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# C1 question help!! Watch

1. Can someone help me answer this question in a past paper! i dont know how to answer plz plz someone help! Mainly looking to get C answerd!
question is in attachment!

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2. (Original post by mujahid_e3)
Can someone help me answer this question in a past paper! i dont know how to answer plz plz someone help! Mainly looking to get C answerd!
question is in attachment!

What have you done so far?
3. Did you get for b) ==> -1/4 < k < 3

for c) draw a parabola, since we know the equation involves k^2

(a parabola is the U curve).

now it mentions to state the coordinates of any intersections with the coordinate axes. Before, it told us that the equation f(x)=0 has no real roots, this means it does not cross the x-axis.

The minimum point can be worked out from the equation (x+p)^2 + q.

I got: p=2k q=(-4k^2 + 3 + 11k)

then it says k=1

so p=2 and q=(-4 + 3 + 11) = 14-4 = 10

p=2 q=10

now then we have (x+2)^2 + 10

so our minimum point is (-2,10)

it is -2 not +2 since, the equation takes the form: (x-a)^2 + b

but in this case "a" must be negative since we have ( x+2) ===> i.e. (x-(-2)) = (x+2)

so you draw a U curve with the minimum point at (-2,10)

Now it crosses the y axis when x=0 ===> so put x=0 into this equation and we get:
(x+2)^2 + 10

(0+2)^2 + 10 = 14

so it crosses the y-axis at (0,14)

I hope this helps!
4. (Original post by sulexk)
Did you get for b) ==> -1/4 < k < 3

for c) draw a parabola, since we know the equation involves k^2

(a parabola is the U curve).

now it mentions to state the coordinates of any intersections with the coordinate axes. Before, it told us that the equation f(x)=0 has no real roots, this means it does not cross the x-axis.

The minimum point can be worked out from the equation (x+p)^2 + q.

I got: p=2k q=(-4k^2 + 3 + 11k)

then it says k=1

so p=2 and q=(-4 + 3 + 11) = 14-4 = 10

p=2 q=10

now then we have (x+2)^2 + 10

so our minimum point is (-2,10)

it is -2 not +2 since, the equation takes the form: (x-a)^2 + b

but in this case "a" must be negative since we have ( x+2) ===> i.e. (x-(-2)) = (x+2)

so you draw a U curve with the minimum point at (-2,10)

Now it crosses the y axis when x=0 ===> so put x=0 into this equation and we get:
(x+2)^2 + 10

(0+2)^2 + 10 = 14

so it crosses the y-axis at (0,14)

I hope this helps!
same, but I had -10 and not 10 in the completed square part (minimum points). But I probably made the mistake
5. (Original post by Beth1234)
What have you done so far?
nothing!!! dont know how to answer it!
6. (Original post by ilyking)
same, but I had -10 and not 10 in the completed square part (minimum points). But I probably made the mistake

Ahh cool
7. (Original post by mujahid_e3)
nothing!!! dont know how to answer it!

I hope its exactly what you are looking for!

Thank you
8. (Original post by sulexk)
Did you get for b) ==> -1/4 < k < 3

for c) draw a parabola, since we know the equation involves k^2

(a parabola is the U curve).

now it mentions to state the coordinates of any intersections with the coordinate axes. Before, it told us that the equation f(x)=0 has no real roots, this means it does not cross the x-axis.

The minimum point can be worked out from the equation (x+p)^2 + q.

I got: p=2k q=(-4k^2 + 3 + 11k)

then it says k=1

so p=2 and q=(-4 + 3 + 11) = 14-4 = 10

p=2 q=10

now then we have (x+2)^2 + 10

so our minimum point is (-2,10)

it is -2 not +2 since, the equation takes the form: (x-a)^2 + b

but in this case "a" must be negative since we have ( x+2) ===> i.e. (x-(-2)) = (x+2)

so you draw a U curve with the minimum point at (-2,10)

Now it crosses the y axis when x=0 ===> so put x=0 into this equation and we get:
(x+2)^2 + 10

(0+2)^2 + 10 = 14

so it crosses the y-axis at (0,14)

I hope this helps!
Yes for b) i got what you wrote. Where you got q=(-4k^2 + 3 + 11k) i got 4k^2-11k-3? is this wrong? how did you get p to = 2k ? could you please explain why it is -2 and not 2??
apart from that helped alot!

Thanks
9. lol wat paper is this from? I think this might have been from my own paper which i sat lol.
10. (Original post by mujahid_e3)
Where you got q=(-4k^2 + 3 + 11k) i got 4k^2-11k-3? is this wrong? how did you get p to = 2k ? could you please explain why it is -2 and not 2??
Well, completing the square, it's:
From the original equation,
(standard way of completing square! )
Rearrange to get:
Comparing this with , we get:
and

As for why it's -2 and not 2, well, the minimum value y can take is y=10. And the point where that occurs is when (x+2)^2=0... so x=-2!

Hope that helps! (And I am known to make mistakes, so if anything confuses you, ask away! )
11. (Original post by peri93)
lol wat paper is this from? I think this might have been from my own paper which i sat lol.
Jan 2010!! do u know how to answer it?
12. (Original post by Beth1234)
Well, completing the square, it's:
From the original equation,
(standard way of completing square! )
Rearrange to get:
Comparing this with , we get:
and

As for why it's -2 and not 2, well, the minimum value y can take is y=10. And the point where that occurs is when (x+2)^2=0... so x=-2!

Hope that helps! (And I am known to make mistakes, so if anything confuses you, ask away! )
Ok so what do i do after that!!
13. (Original post by mujahid_e3)
Ok so what do i do after that!!
has no real roots.
What does this suggest about ?
14. (Original post by TimmonaPortella)
has no real roots.
What does this suggest about ?
Its a quadratic and does not cross x axis?? I dont know there is a reason i ask??
15. (Original post by sulexk)

I hope its exactly what you are looking for!

Thank you
Yes read your working out!! Can u just explain how you got p= 2k??? Apart from that i understood everything!! Thanks for the help
16. (Original post by Beth1234)
Well, completing the square, it's:
From the original equation,
(standard way of completing square! )
Rearrange to get:
Comparing this with , we get:
and

As for why it's -2 and not 2, well, the minimum value y can take is y=10. And the point where that occurs is when (x+2)^2=0... so x=-2!

Hope that helps! (And I am known to make mistakes, so if anything confuses you, ask away! )
can u just explain one thing when you re-arrange how did u get
f(x)=(x+2k)^2+(-4k^2+3k+11k)
?? i dont understand how you got the 3k? aint it just meant to be 3??

Thanks
17. (Original post by mujahid_e3)
Its a quadratic and does not cross x axis?? I dont know there is a reason i ask??
You're very sarcastic for someone who's coming to an online forum to seek help, I must say.

is the discriminant.
The discriminant, as you should have been taught, is

where you have a quadratic equation given as

when
, the equation has one real root.
when , the equation has no real roots
when , the equation has two real roots.
this is because the roots of a quadratic equation are given by the quadratic formula:

so, the discriminant is the bit under the square root. square roots are not defined in real numbers when negative.

so, if has no real roots

solve that inequality for values of k.
18. (Original post by mujahid_e3)
Jan 2010!! do u know how to answer it?
Looks simple Will post workings tomorrow
19. (Original post by TobeTheHero)
Looks simple Will post workings tomorrow

Thanks
20. (Original post by TimmonaPortella)
You're very sarcastic for someone who's coming to an online forum to seek help, I must say.

is the discriminant.
The discriminant, as you should have been taught, is

where you have a quadratic equation given as

when
, the equation has one real root.
when , the equation has no real roots
when , the equation has two real roots.
this is because the roots of a quadratic equation are given by the quadratic formula:

so, the discriminant is the bit under the square root. square roots are not defined in real numbers when negative.

so, if has no real roots

solve that inequality for values of k.

What's with all the . dont get it?

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