Question

Find the roots of 4x^2+4√3x -6 by
Using completing Square method.​

Answer 1

4x^2+4√3x -6 <-------------------Question

BTW bro thats a very basic one you need a lot of practice if you cant getting it :(

=> (2x)^2 + 2(2x * $\sqrt{3}$) + ($\sqrt{3}$)^2 - ($\sqrt{3}$)^2 = 6

in above line I used identity of 8th class that is (a+b)^2 = a^2 + 2ab + b^2

focus in middle term of question 4√3x we know from the above identity that it is equal to 2ab

we know a which is $\sqrt{4x^2}$ yeah 1st term = 2x

there fore,

b =  $\sqrt{3}$

by following identity we can add b^2 so  $\sqrt{3}$^2 and - of it because we modify question so to nullify it :D

solv the 3rd line eqn

(2x)^2 + 2(2x * $\sqrt{3}$) + ($\sqrt{3}$)^2 - ($\sqrt{3}$)^2 = 6

(2x + $\sqrt{3}$)^2 -  ($\sqrt{3}$)^2 = 6

(2x + $\sqrt{3}$)^2 = 9

2x + $\sqrt{3}$ = $\sqrt{9}$

2x + $\sqrt{3}$ =± 3 <-------------because square of +3 = 9 and -3 = 9 also :D

we can separate this

2x + $\sqrt{3}$ = 3  ,  2x + $\sqrt{3}$ = -3

2x = 3-$\sqrt{3}$     ,  2x = -3-$\sqrt{3}$

x = 3-$\sqrt{3}$       ,  x = -3-$\sqrt{3}$

   ----------              ---------

       2                       2

hence you got the answer :)

hope you understand square method :D