Answer:
[tex]\text{D)}\ \ x=\dfrac{2\pm i\sqrt{2}}{3}[/tex]
Step-by-step explanation:
We can solve for x in the quadratic equation:
[tex]9x^2+6=12x[/tex]
by completing the square.
First, get the x-terms on one side.
[tex]9x^2+6=12x[/tex]
↓ subtracting 12x from both sides
[tex]9x^2-12x+6=0[/tex]
↓ subtracting 6 from both sides
[tex]9x^2-12x=-6[/tex]
Next, we can divide both sides by 9 to get the x-squared term's coefficient to be 1:
[tex]\dfrac{9x^2-12x}{9}=\dfrac{-6}{9}[/tex]
[tex]x^2-\dfrac{4}{3}x = -\dfrac{2}{3}[/tex]
Then, we can add half the x-term's coefficient, squared, to both sides:
[tex]x^2-\dfrac{4}{3}x+\left(\dfrac{-4/2}{3}\right)^{\!\!2} = -\dfrac{2}{3} + \left(\dfrac{-4/2}{3}\right)^{\!\!2}[/tex]
[tex]x^2-\dfrac{4}{3}x+\left(\dfrac{-2}{3}\right)^{\!\!2} = -\dfrac{2}{3} + \left(\dfrac{-2}{3}\right)^{\!\!2}[/tex]
[tex]x^2-\dfrac{4}{3}x+\dfrac{4}{9} = -\dfrac{2}{3} + \dfrac{4}{9}[/tex]
[tex]x^2-\dfrac{4}{3}x+\dfrac{4}{9} = -\dfrac{2}{9}[/tex]
This makes it so we can factor the left side as a perfect square:
[tex]\left(x-\dfrac{4/2}{3}\right)^{\!\!2} = -\dfrac{2}{9}[/tex]
[tex]\left(x-\dfrac{2}{3}\right)^{\!\!2} = -\dfrac{2}{9}[/tex]
Finally, we can solve for x by taking the square root of both sides:
[tex]\sqrt{\left(x-\dfrac{2}{3}\right)^{\!\!2}} = \sqrt{-\dfrac{2}{9}}[/tex]
[tex]x-\dfrac{2}{3} =\pm\dfrac{\sqrt{-2}}{3}[/tex]
↓ taking the square root of -1 ... [tex]\sqrt{-1}=i[/tex]
[tex]x-\dfrac{2}{3} = \pm\dfrac{i\sqrt{2}}{3}[/tex]
↓ adding 2/3 to both sides
[tex]\boxed{x=\dfrac{2\pm i\sqrt{2}}{3}}[/tex]
We can identify this as option D.
