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    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
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    (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!

    Thanks in advance
    What have you done so far?
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    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!
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    (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
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    (Original post by Beth1234)
    What have you done so far?
    nothing!!! dont know how to answer it!
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    (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
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    (Original post by mujahid_e3)
    nothing!!! dont know how to answer it!
    did you read my working out for your question?

    I hope its exactly what you are looking for!


    Thank you
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    (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
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    lol wat paper is this from? I think this might have been from my own paper which i sat lol.
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    (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, 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! )
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    (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?
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    (Original post by Beth1234)
    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!!
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    (Original post by mujahid_e3)
    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)?
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    (Original post by TimmonaPortella)
    f(x)=0 has no real roots.
    What does this suggest about (4k)^2-4 \cdot 1 \cdot (3+11k)?
    Its a quadratic and does not cross x axis?? I dont know there is a reason i ask??
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    (Original post by sulexk)
    did you read my working out for your question?

    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
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    (Original post by Beth1234)
    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! )
    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
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    (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.

    (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 &lt;0, the equation has no real roots
    when b^2-4 \cdot a\cdot c &gt;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)&lt;0
    solve that inequality for values of k.
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    (Original post by mujahid_e3)
    Jan 2010!! do u know how to answer it?
    Looks simple Will post workings tomorrow
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    (Original post by TobeTheHero)
    Looks simple Will post workings tomorrow
    Please do will help alot!

    Thanks
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    (Original post by TimmonaPortella)
    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 &lt;0, the equation has no real roots
    when b^2-4 \cdot a\cdot c &gt;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)&lt;0
    solve that inequality for values of k.

    What's with all the . dont get it?
 
 
 
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