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# completing the square watch

1. (Original post by notnek)
There are different ways to think about 'completing the square' but I recommend this method when there are fractions and it's non calculator:

E.g.

Start by halving to get as you say

So you get

Then you need to subtract the square of and then add on 5 (let me know if you can't see where I'm getting these numbers from).

So you end up with :

Now change the so it has the same denominator as

Now see if you can follow this process for a different example. Post your working if you get stuck.
i think i got it

gimme an example n how do u write in that format cox idk it seems easy to follow that way :/
2. The trick is to forget the 'q' or c at the beginning. x^2 + bx + c. Cover the c for now. We start by halving b, lets call this value h. h = 1/2b. whatever sign (+/-ve) is given in the original question, you keep. in your case, you keep it (x-h)^2 as -3x is a NEGATIVE number. now we ALWAYS Subtract h^2 on to the end of what you just got, this miantains 'the square'. YOu should have (x-h)-h^2. Remember q? yh i do. now you can add c to your h^2 value as they're both outside of the brackets. (x-h)-h^2 + c. The reulting value of h^2 + c is 'q'. so Hope this didnt confuse me, i confused myself writing it
3. (Original post by DylanJ42)
as notnek said, try do an example so we can see where you get stuck.

try complete the square on
he has done that one already

i can do a different one but i wanna know how to write in that format so its easy to write
4. (Original post by coolmathametics)
i think i got it

gimme an example n how do u write in that format cox idk it seems easy to follow that way :/
First have a go at which is similar to the last example. Please post your answer or show your working if you get stuck.

I typed it using LaTeX but you don't need to use that - just make sure your working is clear when you post so use brackets

e.g. (x-(3/2))^2 - (9/4) + 5
5. (Original post by notnek)
First have a go at which is similar to the last example. Please post your answer or show your working if you get stuck.

I typed it using LaTeX but you don't need to use that - just make sure your working is clear when you post so use brackets

e.g. (x-(3/2))^2 - (9/4) + 5
ok

(x - 5/2)^2 - 25/4 + 12

(x - 5/2)^2 - 25/4 + 48/4

(x - 5/2)^2 + 23/4

is that right :/
6. (Original post by coolmathametics)
ok

(x - 5/2)^2 - 25/4 + 12

(x - 5/2)^2 - 25/4 + 48/4

(x - 5/2)^2 + 23/4

is that right :/
Correct

Now try and then . Change the decimal to a fraction for the second one.

If you can do them then I think you're done. The topic does get harder when you have something like but that might be for another day!
7. (Original post by coolmathametics)
x^2 - 3x + 5 in the form (x - p)^2 + q
can anyone plx explain how do i do completing the square because i dont know if there is a method to do them
i suck at maths lol
A lot of replies!

We've three tips for completing the square. We are going to make your example x^2 - 3x - 5 as it is a simpler case.

The first tip is to divide through by a whenever a is greater than 1 in ax^2 + bx + c. Not a problem in your example since a = 1 so something to remember for the future.

Working with the example, the second tip is to write x^2 - 3x = 5 and IMMEDIATELY add 9/4 to both sides ie (b/2a)^2. So you have x^2 - 3x + 9/4 = 7.25

Then 'complete the square' : (x - 3/2)^2 = 7.25 (on the LHS - 9/4 + 9/4 cancels out). Then you can square-root both sides:

x - 3/2 = SQRT 7.25

And so x = 3/2 +/- SQRT 7.25

Our third tip is always to make assure you have plus AND minus the square root since it is easy to forget to do this and lose marks.

Lastly, examiners like to ask you to write down the value of x that minimises or maximises the function of x. This is to test that you know one of the useful features of writing ax^2 + bx + c in the form p(x + q)^2 + r.

The very useful thing to know is that if a and p are greater than zero, the function is U-shaped and so has a minimum. Conversely if a and p are less than zero, the function is an upside-down U and so has a maximum. The value of x that gives you the minima or maxima is the one that makes the brackets bit equal zero. So in the example, x = 3/2 gives a minima.
8. (Original post by notnek)
Correct

Now try and then . Change the decimal to a fraction for the second one.

If you can do them then I think you're done. The topic does get harder when you have something like but that might be for another day!
ok so the first one...

(x - 9/2)^2 - 81/4 - 15

(x - 9/2)^2 - 81/4 - 60/4

(x - 9/2)^2 - 141/4 --------> (i feel like its wrong because i end up with a huge number :/ )

2nd one...
(x - 13/2)^2 - 169/4 - 9/2 (45/10 simplifies to 9/2)

(x - 13/2)^2 - 169/4 - 36/4

(x - 13/2)^2 - 205/4

i think both r wrong O_O

A lot of replies!

We've three tips for completing the square. We are going to make your example x^2 - 3x - 5 as it is a simpler case.

The first tip is to divide through by a whenever a is greater than 1 in ax^2 + bx + c. Not a problem in your example since a = 1 so something to remember for the future.

Working with the example, the second tip is to write x^2 - 3x = 5 and IMMEDIATELY add 9/4 to both sides ie (b/2a)^2. So you have x^2 - 3x + 9/4 = 7.25

Then 'complete the square' : (x - 3/2)^2 = 7.25 (on the LHS - 9/4 + 9/4 cancels out). Then you can square-root both sides:

x - 3/2 = SQRT 7.25

And so x = 3/2 +/- SQRT 7.25

Our third tip is always to make assure you have plus AND minus the square root since it is easy to forget to do this and lose marks.

Lastly, examiners like to ask you to write down the value of x that minimises or maximises the function of x. This is to test that you know one of the useful features of writing ax^2 + bx + c in the form p(x + q)^2 + r.

The very useful thing to know is that if a and p are greater than zero, the function is U-shaped and so has a minimum. Conversely if a and p are less than zero, the function is an upside-down U and so has a maximum. The value of x that gives you the minima or maxima is the one that makes the brackets bit equal zero. So in the example, x = 3/2 gives a minima.

thnx for the tips
...
Working with the example, the second tip is to write x^2 - 3x = 5 and IMMEDIATELY add 9/4 to both sides ie (b/2a)^2. So you have x^2 - 3x + 9/4 = 7.25

Then 'complete the square' : (x - 3/2)^2 = 7.25 (on the LHS - 9/4 + 9/4 cancels out). Then you can square-root both sides:

x - 3/2 = SQRT 7.25

And so x = 3/2 +/- SQRT 7.25
...
That's only useful if you're solving though, he/she is just being asked to represent in a different form, where is a quadratic.
10. (Original post by coolmathametics)
ok so the first one...

(x - 9/2)^2 - 81/4 - 15

(x - 9/2)^2 - 81/4 - 60/4

(x - 9/2)^2 - 141/4 --------> (i feel like its wrong because i end up with a huge number :/ )

2nd one...
(x - 13/2)^2 - 169/4 - 9/2 (45/10 simplifies to 9/2)

(x - 13/2)^2 - 169/4 - 16/4

(x - 13/2)^2 - 175/4

i think both r wrong O_O

thnx for the tips
You're very close in both but they both need slight corrections.

For the first one notice that the quadratic I gave you contained instead of so you need to change one of the signs in your answer.

For the second one you said 9/2 is the same as 16/4. Check that again.
11. (Original post by notnek)
You're very close in both but they both need slight corrections.

For the first one notice that the quadratic I gave you contained instead of so you need to change one of the signs in your answer.

For the second one you said 9/2 is the same as 16/4. Check that again.
oh yeah its +9 i always miss the signs so its...
(x + 9/2)^2 - 141/4 ?

i did update my 2nd one again but maybe u didnt got that yet :/
i'll post it here again

2nd one...
(x - 13/2)^2 - 169/4 - 9/2 (45/10 simplifies to 9/2)

(x - 13/2)^2 - 169/4 - 36/4

(x - 13/2)^2 - 205/4
12. (Original post by coolmathametics)
oh yeah its +9 i always miss the signs so its...
(x + 9/2)^2 - 141/4 ?

i did update my 2nd one again but maybe u didnt got that yet :/
i'll post it here again

2nd one...
(x - 13/2)^2 - 169/4 - 9/2 (45/10 simplifies to 9/2)

(x - 13/2)^2 - 169/4 - 36/4

(x - 13/2)^2 - 205/4
The first one is now correct but for the second one 9/2 is not equal to 36/4

I have to go now but hopefully someone else will continue to help tonight if you need it or want to try harder examples. RDKGames is always helpful if he's free.
13. (Original post by notnek)
The first one is now correct but for the second one 9/2 is not equal to 36/4

I have to go now but hopefully someone else will continue to help tonight if you need it or want to try harder examples. RDKGames is always helpful if he's free.
oml i am so stressing out why am i not getting it right arghhh

ok so

the qs is x^2 -13x - 4.5

solution:

(x - 13/2)^2 - 169/4 - 9/2

x - 13/2)^2 - _________ (i am not getting that bit right i think.... do i times 9/2 by 2 to get 18/4???)

RDKGames plx be easy on me
14. (Original post by coolmathametics)
oml i am so stressing out why am i not getting it right arghhh

ok so

the qs is x^2 -13x - 4.5

solution:

(x - 13/2)^2 - 169/4 - 9/2

x - 13/2)^2 - _________ (i am not getting that bit right i think.... do i times 9/2 by 2 to get 18/4???)

RDKGames plx be easy on me
Yeah that's right
15. (Original post by RDKGames)
Yeah that's right
thanks uuuuuuuuu

and a huge thankks to notnek for helping me tremendously n sorting half of my maths out lol
THANKS

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