Completing the square

  • Thread starter Miike012
  • Start date
  • #1
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Problem: -x^2 + 4x

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

Vertex is at (2, -4)

but the vertex is obviously at (2,4)...
what did I do wrong.
 

Answers and Replies

  • #2
34,167
5,782
Problem: -x^2 + 4x

- ( x^2 - 4x + (4/2)^2 ) - 4
The error is here. You have actually added -4 inside the parentheses to complete the square, so to keep the expression equal, you need to add + 4.
= - (x - 2) - 4

Vertex is at (2, -4)

but the vertex is obviously at (2,4)...
what did I do wrong.
 
  • #3
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how did I add -4... then the expression would be x^2 - 4x - 4 which is not a perfect square...? Does it have something to do with x^2 being negative?
 
  • #4
34,167
5,782
Let's go through the steps.

-x2 + 4x
= -(x2 - 4x)
= -(x2 - 4x + 4) + 4
I put in +4 inside the parentheses, but due to the minus sign out front, I have actually added -4, so to balance, I have to add + 4.

= -(x - 2)2 + 4
 
  • #5
HallsofIvy
Science Advisor
Homework Helper
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Another way of looking at it is to do both addition and subtraction inside the parentheses:
[tex]-(x^2- 4x)= -(x^2- 4x+ 4- 4)= -(x^2- 4x+ 4)-(-4)= -(x- 2)^2+ 4[/tex]
 
  • #6
PeterO
Homework Helper
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Let's go through the steps.

-x2 + 4x
= -(x2 - 4x)
= -(x2 - 4x + 4) + 4
I put in +4 inside the parentheses, but due to the minus sign out front, I have actually added -4, so to balance, I have to add + 4.

= -(x - 2)2 + 4
or if you like

-x2 + 4x
= -[x2 - 4x]
= -[x2 - 4x + 4 - 4]
= -[(x - 2)2 - 4]
= -(x - 2)2 + 4
 

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