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Maths and statistics discussion, revision, exam and homework help. 33 Minutes Ago: 17th September 2011 21:00 So i'm going through my C1 book and they say this:

x^2 = 9x
x^2 - 9x = 0
x(x-9) = 0

Then they say x = 0 or 9. I understand how x could be 9, and i understand why they factorised it, but i don't understand why x = 0, could someone explain?

Also, how would i go about doing this question?

3x^2 = 5?

Last edited by zoxe; 24 Minutes Ago at 21:08. 28 Minutes Ago: 17th September 2011 21:05   Re: Could someone explain please? So i'm going through my C1 book and they say this:

x^2 = 9x
x^2 - 9x = 0
x(x-9) = 0

Then they say x = 0 or 9. I understand how x could be 9, and i understand why they factorised it, but i don't understand why x = 0, could someone explain?

Also, how would i go about doing this question?

3x^2 = 5?

BASICALLY - for x(x-9) = 0. X can either be 9 which is obvious or 0 becuase 0 x 0 = 0 and 0 x -9 = 0 so x = 0. lol Last edited by kidoo; 21 Minutes Ago at 21:12. 24 Minutes Ago: 17th September 2011 21:08   Re: Could someone explain please? x(x-9) = 0
x=0/(x-9)
x=0

or x=9

3x^2=5
x^2=5/3
x=SQRT(5/3)

or x=-SQRT(5/3)

there you go armador, happy now?

Last edited by wannabeme; 12 Minutes Ago at 21:20. 20 Minutes Ago: 17th September 2011 21:12   Re: Could someone explain please? Its really simple. so the left hand side of the equation has to equal 0.

So seperate x(x-9)=0 to x=0 and x-9=0. Now rearrange to get x on its own...do this to both of them(x=0 is already done).

this will get you x=0 and x=9.

so either x=0 or x=9 to to make x(x-9)=0

16 Minutes Ago: 17th September 2011 21:17   Re: Could someone explain please? or x=9, which you conveniently glossed over when blindly dividing by something which could be null. OR x=-sqrt(5/3), but you also conveniently glossed over that by assuming that x^2 was an affine function, and therefore, only had one solution. 15 Minutes Ago: 17th September 2011 21:18   Re: Could someone explain please? I'm still pretty unsure about this, like for this one i was solving:

x(6x+11) = 0
x = -11/6 <--- i could get this
x = 0 <--- still not sure why this is possible as in this equation:

2(x^2 - 4) = 0
x = 2
x = -2 and NOT 0

10 Minutes Ago: 17th September 2011 21:22   Re: Could someone explain please? or x=9, which you conveniently glossed over when blindly dividing by something which could be null.the question was how x(x-9) could equal 0, not what it equals, so I merely explained what they were asking for. OR x=-sqrt(5/3), but you also conveniently glossed over that by assuming that x^2 was an affine function, and therefore, only had one solution.Yeah this one is my fault, I'll go change it. 9 Minutes Ago: 17th September 2011 21:23   Re: Could someone explain please? I'm still pretty unsure about this, like for this one i was solving:

x(6x+11) = 0
x = -11/6 <--- i could get this
x = 0 <--- still not sure why this is possible as in this equation:

2(x^2 - 4) = 0
x = 2
x = -2 and NOT 0

The idea is this. Say you know that . Then either or , or both. (Because if neither of and were zero then clearly couldn't be zero!)

So say instead of just and you have some expressions in . In your OP, you have and . Then using this reasoning we have that if then either or . But rearranging the equation just gives . Summarising this, we get that if then either or . We could say "or both" here, but is just a number so it can't be both 0 and 9!

I'll give you another slightly more complicated example. Say that . Then either or or . Rearranging gives that either or or . {Again, we could say "or any combination of the above", but can't take more than one value.}

In the equation , we can factorise to get . Thus either 2=0 (which can't happen, so we discard this) or (so ) or (so ), so we have that if then either or . There's no reason why the solution should appear here {and, indeed, it doesn't}.

We can't say with certainty which of the specific values takes, but you can say with certainty that takes one of those values (which is why we say "or").

Last edited by nuodai; 7 Minutes Ago at 21:26.

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