Year11guy
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#1
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I'm stuck on two questions; the first is:
The equation x²+3px+p=0 where p is a non zero constant, has equal roots.
Find the value of p. I substitued abc into b²-4ac=0 and got 9p²-4p=0.
I don't really know what to do next. I tried factorising but that gives me two different answers. p=0 or p=-4/9. The -4/9 seems more correct than 0 but it doesn't seem right. The final equation is supposed to have two roots and factorise. Can someone show me where I went wrong?

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.

Sorry for the length, any help is much appreciated.
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Sir Cumference
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(Original post by Year11guy)
I'm stuck on two questions; the first is:
The equation x²+3px+p=0 where p is a non zero constant, has equal roots.
Find the value of p. I substitued abc into b²-4ac=0 and got 9p²-4p=0.
I don't really know what to do next. I tried factorising but that gives me two different answers. p=0 or p=-4/9. The -4/9 seems more correct than 0 but it doesn't seem right. The final equation is supposed to have two roots and factorise. Can someone show me where I went wrong?
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.
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ghostwalker
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#3
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#3
(Original post by Year11guy)
I'm stuck on two questions; the first is:
The equation x²+3px+p=0 where p is a non zero constant, has equal roots.
Find the value of p. I substitued abc into b²-4ac=0 and got 9p²-4p=0.
I don't really know what to do next. I tried factorising but that gives me two different answers. p=0 or p=-4/9. The -4/9 seems more correct than 0 but it doesn't seem right. The final equation is supposed to have two roots and factorise. Can someone show me where I went wrong?
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.


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[/latex]
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.
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Sir Cumference
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(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.
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Year11guy
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(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?
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Sir Cumference
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(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?
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Year11guy
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(Original post by notnek)
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?

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.
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Sir Cumference
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(Original post by Year11guy)
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.
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.
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Year11guy
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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?
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Sir Cumference
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(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

Have a look at then end of Ghostwalker's post for the next step.
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Year11guy
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#11
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(Original post by notnek)
No because now you're back to where you started

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
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ghostwalker
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(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 ">"
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Year11guy
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#13
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(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
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ghostwalker
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(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.
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Year11guy
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(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.
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ghostwalker
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(Original post by Year11guy)
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
Should be 16k^2-12-44k=0

The minus sign outside the brackets effects both the "3" and the "11k".
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Year11guy
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(Original post by ghostwalker)
Should be 16k^2-12-44k=0

The minus sign outside the brackets effects both the "3" and the "11k".
So, the answers K<3 or K>-1/4
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ghostwalker
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(Original post by Year11guy)
So, the answers K<3 or K>-1/4
k<3 and k>-1/4 i.e it's in the interval (-1/4,3)

Edit: Corrected - thanks Mr M
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Mr M
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(Original post by ghostwalker)
k>3 and k>-1/4 i.e it's in the interval (-1/4,3)
You don't mean k>3 and at this level they should write -\frac{1}{4}&lt;k&lt;3
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Year11guy
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#20
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(Original post by Mr M)
You don't mean k>3 and at this level they should write -\frac{1}{4}&lt;k&lt;3
If I don't put it in that form, will I lose marks? Also, shouldn't it less than or equal to since they're solutions too?
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