C1 help

Original post by Year11guy

The question says that p is a non-zero constant which is the reason why you should reject p=0.

Your solution p=-4/9 should be p=4/9 - Let me know if you can't find your mistake.

If p=4/9 then you have

x^2+\frac{12}{9}x+\frac{4}{9}=0

Try multiplying this equation by 9 then you should find it easier to factorise.

Reply 2

Original post by Year11guy

You made a sign error; it should be p=4/9.

We can discard the p=0
Edit: As notnek (PRSOM) said - you're told p is non-zero.

What happened to the 3+11k?

It then says f(x)=0 has no real roots, find the set of possible values of k. How do I do this, I can't use the normal method can I? since there's no roots and they want a set of possible values.

Sorry for the length, any help is much appreciated.

If you rearrange your equation to (x+2k)^2= "something"

Then that something must be negative for no real roots, and that gives you the desired condition which you need to investigate.

(edited 9 years ago)

Reply 3

Original post by Year11guy

The next question is: f(x)=x²+4kx+(3+11k) where k is a constant
Express f(x) in the form (x+p)²+q where p and q are constants to be found in terms of k.

I assumed it meant complete the square so I did: (x+4k/2)²-(4k/2)²
It then says f(x)=0 has no real roots, find the set of possible values of k. How do I do this, I can't use the normal method can I? since there's no roots and they want a set of possible values.

You subtracted (4k/2)² but forgot to add on (3+11k). Does this make sense?

Try correcting this part and post your working if you're still stuck.

Reply 4

Original post by notnek

You subtracted (4k/2)² but forgot to add on (3+11k). Does this make sense?

Try correcting this part and post your working if you're still stuck.

Wait, was I right to complete the square? If so, I thought to complete the square it is : (x+b/2)²-(b/2)² and you disregard c, or am I incorrect?

Reply 5

Original post by Year11guy

Wait, was I right to complete the square? If so, I thought to complete the square it is : (x+b/2)²-(b/2)² and you disregard c, or am I incorrect?

Yes, completing the square is correct.

(x+b/2)²-(b/2)² = x² + bx + (b/2)² - (b/2)² = x² + bx

Your method will only give you x² + bx. So you're right that doing this will disregard c but why would you want to do that?

Reply 6

Original post by notnek

I'm sorry, I'm lost. Starting from the original equation:
x²+4kx+(3+11k)

Completing the square would give (x+4k/2)²-(4k/2)² but you want me to add on (3+11k) but the question wants it expressed in the form (x+p)²+q where p and q are constants to be found in terms of k.

Reply 7

Original post by Year11guy

You'll still have something of the form (x+p)²+q if you add on (3+11k) - p and q are in terms of k.

-(4k/2)^2 + (3+11k) is in terms of k so it's fine. You can simplify it though.

Reply 8

Ok, so I got (x+2k)²-(2k)²+(3+11k). What steps do I take to find the set of possible values of k. Am I right in expanding and simplifying to get : x²+4kx+3+11k?

Reply 9

Original post by Year11guy

Ok, so I got (x+2k)²-(2k)²+(3+11k). What steps do I take to find the set of possible values of k. Am I right in expanding and simplifying to get : x²+4kx+3+11k?

No because now you're back to where you started :smile:

Have a look at then end of Ghostwalker's post for the next step.

Reply 10

Original post by notnek

No because now you're back to where you started :smile:

Have a look at then end of Ghostwalker's post for the next step.

I just substituted the original equation into b²-4ac<0 since there's no real roots. I got k<-3 and k<1/4

Reply 11

Original post by Year11guy

I just substituted the original equation into b²-4ac<0 since there's no real roots. I got k<-3 and k<1/4

Your critical points have the right magnitude, but the wrong signs, and one should be ">"

Reply 12

Original post by ghostwalker

Your critical points have the right magnitude, but the wrong signs, and one should be ">"

Should they be: k>-3 and k<1/4

Reply 13

Original post by Year11guy

Should they be: k>-3 and k<1/4

'Fraid not. They both have the wrong sign. Post some working if you don't see why, as you may have the initial inequality incorrect too.

Reply 14

Original post by ghostwalker

'Fraid not. They both have the wrong sign. Post some working if you don't see why, as you may have the initial inequality incorrect too.

Ok, if there's no real roots, then b²-4ac<0. I substituted in a,b and c so it's a=1 b=4kx c=(3+11k). This makes (4k)²-(4*1(3+11k). This simplifies to:
16k²-12+44k=0 This simplifies to 4k²+11k-3=0. This factorises to:
(4k-1)(k+3) The solutions are -3 and 1/4. I don't really get what to do next.

Reply 15