The Student Room Group

Quadratic Query

Hi,

I'm just wondering when is it appropriate to use the complete the square method or the factorisation method where it comes to this for example

x26x+8x^2-6x+8
Reply 1
If b^2-4ac is a perfect square, then that means that the equation y=ax^2+bx+c has rational roots (which means that you can factorize). In this case, b^2-4ac=4, which is a perfect square. If it's not a perfect square, it doesn't have rational roots, which means you it will be extremely difficult to use "the factorisation method."

I usually avoid completing the square or using the quadratic formula, but if you can't see the -2 * -4 = 8 and -2 + -4 = -6, then I think you should give up and complete the square.

In the end, it's all about speed.
(edited 11 years ago)
Reply 2
This one I would factorise.
Reply 3
If you are checking b^2-4ac then you might as well use the formula.
Reply 4
Original post by zed963
Hi,

I'm just wondering when is it appropriate to use the complete the square method or the factorisation method where it comes to this for example

x26x+8x^2-6x+8


You should always be looking for factors in maths. It'll help you incredibly as you move into more advanced algebraic situations.

You should easily be able to spot the factors:
y=(x4)(x2)y = (x - 4)(x - 2)

However, of course you can complete the square:

y=(x3)21y = (x - 3)^2 - 1
Reply 5
Of course, if you're asked for a minimum or maximum value then complete the square.
Reply 6
Original post by BabyMaths
If you are checking b^2-4ac then you might as well use the formula.


But writing (x-6/2±sqrt(4)/2)=(x-3±1)=(x-2)(x-4) is just so...weird! xP
(edited 11 years ago)
Reply 7
Original post by aznkid66
But writing (x-6/2±sqrt(4)/2)=(x-3±1)=(x-2)(x-4) is just so...weird! xP


Yes, that is weird.
You should be able to factorise that one easy.

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