SOLUTION: For this problem I must complete the square, and find the roots of the quadratic equation. x^2

SOLUTION: For this problem I must complete the square, and find the roots of the quadratic equation. x^2 - 2/3x = 0

Question 234944: For this problem I must complete the square, and find the roots of the quadratic equation.
x^2 - 2/3x = 0
Answer by jsmallt9(3758) (Show Source): You can put this solution on YOUR website!
The perfect square trinomials are:

and

Since your equation has a minus in front of 2/3x, we will be using the second form.

The form tells us that the first and third terms are perfect squares themselves. Your first term is x^2 which is obviously the square of x. (Sometimes it is not so obvious. For example, the first term might have been 9x^6. This is the square of 3x^3.) Your equation does not a third term (yet). In fact, we are in the process of trying to figure out what third term we need to make the trinomial a perfect square. Whatever third term we use, the form tells us it will have to be a perfect square.

The middle term in the form is 2ab where "a" is whatever is being squared in the first term and "b" is whatever being squared in the third term. Your middle term is 2/3x. In order to complete the square we need to figure out what "a" and "b" would make 2/3x = 2ab. We already know that your "a" is "x". So 2/3x = 2xb. We can now solve for b by dividing both sides by 2x giving:
1/3 = b

Now that we know that b must be 1/3, we also know what the third term must be. Since the third term in the form is b^2, we need (1/3)^2 = 1/9 for your third term. Now that we have the needed third term we are ready to complete the square of your equation:

Add the missing b^2 term to both sides:

Now that the left side matches our perfect square form, we can rewrite it as a perfect square. Using the "a" and "b" we determined earlier we get:

To solve this we start by finding the square root of each side:

Now we solve this absolute value equation:
or
Solving these, by adding 1/3 to each side, we get:
or

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