why do i need both positive and negative values (completing the square)

OP

I would expand the whole (x-2)^2 out to get:

3\bigg[x^2 -8x + \frac{29}{3}\bigg]

then you'd be halving

-8

to get

3\bigg[(x-4)^2 + \cdots \bigg]

Original post by SeanFM

Neither. To me, it looks like you need to expand the brackets (x-2)^2 first.

oh no this is part of a question sorry, the original question is express 3x^2 -12x + 17 in the form a(x+b)^2 + c

i've got to this stage 3[(x-2)^2 - 4x + 17/3] but im unsure whether to half 4 or -4

what i understand is i got to make it into this form (x±b/2)^2 - (b/2)^2 + C
the b in (b/2)^2 confuses me because im not sure if b in this case is -4 or 4

Reply 21

TSR George

OP

17

Original post by Zacken

I would expand the whole (x-2)^2 out to get:

3\bigg[x^2 -8x + \frac{29}{3}\bigg]

then you'd be halving

-8

to get

3\bigg[(x-4)^2 + \cdots \bigg]

Original post by SeanFM

Neither. To me, it looks like you need to expand the brackets (x-2)^2 first.

actually im fine! thank you i just figured it out myself

Reply 22

Zacken

22

The 'b' is always the number in front of the +x. So if you have -x, this is regarded as +(-1)(x) so b=-1. Also, I think this is a good sign for you to always include the original question in the future. :tongue:

Reply 23

TSR George

OP

17

Original post by Zacken

The 'b' is always the number in front of the +x. So if you have -x, this is regarded as +(-1)(x) so b=-1. Also, I think this is a good sign for you to always include the original question in the future. :tongue:

i know, i will do that from now on, also i need to know how to type mathsy signs on here properly, i can imagine its an eye sore for everyone the way i lay it out :redface:

Reply 24

Zacken

22

No, not really. The way you lay it out is fine.

Reply 25

34908seikj

2

tfw it's 10 to 12 and you complete the square, fighting to keep your eyes open, only to find out OP no longer needs help

Feels bad man.

Reply 26

Zacken

22

Original post by 34908seikj

tfw it's 10 to 12 and you complete the square, fighting to keep your eyes open, only to find out OP no longer needs help

Feels bad man.

If it helps, it's 10 to 3 (a.m) here. :tongue:

Original post by 34908seikj

tfw it's 10 to 12 and you complete the square, fighting to keep your eyes open, only to find out OP no longer needs help

Feels bad man.

I prefer completing the triangle at 2am.

Reply 28

TSR George

OP

17

Original post by Zacken

I would expand the whole (x-2)^2 out to get:

3\bigg[x^2 -8x + \frac{29}{3}\bigg]

then you'd be halving

-8

to get

3\bigg[(x-4)^2 + \cdots \bigg]

HAI!!! just a petty question but in an expression such as 4(2-x)(2-x) how would i FOIL that because i dont know whether the 4 belongs to the first bracket or both brackets? thanks!