SOLUTION: i dont understand how to complete the square on y=.5x^2-3x+19/2. i also have to graph this parabola and i dont know how

Question 134123This question is from textbook Merill Algebra 2 with Trigonometry Aplications and Connections
: i dont understand how to complete the square on y=.5x^2-3x+19/2. i also have to graph this parabola and i dont know how This question is from textbook Merill Algebra 2 with Trigonometry Aplications and Connections

Found 2 solutions by ankor@dixie-net.com, josmiceli:
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
i dont understand how to complete the square on y=.5x^2-3x+19/2. i also have to graph this parabola and i dont know how
:
It's easier to complete the square if the coefficient of x^2 is 1
Multiply equation by 2 to get that. I also gets rid of the fraction, you have:
x^2 - 6x + 19 = 0
We want to have the third term value make it a perfect square, so we:
x^2 - 6x + ___ = -19; subtracted 19 from both sides
:
Find the values of the third term, divide the coefficient of x by 2, and square:
6/2 = 3, square 3 and we have 9, we have to add 9 to both sides.
:
x^2 - 6x + 9 = -19 + 9; (x^2-6x+9 is a perfect square)
:
(x - 3)^2 = -10
:
Take the square root of both sides:
x - 3 = +/-
:
We can't have a neg square root, use i, the square root of -1
x - 3 = +/-
:
x = 3 +/-; add 3 to both sides
:
The two solutions (this equation does not have any real roots)
x = 3 + ;
and
x = 3 - ;
:
;
To graph this, assign values to x and find y, a table of values would look like this:
x | Y
-------
-3 | 23
-2 | 17.5
-1 | 13
0 | 9.5
+1 | 7
+2 | 5.5
+3 | 5
+4 | 5.5
+5 | 7
+6 | 9.5
+7 | 13
You should be able to find +8 and +9, as you can see, it is a parabola
Answer by josmiceli(19441) (Show Source): You can put this solution on YOUR website!
To find the roots, set
You first have to multiply both sides by in
order to get rid of the in front of the term

The rule is, take 1/2 of the coefficient of the x-term, then
square it, then add the result to both sides

Now subtract from both sides

Right away, I see that something squared is supposed
to be negative, so the "something", which is ,
must be an imaginary quantity
Take the square root of both sides

There are 2 answers, and
check answer:

OK
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