C1 question help!!

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!

Thanks in advance

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!

What have you done so far?

same, but I had -10 and not 10 in the completed square part (minimum points). But I probably made the mistake

nothing!!! dont know how to answer it!

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!

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??

lol wat paper is this from? I think this might have been from my own paper which i sat lol.

Well, completing the square, it's:
From the original equation, $f(x)=x^2+4kx+(3+11k)$
$f(x)=(x+2k)^2-(2k)^2+(3+11k)$ (standard way of completing square! )
Rearrange to get: $f(x)=(x+2k)^2+(-4k^2+3k+11k)$
Comparing this with $f(x)=(x+p)^2+q$, we get:
$p=2k$ and $q=-4k^2+3k+11k$

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!!

$f(x)=0$ has no real roots.
What does this suggest about $(4k)^2-4 \cdot 1 \cdot (3+11k)$?

did you read my working out for your question?

I hope its exactly what you are looking for!


Thank you

Well, completing the square, it's:
From the original equation, $f(x)=x^2+4kx+(3+11k)$
$f(x)=(x+2k)^2-(2k)^2+(3+11k)$ (standard way of completing square! )
Rearrange to get: $f(x)=(x+2k)^2+(-4k^2+3k+11k)$
Comparing this with $f(x)=(x+p)^2+q$, we get:
$p=2k$ and $q=-4k^2+3k+11k$

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! )

f(x)=(x+2k)^2+(-4k^2+3k+11k)

Its a quadratic and does not cross x axis?? I dont know there is a reason i ask??

Jan 2010!! do u know how to answer it?

Looks simple Will post workings tomorrow

You're very sarcastic for someone who's coming to an online forum to seek help, I must say.

$(4k)^2-4 \cdot 1 \cdot (3+11k)$ is the discriminant.
The discriminant, as you should have been taught, is
$b^2-4 \cdot a \cdot c$
where you have a quadratic equation given as
$ax^2+bx+c =0$
when
$b^2-4\cdot a\cdot c=0$, the equation has one real root.
when $b^2 - 4\cdot a \cdot c <0$, the equation has no real roots
when $b^2-4 \cdot a\cdot c >0$, the equation has two real roots.
this is because the roots of a quadratic equation are given by the quadratic formula:
$\dfrac{-b \pm \sqrt{b^2 - 4 \cdot a \cdot c}}{2a}$

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

so, if $f(x)=0$ has no real roots
$(4k)^2-4 \cdot 1 \cdot (3+11k)<0$
solve that inequality for values of k.