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iAMfromHuntersBar · Jun 6, 2007, 03:28 PM

Hi guys!

Right, I've got this maths question, and it's the first time in ages I've been honestly stumped as to where to even start! It's like this;

x(squared) − x = 0

now I know that to get rid of the x squared I need to square root it, and thus both sides... but is that where I start?

Cheers everyone!

J

Curlyben · Jun 6, 2007, 03:31 PM

$x^2-x=0$
is what you mean.
rather than rooting this is a quadratic so solve in the form
$(x+a)(x+b)=x^2 +abx +ab=0$

so we're looking at ab = -1
but that isn't going to work, bugger.

Monkey boy or Galactus or Alpha will be along in a bit

Ps try the [ MATH ] tag to get it to display properly with $x^2$ = x^2

galactus · Jun 6, 2007, 04:33 PM

Just lookng at this we can see the answer is 0 or 1.

$x^{2}-x=0\Rightarrow{x(x-1)}=0$

iAMfromHuntersBar · Jun 7, 2007, 12:06 PM

sorry, I still don't understand!

I can see that $x^2 - x = 0$

is the same as $(x * x) - x = 0$

and that 0 and 1 are the only possible answers, but is that right, and is that a good enough explanation?

rockerchick_682 · Jun 7, 2007, 12:16 PM

you factor it out x2-1=0 to x(x-1)=0
and if you're trying to find the roots, then the roots are x=0 and x=1

iAMfromHuntersBar · Jun 7, 2007, 12:24 PM

Lol, sorry, I'm being thick here, how do you get from;

$x^2 - 1 = 0$ to $x(x - 1) = 0$ though, what do you mean factor it?

sorry again!

rockerchick_682 · Jun 7, 2007, 12:42 PM

I'm not sure how to explain it, but if you distributed x(x-1) you'd get x2-x because x*x=x2 and x*-1=-x, does that help?

snoopycool24 · Jun 8, 2007, 06:20 PM

You can use the quadratic formula, too.

hemant_pandey · Aug 15, 2007, 11:53 PM

Originally Posted by iAMfromHuntersBar

Hi guys!

Right, I've got this maths question, and it's the first time in ages I've been honestly stumped as to where to even start! it's like this;

x(squared) − x = 0

now I know that to get rid of the x squared I need to square root it, and thus both sides ... but is that where I start?

Cheers everyone!

J

Dear member this is a basic query of algebra.x^2-x=0 can be solved by taking x common.Hence x(x-1)=0 or x-1 and x=1

Capuchin · Aug 16, 2007, 12:13 AM

I can't believe I missed this when it came about. Hope it's not too late hunter. You just factor by x.

iAMfromHuntersBar · Aug 16, 2007, 12:20 AM

It's too late for the paper I was submitting... but I'd still like to understand it better, which I don't! Lol!

But it's far too early and I haven't had nearly enough coffee to think about anything, let alone algebra! Ha ha!

Capuchin · Aug 16, 2007, 12:21 AM

$\frac{x^2-x}{x} = x-1$

therefore:

$x^2-x = x(x-1)$

Does that make sense?

iAMfromHuntersBar · Aug 16, 2007, 12:32 AM

I think so, because you have removed the "divided by x" on one side you therefore multiply the right-hand side by x?

Capuchin · Aug 16, 2007, 12:33 AM

Yes, I was just trying to demonstrate that the two are equal.

Does the initial division make sense to you?