How do I prove that they DON'T intersect?

I know generally what to do, but I don't know how to use it to conclude this.

**0**

# 2003, Paper I, Q7

Started by herbeey, May 19 2005 01:20 PM

6 replies to this topic

### #1

Posted 19 May 2005 - 01:20 PM

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### #2

Posted 19 May 2005 - 01:57 PM

Make the two equations equal to each other.

Make them = 0

Then use b² - 4ac

If the answer is less than 0, then there's no intersection.

Make them = 0

Then use b² - 4ac

If the answer is less than 0, then there's no intersection.

### #3

Posted 19 May 2005 - 02:01 PM

Right. That's what I did, but I see now that my problem is; what value of 'x' do I use for the discriminant?

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### #4

Posted 19 May 2005 - 02:07 PM

Ok, my solution is

x(sq) + x + 3

So a = 1

b = 1

c = 3

1(sq) - 4*1*3 = 0

= -11

-11<0 So no point of intersection.

x(sq) + x + 3

So a = 1

b = 1

c = 3

1(sq) - 4*1*3 = 0

= -11

-11<0 So no point of intersection.

### #5

Posted 19 May 2005 - 02:10 PM

oh! Sorry! I'm having a bad maths day today,... I totally forgot that it is the

Cheers.

**factors**of x which equal a, b, c.... I was thinking that b=x, not 1!Cheers.

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### #7

Posted 19 May 2005 - 03:11 PM

QUOTE(SomethingTypical @ May 19 2005, 03:10 PM)

No problem, I think most people are having a bad maths day today.

I suppose that's better than having one tomorrow

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