups

= $2\sqrt{3}$ + $4\sqrt{3}$ - $3\sqrt{2}$ - $3\sqrt{2}$

= $\sqrt{3}$(2 + 4) + $\sqrt{2}$(- 3 - 3)

= $6\sqrt{3}$ - $6\sqrt{2}$ or 6($\sqrt{3}$ - $\sqrt{2}$)[Took 6 common]