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hi:) for completing the square , if there’s a negative sign in the bracket with x term (x-2/2)^2 , do I need to put negative in the bracket for (2/2)^2 too?

hi:) for completing the square , if there’s a negative sign in the bracket with x-example-1

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Answer:

Hi there! I don't really understand what you are trying to ask but when I looked at your working for completing the square for x²-2x+3, it looks correct.

Instead of memorizing formulas, I suggest that you understand the process of completing the square.

Below is the steps for completing the squares.

1. Ensure that the coefficient of x² is 1 ✓

2. Notice that the coefficient of the x term here is negative, so what we are trying to achieve is a² -2ab +b², which you could convert to (a-b)² later.

3. Find the values of a and b.

Since your coefficient of x² is a 1, the value of a in a²-2ab+b² is x.

➣x² -2x+3= x² -2(x) ....

4. Now find the value of b in a²-2ab+b².

By observation, you would see that b=1 since when you focus at the x term,

-2x= -2ab

-2x = -2(x)(b)

-2x = -2x (b)

b= 1

➣ So, x²-2x+3= x²-2(x)(1) +1²...

Now here you notice that the constant term is not equal. On the left hand side you have 3 and the right hand side you only have 1 since 1² =1.

So add 2 to the right hand side since 1²+2 would give you 3.

➣ Thus, x²-2x+3

= x²-2(x)(1)+1² +2

= (x-1)² +2

Underlined: since a²-2ab+b²= (a-b)²

And you have completed the square!

And if you were asking what to do if there's a negative sign at the coefficient of the x² term, factorise the negative sign out at the first step.

For example,

-x² +2x -3

= -(x² -2x +3)

= -[(x-1)² +2]

= -(x-1)² -2

But of course if the quadratic equation equals to zero, then divide throughout by -1.

If the coefficient of the x term is positive, then what we are looking at at step 2 is a² +2ab +b²= (a+b)²

Do feel free to ask if you still have any questions :)

User MrVasilev
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