Answer :
Answer:
D) [tex]x^2+\frac{2}{3}x+\frac{1}{9}=1[/tex]
Step-by-step explanation:
Solving the Problem
To be able to choose the correct answer, we complete and solve the equation, from there we cross reference it with the answer choices.
[tex]\dotfill[/tex]
Completing the Square
To complete a square is to make a quadratic become a perfect trinomial.
A perfect trinomial is a quadratic expression/equation where it factors to
(x - a)(x - a) or (x - a)². (I.e. x² + 4x + 4 = (x + 2)²; x² + 6x + 9 = (x + 3)²)
After verifying that the coefficient on the x² term has a value of 1, we add a c-value or constant term, which is the squared value of half of the coefficient on the x term of the given equation.
The c term can be added to both sides of the equation, or the c term and its negative equivalent are added to one side.
[tex]\hrulefill[/tex]
Solving the Equation: Completing the Square
Since there's a term on the right hand side, we subtract 9 both sides to get to the other.
[tex]9x^2+6x+ 1-9 = 0[/tex]
[tex]9x^2+6x-8=0[/tex]
(this matches choice C, so we can eliminate it)
The coefficient on the x² term is not 1, so we divide both sides by its value.
[tex]\dfrac{9x^2+6x-8}{9}=\dfrac{0}{9}[/tex]
[tex]x^2+\dfrac{2}{3}x+\dfrac{8}{9}=0[/tex]
Then, we use the (2/3) value to determine and add our constant.
[tex]c = \left(\dfrac{\dfrac{2}{3} }{2} \right)^2 = \dfrac{1}{9}[/tex]
[tex]x^2+\dfrac{2}{3}x+ \dfrac{1}{9} - \dfrac{1}{9}- \dfrac{8}{9}=0[/tex]
[tex]\left(x^2+\dfrac{2}{3}x+ \dfrac{1}{9}\right) - \dfrac{1}{9}- \dfrac{8}{9}=0[/tex]
[tex]x^2+\dfrac{2}{3}x + \dfrac{1}{9} = 1[/tex]
Thinking of two numbers that add to two-thirds and multiply to one-ninth, we get the values one-third and one-third.
[tex]x^2+\dfrac{2}{3}x + \dfrac{1}{9} = 1[/tex]
[tex]x^2+\dfrac{1}{3}x + \dfrac{1}{3}x + \dfrac{1}{9} = 1[/tex]
[tex]x\left(x+\dfrac{1}{3}\right) + \dfrac{1}{3}\left(x + \dfrac{1}{3}\right) = 1[/tex]
[tex]\left(x+ \dfrac{1}{3}\right)\left(x+ \dfrac{1}{3}\right) = 1[/tex]
[tex]\left(x+ \dfrac{1}{3}\right)^2= 1[/tex]
(this matches with choice A, so we can eliminate it)
[tex]\dotfill[/tex]
Solving the Equation: Solving for x
Now, if we wanted to solve for x, we'd start by taking the square root,
[tex]\sqrt{\left(x+ \dfrac{1}{3}\right)^2} =\sqrt1[/tex]
[tex]\left(x+ \dfrac{1}{3}\right) = \pm 1[/tex]
[tex]x=- \dfrac{1}{3} \pm 1[/tex]
(this aligns with choice B, so we eliminate it)
This leaves choice D as our final answer, usually completing the square is done when the quadratic is equal to 0, regardless we haven't seen a step/equation that looked similar to it.
