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# Completing the square

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1. How do you do it?.
2. (Original post by BTREDinnick)
How do you do it?.
Most questions will be in the general form
ax^2+bx+c
for basic questions in normal maths where there the coefficient of x^2 is 1 you do this;

e.g. x^2+4x+5

1. half the b value and re-write like this
so 4/2 is 2 and this is what you put inside the bracket.

(x+2)^2 +5

2. Then you take the 2 inside the bracket, square it and multiply it by -1 and put it outside the bracket. 2^2 = 4 x -1 = =4
like,
(x+2)^2 -4 + 5

You can do the first two steps all at once to save yourself time.

3. Tidy up the 'normal' numbers at the end so -4+5 = 1

so at the end it will look like this

(x+2)^2 +1

For when there is a coefficient of x^2 which is larger than 1, you do this
Question : 2x^2 + 8x +9
1. Factorise the first two terms, but DO NOT TAKE OUT ANY X'S!
2(x^2 + 4x) + 9

2. Complete the square on the middle bit like before and multiply the number inside by itself and then -1 again.
You will need 2 sets of brackets
2((x^2 + 2) -4) +9

3. Expand the first bracket
The minus 8 is from the 2 x -4
2(x^2 + 2) -8 +9

4. And tidy tidy,

2(x^2 + 2) + 1
3. Thanks. That helps me alot.
4. (Original post by zara_ruby)
Most questions will be in the general form
ax^2+bx+c
for basic questions in normal maths where there the coefficient of x^2 is 1 you do this;

e.g. x^2+4x+5

1. half the b value and re-write like this
so 4/2 is 2 and this is what you put inside the bracket.

(x+2)^2 +5

2. Then you take the 2 inside the bracket, square it and multiply it by -1 and put it outside the bracket. 2^2 = 4 x -1 = =4
like,
(x+2)^2 -4 + 5

You can do the first two steps all at once to save yourself time.

3. Tidy up the 'normal' numbers at the end so -4+5 = 1

so at the end it will look like this

(x+2)^2 +1

For when there is a coefficient of x^2 which is larger than 1, you do this
Question : 2x^2 + 8x +9
1. Factorise the first two terms, but DO NOT TAKE OUT ANY X'S!
2(x^2 + 4x)^2 + 9

2. Complete the square on the middle bit like before and multiply the number inside by itself and then -1 again.
You will need 2 sets of brackets
2((x^2 + 2)^2 -4) +9

3. Expand the first bracket
The minus 8 is from the 2 x -4
2(x^2 + 2)^2 -8 +9

4. And tidy tidy,

2(x^2 + 2)^2 + 1
I forgot to put some of the x^2's outside the brackets!
5. (Original post by zara_ruby)
Most questions will be in the general form
ax^2+bx+c
for basic questions in normal maths where there the coefficient of x^2 is 1 you do this;

e.g. x^2+4x+5

1. half the b value and re-write like this
so 4/2 is 2 and this is what you put inside the bracket.

(x+2)^2 +5

2. Then you take the 2 inside the bracket, square it and multiply it by -1 and put it outside the bracket. 2^2 = 4 x -1 = =4
like,
(x+2)^2 -4 + 5

You can do the first two steps all at once to save yourself time.

3. Tidy up the 'normal' numbers at the end so -4+5 = 1

so at the end it will look like this

(x+2)^2 +1

For when there is a coefficient of x^2 which is larger than 1, you do this
Question : 2x^2 + 8x +9
1. Factorise the first two terms, but DO NOT TAKE OUT ANY X'S!
2(x^2 + 4x) + 9

2. Complete the square on the middle bit like before and multiply the number inside by itself and then -1 again.
You will need 2 sets of brackets
2((x + 2)^2 -4) +9

3. Expand the first bracket
The minus 8 is from the 2 x -4
2(x + 2)^2 -8 +9

4. And tidy tidy,

2(x + 2)^2 + 1
6. Since you've been helped already I'll be facetious

7. (Original post by 13 1 20 8 42)
Since you've been helped already I'll be facetious

Knew it'd be you

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