Completing the square / Changing format of equation

  • Thread starter Thread starter danielle36
  • Start date Start date
  • Tags Tags
    Format Square
AI Thread Summary
To rewrite equations in the form y = a(x - p)² + q using the method of completing the square, start by rearranging the quadratic equation. For y = x² + 6x, complete the square by adding and subtracting 9, resulting in y = (x + 3)² - 9. For the equation y = -2x² + 8x + 5, factor out -2 to simplify, then complete the square for the expression inside the parentheses. This involves determining the necessary value to add and subtract to maintain equality, ultimately leading to the desired format. Understanding these steps will clarify the process of converting quadratic equations into vertex form.
danielle36
Messages
29
Reaction score
0
I'm supposed to use the method of completeing the square to write an equation in the form y = a(x - p)^{2} + q.

Here's one of the equations: y = x^{2} + 6x
and another: y = -2x^{2} + 8x + 5

I really don't know where to start, which is why I can't include any work... The main problem here is I don't know how to get the equation in the (x - p)^{2} format. I can complete the square but its the - sign that's throwing me off.
If someone could just explain how to go about doing this it would be greatly appreciated.
 
Physics news on Phys.org
danielle36 said:
I'm supposed to use the method of completeing the square to write an equation in the form y = a(x - p)^{2} + q.

Here's one of the equations: y = x^{2} + 6x
and another: y = -2x^{2} + 8x + 5

I really don't know where to start, which is why I can't include any work... The main problem here is I don't know how to get the equation in the (x - p)^{2} format. I can complete the square but its the - sign that's throwing me off.
If someone could just explain how to go about doing this it would be greatly appreciated.

theres a completeing the square formula but the first one is very obvious so i'll do it for you

it'll be easier to see if you rewrite this
y = (ax^2-2xp +p^2) -p^2
then your first problem is

y = x^{2} + 6x + 9 - 9 = (x^2 +3)^2 - 9
 
danielle36,

Other messages have mentioned this; but if you could find a graphical representation of Completing The Square, you will probably understand the process very well. The idea of converting the equation from general form into standard form will require subtracting the "piece" that you add in order for the expression to be equivalent to the original expression.
 
ice109 said:
y = x^{2} + 6x + 9 - 9 = (x^2 +3)^2 - 9

...just fixing a typo. the last x should not be squared on the far RHS
<br /> (x+3)^2-9<br />
 
(x-a)2= x2- 2ax+ a2

If you want to change x2- 6x to a perfect square what must "a" be? You know that -6x= -2ax.

-2x2+ 8x+ 5 is a little harder. Write it as -2(x2- 4x)+ 5. Now look just at the x2- 4x. Remembering that -2ax= -4x, what is a? What is a2? Add and subtract a2 inside the parentheses and then take part outside the parentheses.
 
Last edited by a moderator:
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...

Similar threads

Replies
13
Views
2K
Replies
5
Views
2K
Replies
6
Views
2K
Replies
11
Views
2K
Replies
11
Views
2K
Replies
8
Views
2K
Back
Top