complete the square for the equation
x^2-3x+____
how do I figure out what goes there
well..I can't remember the formular for that one. But, you could figure it out by looking at equation.
x^2-3x+(?)
if you have to factor the equation, it will be like this: (x- ?) (x-?)
Because of the middle number is -3x and the (?) is positive, it has to be (-) something * (-) something. When you add those two, you have to get -3x. IF you multiply them, you 'll get positve (+) number.
So,
(x-2) (x-1)= x^2-x-2x+2
=x^2-3x+2
Sorry, but that is not correct
The question was to complete the square, so the first and last terms have to be perfect squares.
in general if the expression starts with 1x^2, take onehalf of the middle term coefficient and square that.
so 3/2 squared is 9/4
then x^2-3x+9/4 = (x-3/2)^2
Yea..I remembered now. Thanks 4 correcting !
To complete the square for the equation x^2 - 3x + ____, you need to follow these steps:
Step 1: Take the coefficient of the x-term, which is -3, and divide it by 2. Square the result.
- (-3 / 2) = -3/2
- (-3/2)^2 = 9/4
Step 2: Add the number you obtained in Step 1 to the expression, but also subtract the same number outside the parentheses to maintain the equation's balance.
- x^2 - 3x + 9/4 - 9/4
Step 3: Simplify the expression inside the parentheses.
- x^2 - 3x + (9/4) - (9/4)
Step 4: Combine like terms.
- x^2 - 3x + (9/4) - (9/4)
- x^2 - 3x + (9/4) - (9/4)
- x^2 - 3x + 1/2
So, the completed square for the equation x^2 - 3x is x^2 - 3x + (9/4) - (9/4), or simplified further, x^2 - 3x + (1/2).