ASAP! I really need help with this question! No nonsense answers, and please attach the solution.

Step-by-step explanation:

Subtract x from both sides.

Square both sides.

Subtract x²-6x+9 from both sides.

Factor left side of the equation.

Set factors equal to 0.

Check if the solutions are extraneous or not.

Plug x as 2.

x = 2 works in the equation.

Plug x as 6.

x = 6 does not work in the equation.

option d

Step-by-step explanation:

Square both sides

  {Add like terms}

Sum = -8

Product = 12

Factors =  -2 , - 6

x² - 2x - 6x + (-2) * (-6) = 0

x(x -2) - 6(x -2) = 0

(x -2) (x - 6) = 0

x - 2 =0   ; x - 6 = 0

x = 2      ; x = 6

roots of the equation : 2  , 6

But when we put x = 6, it doesn't satisfies the equation.

When x = 6,

3 + 6≠ 3

Therefore, x = 2  but x = 6 is extraneous

it would be 6xy and -16xy.

step-by-step explanation:

the shape,the size and the distance