Question

x square - bracket open 2 root 3 + 3 Y bracket close X + 6 root 3 I is equals to zero ​

Answer 1

Answer:

x = 1z

    y = ((root3 * 4 + 1)/3)z                                                                      

(z = Natural no.)so that the value of both will be equal

see below

Step-by-step explanation:

x^2 - (2root3 + 3y)x + 6root3 = 0

x^2 - 2x root3 - 3xy + 6 root3 = 0

x^2 - 3xy = 2x root3 - 6 root3

x^2 - 3xy = root3 (2x -6)

substituting x as 1 then

1^2 -3(1)y = root3 (2(1)-6)

1 -3y = root3(-4)

1 -3y = -(root3 * 4)

-3y = -(root * 4) -1

-3y = -(root * 4 + 1)

3y = root3 * 4 + 1

y = (root3 * 4 + 1)/3              

so x = 1z

    y = ((root3 * 4 + 1)/3)z                                                                      

(z = Natural no.)so that the value of both will be equal

eg,  let z = 3/(root3 * 4 + 1) then,

then, y = (root3 * 4 + 1)/3 * 3/(root3 * 4 + 1) = 1

and  x = 1 * 3/(root3 * 4 + 1) =  3/(root3 * 4 + 1)                    ...............(1)

or z = 2

x = 1 * 2 = 2

y = (root3 * 4 + 1)/3 * 2 =  ((root3 * 4 + 1)2)/3                        .................(2)

using both (1) and (2) in the equation -- x^2 - (2root3 + 3y)x + 6root3 = 0

in both the cases the answer will be same