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completing the square

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

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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


ax2+bx+c=a(x2+bax+ca)=a((x+b2a)2b24a2+ca) \displaystyle ax^2 + bx + c = a(x^2 + \frac{b}{a}x + \frac{c}{a}) = a \left( (x+\frac{b}{2a})^2 - \frac{b^2}{4a^2} + \frac{c}{a} \right)
Original post by DylanJ42
ax2+bx+c=a(x2+bax+ca)=a((x+b2a)2b24a2+c) \displaystyle ax^2 + bx + c = a(x^2 + \frac{b}{a}x + \frac{c}{a}) = a \left( (x+\frac{b}{2a})^2 - \frac{b^2}{4a^2} + c \right)


oml that looks way confusing lol
would u be able to make up an example n substitute the values plx so that way i know what to substitute n what is actually happening
Reply 3
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

I recommend you watch a few videos on the topic and then come back to us if you're still stuck. You could try this video for example.
Reply 4
Original post by DylanJ42
ax2+bx+c=a(x2+bax+ca)=a((x+b2a)2b24a2+ca) \displaystyle ax^2 + bx + c = a(x^2 + \frac{b}{a}x + \frac{c}{a}) = a \left( (x+\frac{b}{2a})^2 - \frac{b^2}{4a^2} + \frac{c}{a} \right)

Is that really going to help someone who can't complete the square for simple quadratic expressions?
(edited 7 years ago)
Original post by coolmathametics
oml that looks way confusing lol
would u be able to make up an example n substitute the values plx so that way i know what to substitute n what is actually happening


yea sure, so something like x2+4x+7 \displaystyle x^2 + 4x + 7

to complete the square we write this as (x+2)24+7 \displaystyle (x + 2)^2 -4 + 7

this is because if you multiply out the squared bracket you get x2+2x+2x+4=x2+4x+4 \displaystyle x^2 + 2x + 2x + 4 = x^2 + 4x + 4 , however we only want x2+4x \displaystyle x^2 + 4x so we need to subtract the 4
Original post by notnek
I recommend you watch a few videos on the topic and then come back to us if you're still stuck. You could try this video for example.


i can do those ones but if the middle term is 7 or 11 then its a bit difficult :s-smilie:

Original post by notnek
Is that really going to help someone who can't complete the square for simple quadratic expressions?


lol i did i say i suck at maths but not that much, i can do simple ones but difficulty with the ones i explained above
(edited 7 years ago)
Original post by DylanJ42
yea sure, so something like x2+4x+7 \displaystyle x^2 + 4x + 7

to complete the square we write this as (x+2)24+7 \displaystyle (x + 2)^2 -4 + 7

this is because if you multiply out the squared bracket you get x2+2x+2x+4=x2+4x+4 \displaystyle x^2 + 2x + 2x + 4 = x^2 + 4x + 4 , however we only want x2+4x \displaystyle x^2 + 4x so we need to subtract the 4


what is the middle term is 11 instead
Original post by coolmathametics
what is the middle term is 11 instead


are you comfortable with the idea of comparing coefficients

eg if 3x2+4xax2+bx \displaystyle 3x^2 + 4x \equiv ax^2 +bx then a=3 and b=4

does that make sense?
can some1 help me plx
sorry if i seems desperate
Reply 10
Original post by coolmathametics
i can do those ones but if the middle term is 7 or 11 then its a bit difficult :s-smilie:

You should have mentioned that in your initial post :smile:

Can you try completing the square for x^2 - 3x + 5?

It will be useful if you show us where you get stuck so please post all your working / thoughts up to the point you get stuck.

Also, is this for A Level or GCSE? Will you be using a calculator to do these types of questions?
Original post by coolmathametics
can some1 help me plx
sorry if i seems desperate


If you have x2+bx+cx^2+bx+c then it can be written in the form (x+b2)2(b2)2+c(x+\frac{b}{2})^2-(\frac{b}{2})^2+c

So in your question you have b=3b=-3 and c=5c=5 - all it means is that an odd number being halved will give you a fraction to work with instead.

To understand why this works, you should watch a video on it and see how this is derived.
Original post by notnek
You should have mentioned that in your initial post :smile:

Can you try completing the square for x^2 - 3x + 5?

It will be useful if you show us where you get stuck so please post all your working / thoughts up to the point you get stuck.

Also, is this for A Level or GCSE? Will you be using a calculator to do these types of questions?


ooops sorry but i can do e.g. x^2 + 4x + 6
because i k 2 + 2 would be 4
so its gonna be (x+2)^2 + 2 i think :/
regards to ur question, i dunno how to start like the first step cox i can't find the half of 3x so its gonna be like 3/2 which is gonna be difficult to see how to work out the part that goes outside of the bracket... if u see what i mean

its for A level - core 1 non cal
Original post by DylanJ42
are you comfortable with the idea of comparing coefficients

eg if 3x2+4xax2+bx \displaystyle 3x^2 + 4x \equiv ax^2 +bx then a=3 and b=4

does that make sense?


i am so confused
ok so lets say ---> x^2 + 7x - 3
what i need to know is how i get half of 7 cox it cant be divided by 2
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


http://www.mathsgenie.co.uk/completing-the-square.html
Watch this ^ the guy explains it quickly and easily.
Original post by coolmathametics
i am so confused
ok so lets say ---> x^2 + 7x - 3
what i need to know is how i get half of 7 cox it cant be divided by 2


its just 72 \displaystyle \frac{7}{2} , fractions are allowed in completing the square
Original post by DylanJ42
its just 72 \displaystyle \frac{7}{2} , fractions are allowed in completing the square


alright but how am i gonna work out + q (the outside bit)
Original post by Anonymous1502
http://www.mathsgenie.co.uk/completing-the-square.html
Watch this ^ the guy explains it quickly and easily.


i can do those ones pretty easily cox there ain't any fractions in the answer *_*
Reply 18
Original post by coolmathametics
ooops sorry but i can do e.g. x^2 + 4x + 6
because i k 2 + 2 would be 4
so its gonna be (x+2)^2 + 2 i think :/
regards to ur question, i dunno how to start like the first step cox i can't find the half of 3x so its gonna be like 3/2 which is gonna be difficult to see how to work out the part that goes outside of the bracket... if u see what i mean

its for A level - core 1 non cal

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. x23x+5x^2 - 3x + 5

Start by halving 3-3 to get 32-\frac{3}{2} as you say

So you get (x32)2\left(x-\frac{3}{2}\right)^2

Then you need to subtract the square of
Unparseable latex formula:

-\frac{3}{2}\right

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 :

(x32)2(32)2+5\displaystyle \left(x-\frac{3}{2}\right)^2 - \left(-\frac{3}{2}\right)^2 + 5

=(x32)294+5\displaystyle =\left(x-\frac{3}{2}\right)^2 - \frac{9}{4} + 5


Now change the 55 so it has the same denominator as 94\frac{9}{4}

=(x32)294+204\displaystyle =\left(x-\frac{3}{2}\right)^2 - \frac{9}{4} + \frac{20}{4}

=(x32)2+114\displaystyle =\left(x-\frac{3}{2}\right)^2+\frac{11}{4}


Now see if you can follow this process for a different example. Post your working if you get stuck.
(edited 7 years ago)
Original post by coolmathametics
alright but how am i gonna work out + q (the outside bit)


as notnek said, try do an example so we can see where you get stuck.

try complete the square on x23x+5 \displaystyle x^2 - 3x + 5

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