Question

The lesson is about completing the square. help.
Solve: x^2 - 4/3x = 4/9 = 0

Answer 1

Do you have to answer is fractions?

Answer 2

You have two equal signs.

Answer 3

should read - 4/9 not = 4/9

Answer 4

\(x^2 - \dfrac{4}{3}x - \dfrac{4}{9} = 0\)

Answer 5

First step, move constant term to right side.
Add 4/9 to both sides.

Answer 6

\(x^2 - \dfrac{4}{3}x - \dfrac{4}{9} = 0\)

\(x^2 - \dfrac{4}{3}x ~~~~~~~~~~~= \dfrac{4}{9}\)

Answer 7

Since the coefficient of the x^2 term is one, we can do the next step.

Answer 8

Take the coefficient of the x-term, -4/3. Divide it by 2. Then square it. Then add it to both sides.

Answer 9

Bring everything over to one side
assuming the 1st equals is a plus.....
x^2 - 4/3x + 4/9 = 0

Add and subtract the quantity (b/2)^2 or ((4/3)/2)^2
x^2 - 4/3x - 4/9 + 4/9 - 4/9 = 0

Then you can make a perfect square with the
"x^2 - 4/3x + 4/9" -4/9 - 4/9 = 0

Making the perfect square you take the "b" term and divide it by two and make that the constant term inside the quantity squared.
(x-2/3)^2 - 8/9 = 0

Answer 10

\(x^2 - \dfrac{4}{3}x ~~~~~~~~~~~= \dfrac{4}{9}\)

Half of -4/3 is -4/6 = -2/3.
Squaring -2/3 = 4/9
Now we add 4/9 to both sides.

\(x^2 - \dfrac{4}{3}x + \dfrac{4}{9} = \dfrac{4}{9} + \dfrac{4}{9}\)

The left side is now the square of a binomial:

\( \left( x - \dfrac{2}{3} \right)^2 = \dfrac{8}{9} \)

Now you can solve the equation by taking the square root of both sides.
Remember that if
\(x^2 = a\), then
\( x = \pm \sqrt{a} \)