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
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
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
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.
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
to complete the square we write this as (x+2)2−4+7
this is because if you multiply out the squared bracket you get x2+2x+2x+4=x2+4x+4, however we only want x2+4x so we need to subtract the 4
You should have mentioned that in your initial post
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
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
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. x2−3x+5
Start by halving −3 to get −23 as you say
So you get (x−23)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 :
(x−23)2−(−23)2+5
=(x−23)2−49+5
Now change the 5 so it has the same denominator as 49
=(x−23)2−49+420
=(x−23)2+411
Now see if you can follow this process for a different example. Post your working if you get stuck.