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OCR Core Maths

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Original post by PeaceTreaty
I thought it was two roots, doesn't (x+n)^2 mean that there's a repeated root at x=-n?


One root was x=3/2 whilst two roots were x=-2 so yes it does have a repeated root but you didnt need to find the maximum point.

By 3 roots I meant 2 being the same, sorry for any confusion. It sounded like you calculated the maximum point but you may of just made that observation rather than working out. If you did, I apologise.
Original post by xx_Dan_xx
You had to use the Point P and the Centre in Pythag when it was less than the radius^2. This introduces the inequalities giving you a ranges. K can be any coord between whatever the answers were. Because between those values the length was still inside the circle.

You were never given a length but only know it lies inside the circle which means the you had to add the inequality which was less than the radius squared.

I just said that the radius was 5, the centre was -5, therefore it reaches the x-axis but never goes above it.
Original post by Minecraft27
I just said that the radius was 5, the centre was -5, therefore it reaches the x-axis but never goes above it.


You should get the marks.
Original post by xx_Dan_xx
One root was x=3/2 whilst two roots were x=-2 so yes it does have a repeated root but you didnt need to find the maximum point.

By 3 roots I meant 2 being the same, sorry for any confusion. It sounded like you calculated the maximum point but you may of just made that observation rather than working out. If you did, I apologise.


That's okay, I googled the sketch and everything was right but the minimum point which I drew at the y intercept (it was slightly right of that) was I supposed to work it out?
Finished the paper in 30 minutes haha, easiest paper I've done in my life! The only slight problem I had was the question about point Q which touches the y-axis, was the answer (0,-2)? If so, 100% achieved lol
Original post by PeaceTreaty
That's okay, I googled the sketch and everything was right but the minimum point which I drew at the y intercept (it was slightly right of that) was I supposed to work it out?


You didn't need to work it out. Given its a +xe cubic and it had a turning point at x=-2, you knew it was a max but you didnt need to state its nature but just had to draw the turning point at y=0 x = -2.
Original post by RhymeAsylumForever
Finished the paper in 30 minutes haha, easiest paper I've done in my life! The only slight problem I had was the question about point Q which touches the y-axis, was the answer (0,-2)? If so, 100% achieved lol


Yes it was (0, -2) I got that.
Reply 87
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newgu
Fairly sure I got 70/72. What UMS do we think that will be? 96?
Reply 88
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Abi790
13
Original post by manchestermadnes
It didn't ask for a range, it said a set of values. If the circle already has seet coordinates then you know exactly what k is

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It definitely didn't ask for a range, because I was at there trying to work out if it wanted inequalities or set values for ages :L
Not sure newgu.

Overall, I did every question and did all the working out right. I should only lose a few marks because I forgot to put +- the root in some equations. Stupid me xD
Can somebody work through how they got the discriminant as -72 please? I got -304 somehow, heres how I remember answering it:
x = 2y
y = 1/2*x
(x-2)^2 + (y+5)^2 = 25
(x-2)^2 + (1/2*x+5)^2 = 25
x^2 - 4x + 4 + 1/4x^2 + 5x + 25 = 25
5/4*x^2 + x + 4 = 0
5x^2 + 4x + 16 = 0
b^2 - 4ac = 4^2 - 4*5*16 = 16 - 320 = -304
Original post by xx_Dan_xx
Yes it was (0, -2) I got that.


How did you get it?
It was only 2 marks and I did it a really long way which somehow ended up as me getting x=1 (I think) as I got 9x^4-8x^2-1=0 (iirc) and subbing it back in and getting -2. Don't know why I got the right answer, might have just been lucky that the numbers worked out.
Original post by Abi790
It definitely didn't ask for a range, because I was at there trying to work out if it wanted inequalities or set values for ages :L


You knew Point P was inside the circle. Therefore the length from the Centre to Point P had to below the radius ^2.

If it wanted a set length then you can have the set values as answers but you don't know the length so you can know the value of p only the ranges between them.
Original post by Rifleboy123
Can somebody work through how they got the discriminant as -72 please? I got -304 somehow, heres how I remember answering it:
x = 2y
y = 1/2*x
(x-2)^2 + (y+5)^2 = 25
(x-2)^2 + (1/2*x+5)^2 = 25
x^2 - 4x + 4 + 1/4x^2 + 5x + 25 = 25
5/4*x^2 + x + 4 = 0
5x^2 + 4x + 16 = 0
b^2 - 4ac = 4^2 - 4*5*16 = 16 - 320 = -304


u dun goofed
Original post by Rifleboy123
Can somebody work through how they got the discriminant as -72 please? I got -304 somehow, heres how I remember answering it:
x = 2y
y = 1/2*x
(x-2)^2 + (y+5)^2 = 25
(x-2)^2 + (1/2*x+5)^2 = 25
x^2 - 4x + 4 + 1/4x^2 + 5x + 25 = 25
5/4*x^2 + x + 4 = 0
5x^2 + 4x + 16 = 0
b^2 - 4ac = 4^2 - 4*5*16 = 16 - 320 = -304


The discriminant was -76

The corresponding quadratic in y was 5y^2 - 2y + 4 = 0 I think.

You went the hard way by substituting in x instead of y...

...looks like you did it correctly though so should be okay.
(edited 9 years ago)
dam just realised i got the discriminant round and got -276 instead of -76, do you reckon id only lose one mark or two?
Original post by RhymeAsylumForever
How did you get it?
It was only 2 marks and I did it a really long way which somehow ended up as me getting x=1 (I think) as I got 9x^4-8x^2-1=0 (iirc) and subbing it back in and getting -2. Don't know why I got the right answer, might have just been lucky that the numbers worked out.


I cant remember the exact details of the question so if you do can you post them. I just remember I definitely got that answer.
Original post by RhymeAsylumForever
How did you get it?
It was only 2 marks and I did it a really long way which somehow ended up as me getting x=1 (I think) as I got 9x^4-8x^2-1=0 (iirc) and subbing it back in and getting -2. Don't know why I got the right answer, might have just been lucky that the numbers worked out.


Because it was a tangent at a minimum point, then the line had gradient 0 so the y value must be the same as at minimum point (and x=0 because it interescts y axis)
Original post by Rifleboy123
Can somebody work through how they got the discriminant as -72 please? I got -304 somehow, heres how I remember answering it:
x = 2y
y = 1/2*x
(x-2)^2 + (y+5)^2 = 25
(x-2)^2 + (1/2*x+5)^2 = 25
x^2 - 4x + 4 + 1/4x^2 + 5x + 25 = 25
5/4*x^2 + x + 4 = 0
5x^2 + 4x + 16 = 0
b^2 - 4ac = 4^2 - 4*5*16 = 16 - 320 = -304


I remember getting -76, I think a=1, b=2, c=20

b^2-4ac
2^2-4(1)(20)
4-80
=-76

so the original equation must have been x^2+2x+20? Or something like that, I definitely remember getting 4-80=-76
Original post by Wiggledinho
dam just realised i got the discriminant round and got -276 instead of -76, do you reckon id only lose one mark or two?


I think 1 because you still can make the conclusion it has no roots.

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