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Question regarding quadratic equation roots

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I know, I'm sorry lol. I keep doing that, I just don't understand how you factorise it.

EDIT: My friend kind of helped me out here.

x-1[(x-1)^2-(1+x)] = 0[br]x-1[x^2-2x+1-(1+x)] = 0[br]x-1[x^2-3x]=0[br](x-1)(x)(x-3) = 0

(edited 8 years ago)

Reply 21

Kevin De Bruyne

Original post by Chittesh14

I know, I'm sorry lol. I keep doing that, I just don't understand how you factorise it.

(x-1)^3 - (x-1)(x+1) = 0.

If we call (x-1) a and (x+1) b then you have a^3 - ab = 0.

So how do you factorise that?

Spoiler

Then put it back in then you have (x-1)((x-1)^2 - (x+1)) ... and then all of the lines from your V2 follow.

Reply 22

Chittesh14

Original post by SeanFM

(x-1)^3 - (x-1)(x+1) = 0.

If we call (x-1) a and (x+1) b then you have a^3 - ab = 0.

So how do you factorise that?

Spoiler

Then put it back in then you have (x-1)((x-1)^2 - (x+1)) ... and then all of the lines from your V2 follow.

I've got it, thanks buddy!
I don't know why, sometimes, my brain just can't do this... lol
Sometimes, I just forget how to factorise equations which are raised to a power greater than 2.

Reply 23

Kevin De Bruyne

Original post by Chittesh14

I've got it, thanks buddy!
I don't know why, sometimes, my brain just can't do this... lol
Sometimes, I just forget how to factorise equations which are raised to a power greater than 2.

It's okay, it may be tough at first but you've seen it done now so hopefully you should be okay :borat:

Reply 24

Chittesh14

Original post by SeanFM

It's okay, it may be tough at first but you've seen it done now so hopefully you should be okay :borat:

No, it's actually easy lol.
It's just that I haven't done it for a long time lol.

Reply 25

Kevin De Bruyne

Original post by Chittesh14

No, it's actually easy lol.
It's just that I haven't done it for a long time lol.

I was trying to make you feel better :redface:

Don't be too hard on yourself :wink:

Reply 26

Chittesh14

Original post by SeanFM

I was trying to make you feel better :redface:

Don't be too hard on yourself :wink:

Oh right lol, thanks :biggrin:

.

Here's another question :redface:

.

x(x-2) = x(x-2)(x-3)

I can't factorise here right?

Reply 27

Kevin De Bruyne

Original post by Chittesh14

Oh right lol, thanks :biggrin:

.

Here's another question :redface:

.

x(x-2) = x(x-2)(x-3)

I can't factorise here right?

Yes you can.

x(x-2) - x(x-2)(x-3) = 0, so..

Whenever you want to 'cancel out' something from both sides, like x(x-2) = x(x-2)(x-3)
those are the things you need to factor out.

(edited 8 years ago)

Reply 28

Chittesh14

Original post by SeanFM

x(x-2)[1-(x-3)] = 0[br]x(x-2)[-x+4] = 0[br]x(x-2)(-x+4) = 0[br]x(x-2)(x-4) = 0

Lol, why do I always get it wrong when I do it in my book. I just can't think of applying this same principle lol.
Well, I've learnt something new :smile:

!

I hope this is right :biggrin:

Reply 29

Kevin De Bruyne

Original post by Chittesh14

x(x-2)[1-(x-3)] = 0[br]x(x-2)[-x+4] = 0[br]x(x-2)(-x+4) = 0[br]x(x-2)(x-4) = 0

Lol, why do I always get it wrong when I do it in my book. I just can't think of applying this same principle lol.
Well, I've learnt something new :smile:

!

I hope this is right :biggrin:

Correct. Well done :borat:

So just think, like a^2 + ab = a(a+b) and a+ ab = a(1 +b) etc.

Reply 30

Chittesh14

Original post by SeanFM

Correct. Well done :borat:

So just think, like a^2 + ab = a(a+b) and a+ ab = a(1 +b) etc.

Yeah, I'll try to remember it in that way, or a way that I can remember it :smile:

.
Thank you so much though :P

Reply 31

Chittesh14

Hey, I keep making mistakes when drawing curves.

This graph is

x^3 - 3x^2 - 4x = 0

Why can't I draw the graph in the way I've drawn it in the second picture?
Is there something that will tell me that the curve goes downwards and not upwards after it crosses the y-axis? If so, what is it?