# Completing The Square in Form (2x - a)^2 - b

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#1
Hi! I've got a level three algebra exam tomorrow and was doing some practice papers when I stumbled across a completing the square question (for the function 4x^2 - 28x) that asks you to factorise in the form (2x - a)^2 - b

I know how to deal with a coefficient of x^2 that is larger than one normally (factorise it out) but not sure when I'm being asked for the 2x inside the brackets.

I can see how you work this out with a clumsy method of finding a number that when multiplied by 2x and then multiplied by 2 gives -28x (which would be -7) and then finding that (-7)^2 = 49 so you need to take this away to give

(2x - 7)^2 - 49

but is there a more logical way to do this? Thank you!!
0
4 months ago
#2
You could expand (2x-a)²-b and then compare coefficients for a more algebraically grounded way of doing things
1
4 months ago
#3
(Original post by apolaroidofus)
Hi! I've got a level three algebra exam tomorrow and was doing some practice papers when I stumbled across a completing the square question (for the function 4x^2 - 28x) that asks you to factorise in the form (2x - a)^2 - b

I know how to deal with a coefficient of x^2 that is larger than one normally (factorise it out) but not sure when I'm being asked for the 2x inside the brackets.

I can see how you work this out with a clumsy method of finding a number that when multiplied by 2x and then multiplied by 2 gives -28x (which would be -7) and then finding that (-7)^2 = 49 so you need to take this away to give

(2x - 7)^2 - 49

but is there a more logical way to do this? Thank you!!
As well as the previous good idea, do what you've been doing
4(x^2 - 7x)
= 2^2*(x - 7/2)^2 - ...
= (2x - 7)^2 - ...
1
#4
(Original post by 3pointonefour)
You could expand (2x-a)²-b and then compare coefficients for a more algebraically grounded way of doing things
Ah that's brilliant thank you!!
1
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