Question

$\sqrt{12 + 6 \sqrt{3} } + \sqrt{12 - 6 \sqrt{3} }$
simplify the expression given above.

Answer 1

Hope this helps.....

Answer 2

Step-by-step explanation:

Given : $\sqrt{12+6\sqrt{3} } +\sqrt{12-6\sqrt{3} }$

To find : The answer of given term

Formula used : We use algebraic identities,

1. $(a+b)^2=a^2+b^2+2ab$
2. $(a-b)^2=a^2+b^2-2ab$

• Calculation for given term

We have,

$\sqrt{12+6\sqrt{3} } +\sqrt{12-6\sqrt{3} }$

for simplifying the terms we write,

$\sqrt{A}+\sqrt{B}$                   --(1)

Where $A=12+6\sqrt3$    and  $B=12-6\sqrt3$

taking A and B (which are inside the square roots) separately,

$A=12+6\sqrt{3}$

we can write A  as,

⇒  $(3)^2+(\sqrt3)^2 + 2*3*\sqrt{3}$

now we get the equation just similar to formula (1) where $a=3$ and $b=\sqrt3$ so the term is ,

$A=(3+\sqrt3)^2$

Now take B,

$B=12-6\sqrt{3}$

B can be written as,

⇒  $(3)^2+(\sqrt3)^2 - 2*3*\sqrt{3}$

it is just like the formula (2) where  $a=3$ and $b=\sqrt3$ so the term is,

$B=(3-\sqrt3)^2$

now we have the values of A and B and by putting these values in equation (1) we get,

⇒  $\sqrt{A}+\sqrt{B}$    

$A=(3+\sqrt3)^2$   and   $B=(3-\sqrt3)^2$

⇒  $\sqrt{(3+6\sqrt3)^2}+\sqrt{(3-6\sqrt3)^2}$

⇒  $3+6\sqrt3+3-6\sqrt3$

⇒  $3+3$

⇒  $6$

Hence we get the answer as 6.