How Do You Complete the Square for x2+y2=2x?

[christian0710]

Hi, how do you complete this square x²+y²=2x

To get this result (x-1)²+y²=1

 

[LCKurtz]

Hi, how do you complete this square x²+y²=2x

To get this result (x-1)²+y²=0

Write it as $(x^2 - 2x\quad\quad)+y^2 = 0$ Then figure out what number you can add to both sides to fill in the blank and make the quantity in parentheses a perfect square.

 

[symbolipoint]

Hi, how do you complete this square x²+y²=2x

To get this result (x-1)²+y²=1

Using number properties to have each variable on either one side or the other, you obtain:

x²-2x=y²

The expression on the left side is not a square; you WANT a square. The expression can be factored,

x(x-2) = y².

Notice you can represent the left hand side as a rectangular area with a picture (if you wanted). One side is x long and the other side is x-2 long. Nevermind the negative or subtraction; but if you could split the "-2" in half and reposition one of those halves along the other part of the length, x, (this really needs a picture for you to see), then you would see a missing square piece on a corner. The area of this missing square piece is (2/2) by (2/2), or 1 by 1 square units. This is 1 square unit, to "complete the square".

Continuing then, add 1 to both sides of the equation:

x²-2x+1=y²+1

As I said, a picture will help to show this completion of the square. Can you finish the problem from here?
EDIT: I posted this too quickly and I believe I made a sign mistake.

 

[christian0710]

Ahh Yess If you add one one both sides it works :)
thank you!

 

[symbolipoint]

Write it as $(x^2 - 2x\quad\quad)+y^2 = 0$ Then figure out what number you can add to both sides to fill in the blank and make the quantity in parentheses a perfect square.

That's it. When you find the quantity, be sure to add it to both sides of the equation.
EDIT: Good, you found it already.

 

[christian0710]

By the way, i always feel like i get competent understandable explanations in this forum. Are some of you teachers? Or just very devoted in helping others understand?

 

[Bonaparte]

Although this already has been answered, I would just like to give something I believe is missing:
As said: x^2-2x = -y^2.
As said also, x^2-2x = x(x-2), now were looking for a perfect square, you can do any number, but easy ones will be (x-2)^2. Now we calculate (x-2)^2 = x^2-2x+4. So we need to add 4 to both sides, that is x^2-2x+4 = 4-y^2. Taking the square root (this is what we planned everything for, we made sure the left side will be a nice root):
(x-2)^2 = (2+y)(2-y), taking the square root yields:
x-2 = sqrt((2+y)(2-y)
so x = 2+ sqrt((2+y)(2-y)

Bonaparte

 

[LCKurtz]

By the way, i always feel like i get competent understandable explanations in this forum. Are some of you teachers? Or just very devoted in helping others understand?

You have some of everything: Other students, High School teachers, graduate students, and both active and retired university professors as well as working professionals.

 

[Mentallic]

As said also, x^2-2x = x(x-2), now were looking for a perfect square, you can do any number, but easy ones will be (x-2)^2. Now we calculate (x-2)^2 = x^2-2x+4.

$$(x-2)^2=x^2-4x+4$$

So we need to add 4 to both sides, that is x^2-2x+4 = 4-y^2. Taking the square root (this is what we planned everything for, we made sure the left side will be a nice root):
(x-2)^2 = (2+y)(2-y), taking the square root yields:
x-2 = sqrt((2+y)(2-y)
so x = 2+ sqrt((2+y)(2-y)

Bonaparte

These steps aren't needed because the problem was to complete the square, not to solve for x.